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GRADES 7−12+10612BEP

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Algebra I & IIKey Concepts,

Practice, and Quizzes

Meizhong Wang

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Algebra I & II Key Concepts, Practice, and Quizzes Table of Contents

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v-xIntroduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .vReview and Test-Taking Tips . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiAbout the Author . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x

Unit 1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-221-1 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1

the real number system, sets, basic mathematic symbols, absolute value1-2 Operations With Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61-3 Exponents & Order of Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .81-4 Algebraic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10

evaluating expressions, translating words into algebraic expressions, properties of addition and multiplication1-5 Simplifying Algebraic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13

equivalent expressions, combining like terms, removing parentheses1-6 Exponents&ScientificNotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21

Unit 2 Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23-582-1 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23

linear equations, properties of equality, procedure for solving equations, equations involving decimals/fractions2-2 Linear Equations and Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28

geometry formulas, consecutive integers, business problems, motion problems, concentration/mixture problems2-3 Sets and Inequalities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .39

intervals, properties of inequalities, solving inequalities2-4 Intersections and Unions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .452-5 Absolute-Value Equations & Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .48 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57

Unit 3 Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59-823-1 Graphing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59

the coordinate plane, graphs of linear equations, graphing nonlinear equation with two variables3-2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .62

findingfunctionvalues,graphingafunction,theverticallinetest3-3 Domain, Range, and Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .663-4 Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .68

slope-intercept function of a line, slope 3-5 Graphing Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .71

graphing using the slope and the y-intercept, vertical and horizontal lines, perpendicular and parallel lines3-6 Straight Line Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .75

point-slopeequationofaline,findinganequationofaline Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .79 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .81

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Unit 4 Systems of Equations & Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83-964-1 Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83

solving linear systems by graphing, properties of a linear system4-2 Solving Systems by Substitution or Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .86

systems involving decimals or fractions, applications4-3 Systems of Linear Inequalities in Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .90 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .94 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .96

Unit 5 Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .97-1165-1 Addition & Subtraction of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

polynomials, degree of polynomial, evaluating polynomial functions5-2 Multiplying Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

FOIL method to multiply binomials, special binomial products5-3 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

greatest common factor, factoring polynomials by grouping, factoring x2 + bx + c5-4 Factoring ax2 + bx + c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

factoring trinomials, AC method5-5 Factoring Special Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

sum&differenceofcubes Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Unit 6 Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117-1416-1 Rational Expressions & Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

rational functions, multiplying and dividing rational expressions6-2 Adding & Subtracting Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1206-3 Polynomial Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

long division of polynomials, synthetic division6-4 Complex Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1286-5 Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1306-6 Applications of Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

number problems, work problems, proportions, motion problems Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Unit 7 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142-1687-1 Roots and Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

square roots, square root functions, odd and even roots7-2 Rational Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

powers of roots7-3 Simplify Radicals Using Product & Quotient Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1497-4 Operations With Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

adding and subtracting radicals, multiplying radicals7-5 Dividing Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

rationalizing denominators 7-6 Solving Equations With Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

square root equations, extraneous solutions, equations with two radicals7-7 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

imaginary unit i, operations with complex numbers, complex conjugates, complex division Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167



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Unit 8 Quadratic Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169-1938-1 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1698-2 Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1718-3 The Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1748-4 Applications of Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1778-5 Discriminant of Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

writing equation from solutions 8-6 Solving Equations in Quadratic Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1838-7 Quadratic and Rational Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

solving quadratic inequalities, solving rational inequalities Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

Unit 9 Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194-2229-1 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

the distance formula 9-2 Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1969-3 Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2049-4 Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2089-5 The General Conic Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

function transformations, general-form conic equations9-6 Nonlinear Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

Unit 10 Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223-25410-1 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22310-2 Inverse and Composite Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22710-3 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23510-4 Rules of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24010-5 Common and Natural Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

changing the base of a logarithm10-6 Exponential and Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

Unit 11 Determinants and Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255-28111-1 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

expansion by diagonals, expansion by minors, expansion by any row/column11-2 Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25911-3 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

matrix addition & subtraction, matrix multiplication11-4 Matrix Inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268

identitymatrix,Gauss-JordaneliminationmethodtofindA-1, solving a linear system, using a graphing calculator (TI-83 Plus) Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 Practice Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

Answers & Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283-293Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290



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Algebra I & II Key Concepts, Practice, and Quizzes Introduction

Introduction

If you are looking for a quick exam, homework guide, and review book in algebra, “Algebra I & II Key Concepts, Practice, and Quizzes” is an excellent source . Skip the lengthy and distracting books and instead use this concise book as a guideline for your studies, quick reviewing, tutoring, or helping children with homework .

This unique and well-structured book is an excellent supplement and convenient reference book for algebra textbooks . It provides a concise, understandable, andeffectiveguideonbasicalgebraplusthe following topics: factoring, radicals, exponents, graphing, linear equations, quadratic equations, inequalities, functions, conics, logarithms, determinants, matrices, and more .

Lecture notes that built the foundation of this book have been class-tested for many years, and received very good response from students . The following are some sample evaluation comments from students:

• “Excellentabilitytomakedifficultmaterialunderstandable.”

• “I feel Mei is an excellent teacher . She makes everything seem black and white and straight to the point .”

• “The material is presented in a manner that is readily understood .”



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Key Features

As an aid to readers, the book provides some noteworthy features:

• An excellent supplemental and convenient reference book for any algebra textbook . Each topic, concept,term,andphrasehasacleardefinitionfollowedbyexamplesoneachpage.

• A concise study guide, quickly getting to the heart of each particular topic, helping students with a quick review before doing mathematics homework as well as preparation for tests .

• Keyterms,definitions,properties,phrases,concepts,formulas,rules,equations,etc.areeasilylocated . Clear step-by-step procedures for applying theorems .

• Clear and easy-to-understand written format and style . Materials presented in visual and color format with less text and more outlines, tables, boxes, charts, etc .

• Tables that organize and summarize procedures, methods, and equations; clearly presenting informationandmakingstudyingmoreeffective.

• Procedures and strategies for solving word problems, using realistic real-world application examples .

• Summary at the end of each unit to emphasize the key points and formulas in the chapter, which is convenient for students reviewing before exams .

• Quizzes at the end of each unit test students’ understanding of the material . Students can take the quiz before beginning the unit to determine how much they know about the topic . Those who do well may decide to move on to the next unit .

• “Reviewing and Test Taking Tips” to help students improve their test score .



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Suitable Readers

This book can be used for:

• Adult Basic Education programs at colleges .

• Students in community colleges, high schools, tutoring, or resource rooms .

• Self-study readers, including new teachers to brush up on their mathematics .

• Professionals as a quick review of some mathematic formulas and concepts, or parents to help their children with homework .

There are many algebra books on the market, but this unique Algebra I & II Key Concepts, Practice, and Quizzesprovidesaconcise,understandable,andeffectiveguidetoalgebra.

Acknowledgements

I want to thank Michael O . Baker, the president of The Critical Thinking Co .™, for his support in publishing this book .

Special thanks to Patricia Gray, the editorial coordinator of The Critical Thinking Co .™, for her hard work in helping and supporting me throughout the entire process .

I would also like to express my sincere gratitude for the math editors of The Critical Thinking Co .™, Joe Walker and Chip Dombrowski, for their accuracy in reviewing the book and checking all the answers.Theirthoughtfulandinvaluablecorrectionsandsuggestionshavehelpedtorefinethewritingof this book .

I also appreciate my daughter Alice Wang, who deserves an acknowledgment for proofreading this book .



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Reviewing and Test Taking Tips

Test Taking Tips

• Scan the entire test as soon as you receive it.

Mark the easy questions you know that you can answer quickly .

Mark the hard questions and do those later .

Mark the questions that you don’t know .

• Skip the questions you don’t know.

Don’t waste time on tough questions that you’re not sure about . You can always go back

if there is time .

• Keep an eye on the time.

Make sure you are sticking to some kind of time schedule, so you can finish the entire

test within the time limit . Allow a few minutes to check your work at the end of the

testing time .

• Read each question carefully.

Don’t make some silly errors by misreading the information in the question .

• Ask questions.

If you’re confused about the wording or meaning of a question, ask your teacher .

Don’t risk getting a question wrong because you misunderstood it .

• Write each step of the answer neatly.

It will make it easier to check, and you may get partial credit for correct steps, even if

your final answer is wrong .

• Check your answers.

Always check your work after you’ve finished the test . Make sure you didn’t make any

careless mistakes .

Reviewing and Test-Taking Tips



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Reviewing Tips

• Make a review schedule.

Make a review schedule and make sure you have enough time to review all contents

before the test .

• Go over materials and make notes.

Go over lecture notes, homework, practice tests, review material, the textbook, and

exams of a previous class if available, etc . and make key notes .

• Make a summary sheet.

Write the key concepts/principles/rules/formulas etc . on a sheet of paper and do a quick

review before the test .

• Form a study group.

A study group is a good way for students to help each other, review material quickly, and

benefit from other’s strengths .

• Go to review sessions.

Ask your teacher about concepts and problems that you are not sure about . Also ask

which topics will be included on the test .

• Get a good night’s sleep.

Try to study earlier (a little bit each night) before the test, do a quick review on the last

night, and get a good night’s sleep before the exam .



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Algebra I & II Key Concepts, Practice, and Quizzes About the Author

About the Author

Meizhong Wang (Mei) has been an instructor at the College of New Caledonia (CNC) in Canada for more than 23 years . She teaches algebra I and II for students of Adult Basic Education . She has also taughtprobabilityandstatistics,finitemath,calculus,technologymath,electronicsmath,fundamentalmath,andothermathematicalcoursesindifferentprogramsatCNC.

In addition to being an instructor of math, Mei teaches computer studies at CNC, and has taught physics, electronics, and Mandarin at colleges and universities in Canada and China .

The Higher Education Press, one of the largest and most prominent publishers of educational books in China, published the Chinese version of Mei’s book “简明电路基础” (Understandable Electric Circuits) in 2005, and reprinted it in 2009 .

Michael Faraday House of the Institution of Engineering and Technology (IET), one of the world’s leading professional societies for the engineering and technology community, published the English version of Mei’s book Understandable Electric Circuits in 2010 .

CNC Press published Math Made Easy – Essential Math Concepts Review in Canada in July 2011, and issued the second edition in April 2013 .

Lily Chow and Meizhong Wang published the English and Chinese versions of the book Legends of Four Chinese Sages in 2007 .



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Algebra I & II Key Concepts, Practice, and Quizzes Unit 1 – Fundamental Concepts

UNIT 1 FUNDAMENTAL CONCEPTS1-1 THE REAL NUMBERS

The Real Number System

• Natural numbers are the numbers used for counting . {1,2,3,4,5,6, …}

• Whole numbers are the natural numbers including 0 . {0,1,2,3,4,5,6, …}

• Integers are all the whole numbers and their negatives . {… -3, -2, -1, 0, 1, 2, 3, …}

• A number line is a straight line on which every point corresponds to an integer .

Negative numbers Origin Positive numbers

• Rational numbers are numbers that can be expressed as a fraction 𝑎𝑎 𝑏𝑏

, where a and b are integers and b ≠ 0. Rational numbers can be expressed as decimal terminates or repeats .Example: 3

4 = 0 .75 Terminating

2 3

= 0 .66666… = 0. 6� Repeating

Example of rational numbers: 0.52 = 52100

, -4 .5 = -92

, 07

, -11 = -111

• Irrational numbers are real numbers that cannot be represented by a fraction (or the

ratio of two integers) .

Irrational numbers can be expressed as non-terminating, non-repeating decimals .

Example: π = 3 .1415926 … Non-terminating

√2 = 1 .1414213562 Non-repeating

Example of irrational numbers: ,7 2π, - ,19 135

• Real numbers are the rational numbers plus irrational numbers .

• The real number system a

Real Numbers Rational Numbers: 3

4 , -2 .13

Integers: … -2, -1, 0, 1, … Irrational Numbers Whole Numbers: 0 . 1, 2, 3 … √5, π , … Natural Numbers: 1,2, 3 …

&

Page 1- 1

Rational Numbers: 34, -2 .13

Integers: … -2, -1, 0, 1, …Whole Numbers: 0, 1, 2, 3 …Natural Numbers: 1, 2, 3 …



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Sets

• A set is a group of elements or numbers (in mathematics) .

Example: “the set of things in an emergency box” can be written as:

{A bottle of water, cookies, flash light, bandages, blanket, …} The curly braces { } represent a set .

• Roster Notation { }

Roster Notation { } ExampleList all the elements or numbers of the set .

The set of odd numbers between 3 and 17: { 5, 7, 9, 11, 13, 15}Note: “between” not including 3 and 17 .

• Set-builder notation: a mathematic form { x | x …} used to represent a set of numbers .

Set-Builder Notation Example{ x | x … }

The set of x such as condition of x

{ x | x > 3 }

The set of x such as x greater than 3

Example: Write in roster notation .

1. The set of all letters in the word “hope .”

{ h, o, p, e}

2. A = { x | x is a number between -4 and 6 }

A = {-3, -2, -1, 0, 1, 2, 3, 4, 5} “Between” — not including -4 and 6 .

Example: Write the following in a. set- builder notation and b. roster notation .

“The set of numbers between -3 and 4 .”

1. A = { x | x is a number between -3 and 4} Set-builder notation

2. A = {-2, -1, 0, 1, 2, 3} Roster notation

• Elements of a set: In set notation, x ∈ Z means x belongs to the set Z .

Page 1- 2

Sets

• A set is a group of elements or numbers (in mathematics) .

Example: “the set of things in an emergency box” can be written as:

{A bottle of water, cookies, flash light, bandages, blanket, …} The curly braces { } represent a set .

• Roster Notation { }

Roster Notation { } ExampleList all the elements or numbers of the set .

The set of odd numbers between 3 and 17: { 5, 7, 9, 11, 13, 15}Note: “between” not including 3 and 17 .

• Set-builder notation: a mathematic form { x | x …} used to represent a set of numbers .

Set-Builder Notation Example{ x | x … }


{ x | x > 3 }


Example: Write in roster notation .

1. The set of all letters in the word “hope .”

{ h, o, p, e}

2. A = { x | x is a number between -4 and 6 }

A = {-3, -2, -1, 0, 1, 2, 3, 4, 5} “Between” — not including -4 and 6 .

Example: Write the following in a. set- builder notation and b. roster notation .


1. A = { x | x is a number between -3 and 4} Set-builder notation

2. A = {-2, -1, 0, 1, 2, 3} Roster notation

• Elements of a set: In set notation, x ∈ Z means x belongs to the set Z .

Page 1- 2





Real Numbers and Sets

• Real numbers summaryName Set of Numbers Example

natural numbers {1,2,3,4,5,6, …} 2, 7, 11, 35, 167whole numbers {0,1,2,3,4,5,6, …} 0, 4, 8, 23, 2009

integers {… -4, -3, -2, -1, 0, 1, 2, 3, -4 …} -215, -31, -6, 0, 8, 24, 190

rational numbers(or fractions) { 𝒂𝒂

𝒃𝒃 | a and b are integers, b ≠ 0}

terminating: 45

= 0 .8

repeating: - 89 = - 0 .8888 … = 0 .8�

irrational numbers {x | x cannot be expressed as a fraction for any integers}

nonterminating, nonrepeating 236 .25 ≈ π ≈ 3.1416

• A real number line is a straight line on which every point corresponds to a real number .

Example: Put the following numbers in order from least to greatest on the real number line .

34

, -2 13 , -0 .67 , ,73 .13 ≈ π ≈ 3.1416

• A prime number is a whole number that only has two factors, 1 and itself .

Example: 2, 3, 5, and 7 are prime numbers . They all have two factors: 1 and itself .

• A composite number is a whole number that has more than two factors .

Example: 4, 6, 8, 9, and 10 are composite numbers .

Example: Set-Builder Notation Roster Notation

{x | x is a natural number greater than 2 and less than 7} . {3, 4, 5, 6} {y | y is a prime number between 2 and 9} . {3, 5, 7} {a | a is an integer greater than -2 and less than 4} . {-1, 0, 1, 2, 3}

Example: Given A = {-2, π, √7, 45

} . Specify the following sets . {x | x is a negative number} . {-2}

{b | b is an irrational number} . { π, √𝟕𝟕}

Example: If I = irrational numbers, Z = integers, and N = natural numbers, then list the

numbers in the following sets .

1. A = { x | x ∈ Z, x is less than or equal to -4 and greater than -7} . {-7 < x ≤ -4}

2. A = { a | a ∈ N, a is a prime number between 15 and 25} . {17, 19, 23}3. A = { y | y ∈ I, y is greater than 7

8and less than 3

8} . ∅

∅ is an empty set that has no numbers .

34

-213

π ≈ 3.1416 -0.67 √3 ≈ 1.73

Page 1- 3

-5 -4 -3 -2 -1 0 1 2 3 4 5





Basic Mathematic Symbols

• Basic mathematic symbolsSymbol Meaning Example

= equal a = b, 2 + 1 = -1 + 4≠ not equal a ≠ b, 2 + 7 ≠ 6≈ approximately a ≈ b, 3.667 ≈ 4> is greater than a > b, 4 > -2< is less than a < b, -3 < 0≥ is greater than or equal to a ≥ b, 4 ≥ 3≤ is less than or equal to a ≤ b, 8 ≤ 9± plus or minus a ± b, 3 ± 2 = 5 and 1∓ minus or plus a ∓ b, 2 ∓ 7 = -5 and 9

( ) or ○ open (empty) circle:the point is not included

orx < 3 3 x < 3 3

[ ] or ● closed (filled) circle: the point is included

or -5 x ≥ -5 -5 x ≥ -5

Example: Sketch the graphs of the following inequalities .

1. x < - 4 or ) -4 -4

2. - 2 .3 ≤ x ∙ or [ -2.3 -2.3

3. { y | -4 ≤ y < 8 } [ ) -4 0 8

• A positive real number > 0, and a negative real number < 0. x

Example: 3 > 0 -2 < 0

• Memory aid for > and < x

bigger > smaller smaller < bigger

• x > y also means y < x x

Example: Write an equivalent inequality . Answer

-3 < y y > -3

x ≥ 2 57

2 𝟓𝟓𝟕𝟕

≤ x

Example: Insert an appropriate symbol for the following numbers . Answer-2 .3 and -9 .6 -2.3 > -9.6 or -9.6 < -2.354

and 34

𝟓𝟓𝟒𝟒

> 𝟑𝟑𝟒𝟒

or 𝟑𝟑𝟒𝟒

< 𝟓𝟓𝟒𝟒

Page 1- 4

)

[





Absolute Value

• Absolute value: geometrically, it is the distance (how far) of a number x from zero on the

number line . It is symbolized by “|𝑥𝑥|” .

Example: |5| is 5 units away from 0 .

• No negatives for absolute value: Distance is always positive, and absolute value is

distance, so the absolute value is never negative .

Example: |2| is 2 units away from 0 . 2 units

-2 0 2

�-2� is also 2 units away from 0 . 2 units

-2 0 2

Example: Evaluate the following .

�-18� = 18

|0 − 11| = �-11� = 11

�- 5

7� = 𝟓𝟓

𝟕𝟕

-�-5� = - (5) = -5

�-2 ∙ 7� = �-14� = 14

|7 − 5| − |3 − 8| = 2 − 5 = -3

Page 1- 5

Absolute Value

• Absolute value: geometrically, it is the distance (how far) of a number x from zero on the

number line . It is symbolized by “|𝑥𝑥|” .


• No negatives for absolute value: Distance is always positive, and absolute value is

distance, so the absolute value is never negative .


-2 0 2


-2 0 2


�-18� = 18

|0 − 11| = �-11� = 11

�- 5

7� = 𝟓𝟓

𝟕𝟕

-�-5� = - (5) = -5

�-2 ∙ 7� = �-14� = 14

|7 − 5| − |3 − 8| = 2 − 5 = -3

Page 1- 5





1-2 OPERATIONS WITH REAL NUMBERS

Operations With Signed Numbers

• Terms of operationsOperations

addition addend + addend = sumsubtraction subtrahend – minuend = difference

multiplication multiplicand × multiplier (factor) (factor)

= product

division dividend ÷ divisor = quotient

• Examples of positive and negative numbers (signed numbers)Meaning Example

temperature + °C: above 0 degree – °C: below 0 degree

+20°C-5°C

money + $: gain or own – $: loss or owe

own +$10,000 owe -$500

sports + points : gain – points: loss

gain 3 points: +3lost 2 points: -2

• Adding signed numbers Example

To add two numbers with the same sign: add their values, 3 + 4 = 7

and keep their common sign . (-2) + (-3) = -5

To add two numbers with different signs: subtract their values, 2 + (-5) = -3

and keep the sign of the larger absolute value . (-3) + 7 = 4

• Subtracting signed numbers Example

Subtract a number by adding its opposite . 3 – (-4) = 3 + (4) = 7-5 – 3 = -5 + (-3) = -8

• Multiplying signed numbersSigns Multiplication Example

(+)(+) = (+) (a)(b) = ab 5 ∙ 6 = 30(–)(+) = (–) (-a)(b) = -ab (-5)(6) = -30(+)(–) = (–) (a)(-b) = -ab (5)(-6) = -30(–)(–) = (+) (-a)(-b) = ab (-5)(-6) = 30

Example

the same sign + (-2) (-5) = 10

different signs – (-3) (2) = -6

even number of negative numbers + (-2)(-3)(-1)(-4) = 24

odd number of negative numbers – (-3)(-1)(-5) = -15

Page 1- 6





Dividing Signed Numbers

• Dividing signed numbers Signs Division Example

++

= +𝑎𝑎𝑏𝑏

= 𝑐𝑐 93

= 3

−−

= +–𝑎𝑎–𝑏𝑏

= 𝑐𝑐 -9-3

= 3

+−

= −𝑎𝑎

–𝑏𝑏= −𝑐𝑐 9

-3= −3

−+

= −–𝑎𝑎𝑏𝑏

= −𝑐𝑐 -93

= −3

Note: - 𝑎𝑎𝑏𝑏

= −𝑎𝑎𝑏𝑏

= 𝑎𝑎−𝑏𝑏

• Properties of zero in division x The number 0 divided by any nonzero number is 0 . 𝟎𝟎

𝑨𝑨= 𝟎𝟎 (A ≠ 0)

Example: 09

= 0 0 apples divided by 9 kids, each kid gets 0 apples .

A number divided by 0 is undefined (not allowed) . 𝑨𝑨𝟎𝟎

𝐢𝐢𝐢𝐢 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐢𝐢𝐮𝐮𝐮𝐮𝐮𝐮

Example: 90

=? 9 apples shared by zero kids has no meaning .

dividing fractions

- Change the divisor to its reciprocal (switch the numerator and denominator) .

- Multiply the resulting fractions . 2110

3752

35

72

53

72

=××

=×=÷

- Signed numbers summary

Operation Method

adding signed numbers

- Add two numbers with the same sign:add their values, and keep their common sign .

- Add two numbers with different signs:subtract their values, and keep the sign of the larger number .

subtracting signed numbers Subtract a number by adding its opposite .multiplying signed numbers (+)(+) = (+), (-)(-) = (+), (-)(+) = (-), (+)(-) = (-)

dividing signed numbers++ = + ,

−− = + ,

+− = − ,

−+ = −

Note: 0𝐴𝐴

= 0 , 𝐴𝐴0

is undefined

• Opposite (or additive or negative inverse): the opposite of a number (two numbers whose

sum is 0) .

Example: 1. The additive inverse of 5 is -5 . 5 + (-5) = 0

2. The additive inverse of - 34

is𝟑𝟑𝟒𝟒

. - 34

+ 34

= 0

3. The additive inverse of 0 is 0 . 0 + 0 = 0

Page 1- 7

Dividing Signed Numbers

• Dividing signed numbers Signs Division Example

++

= +𝑎𝑎𝑏𝑏

= 𝑐𝑐 93

= 3

−−

= +–𝑎𝑎–𝑏𝑏

= 𝑐𝑐 -9-3

= 3

+−

= −𝑎𝑎

–𝑏𝑏= −𝑐𝑐 9

-3= −3

−+

= −–𝑎𝑎𝑏𝑏

= −𝑐𝑐 -93

= −3

Note: - 𝑎𝑎𝑏𝑏

= −𝑎𝑎𝑏𝑏

= 𝑎𝑎−𝑏𝑏

• Properties of zero in division x The number 0 divided by any nonzero number is 0 . 𝟎𝟎

𝑨𝑨= 𝟎𝟎 (A ≠ 0)

Example: 09

= 0 0 apples divided by 9 kids, each kid gets 0 apples .

A number divided by 0 is undefined (not allowed) . 𝑨𝑨𝟎𝟎

𝐢𝐢𝐢𝐢 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐢𝐢𝐮𝐮𝐮𝐮𝐮𝐮

Example: 90

=? 9 apples shared by zero kids has no meaning .

dividing fractions

- Change the divisor to its reciprocal (switch the numerator and denominator) .

- Multiply the resulting fractions . 2110

3752

35

72

53

72

=××

=×=÷

- Signed numbers summary

Operation Method


- Add two numbers with the same sign:add their values, and keep their common sign .

- Add two numbers with different signs:subtract their values, and keep the sign of the larger number .

subtracting signed numbers Subtract a number by adding its opposite .multiplying signed numbers (+)(+) = (+), (-)(-) = (+), (-)(+) = (-), (+)(-) = (-)

dividing signed numbers++ = + ,

−− = + ,

+− = − ,

−+ = −

Note: 0𝐴𝐴

= 0 , 𝐴𝐴0

is undefined

• Opposite (or additive or negative inverse): the opposite of a number (two numbers whose

sum is 0) .

Example: 1. The additive inverse of 5 is -5 . 5 + (-5) = 0

2. The additive inverse of - 34

is𝟑𝟑𝟒𝟒

. - 34

+ 34

= 0

3. The additive inverse of 0 is 0 . 0 + 0 = 0

Page 1- 7





1-3 EXPONENTS & ORDER OF OPERATIONS

Exponential Notation

• Exponent (power) is a number repeatedly multiplied by itself .

• Exponent review

• Basic properties

Name Property Examplezero exponent a0 a0 = 1 (a ≠ 0, 00 is undefined) 150 = 1one exponent a1 a1 = a (But 1n = 1) 71 = 7 , 113 = 1

negative exponent 𝑎𝑎-𝑛𝑛 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛 (a ≠ 0) 4–2 = 1

42= 1

16 1

𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛 or 𝑎𝑎-𝑛𝑛 ∙ 𝑎𝑎𝑛𝑛 = 11

4-2 = 42 = 16

an and a-n are reciprocals: 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎-𝑛𝑛 ∙ 𝑎𝑎𝑛𝑛 = 1𝑎𝑎𝑛𝑛∙ 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛

𝑎𝑎𝑛𝑛= 1

Examples: Evaluate the following .

(-3)2 = (-3) (-3) = 9 an = a · a · a … a

(- 0 .3)3 = - 0.027

-52 = - (52) = -25n ∙ n ∙ n ∙ n ∙ n = n5

�-13�3

= �-13� �-1

3� �-1

3� = -𝟏𝟏

𝟐𝟐𝟐𝟐

�-13�1

= -𝟏𝟏𝟑𝟑

a1 = a

(-111)0 = 1 a0 = 1

�32�

-2= 1

�32�2 = 1

�94�1 = 1 ÷ 9

4= 1 × 4

9= 𝟒𝟒

𝟗𝟗 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

13-2 = 32 = 9 1

𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛

Exponential Notation Example Exponent or power

an = a · a · a · a … a Base Read “a to the nth”

or “the nth power of a .”

24 = 2 ∙ 2 ∙ 2 ∙ 2 = 16

Read “2 to the 4th .”

Page 1- 8

1-3 EXPONENTS & ORDER OF OPERATIONS

Exponential Notation

• Exponent (power) is a number repeatedly multiplied by itself .

• Exponent review

• Basic properties


negative exponent 𝑎𝑎-𝑛𝑛 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛 (a ≠ 0) 4–2 = 1

42= 1

16 1

𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛 or 𝑎𝑎-𝑛𝑛 ∙ 𝑎𝑎𝑛𝑛 = 11

4-2 = 42 = 16

an and a-n are reciprocals: 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎-𝑛𝑛 ∙ 𝑎𝑎𝑛𝑛 = 1𝑎𝑎𝑛𝑛∙ 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛

𝑎𝑎𝑛𝑛= 1

Examples: Evaluate the following .

(-3)2 = (-3) (-3) = 9 an = a · a · a … a

(- 0 .3)3 = - 0.027

-52 = - (52) = -25n ∙ n ∙ n ∙ n ∙ n = n5

�-13�3

= �-13� �-1

3� �-1

3� = -𝟏𝟏

𝟐𝟐𝟐𝟐

�-13�1

= -𝟏𝟏𝟑𝟑

a1 = a

(-111)0 = 1 a0 = 1

�32�

-2= 1

�32�2 = 1

�94�1 = 1 ÷ 9

4= 1 × 4

9= 𝟒𝟒

𝟗𝟗 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

13-2 = 32 = 9 1


Exponential Notation Example Exponent or power

an = a · a · a · a … a Base Read “a to the nth”

or “the nth power of a .”

24 = 2 ∙ 2 ∙ 2 ∙ 2 = 16

Read “2 to the 4th .”

Page 1- 8





Order of Operations

• Order of operations

Order of Operations 1 . brackets or parentheses and absolute values

(innermost first) ( ), [ ], { },

2 . exponent an

3 . multiplication or division (from left-to-right) × and ÷ 4 . addition or subtraction (from left-to-right) + and -

• Memory aid - BEDMAS

B E D M A SBrackets Exponents Divide or Multiply Add or Subtract

• Grouping symbols: If parentheses are inside one another, calculate the inside set first .

Parentheses ( ) are used in the inner most grouping .

Square brackets [ ] are used in the second higher level grouping .

Braces { } are used in the most outer grouping .


1. 4 × 32 + {[5 + (2+1)] - 3} = 4 × 32 + {[5 + 3] - 3} ( ), [ ]

= 4 × 32 + {8 - 3} { }

= 4 × 32 + 5 an

= 4 × 9 + 5 ×

= 36 + 5 +

= 41

2. |𝟒𝟒−𝟔𝟔|+𝟑𝟑∙𝟐𝟐𝟐𝟐𝟐𝟐+𝟔𝟔

= 2+3∙24+6

| | and an

= 2+64+6

×

= 810

= 𝟒𝟒𝟓𝟓

÷ and simplify

3. 𝟐𝟐𝟑𝟑+𝟒𝟒𝟐𝟐−𝟑𝟑∙𝟓𝟓𝟐𝟐|𝟑𝟑−𝟓𝟓|÷(−𝟒𝟒)

= 23+42−3∙52∙2÷(−4)

= 8+16−3∙52∙2÷(-4)

| | and an

= 8+16−15-1

= 9-1

= -𝟗𝟗 × and ÷

Page 1- 9

Order of Operations

• Order of operations

Order of Operations 1 . brackets or parentheses and absolute values

(innermost first) ( ), [ ], { },

2 . exponent an

3 . multiplication or division (from left-to-right) × and ÷ 4 . addition or subtraction (from left-to-right) + and -

• Memory aid - BEDMAS

B E D M A SBrackets Exponents Divide or Multiply Add or Subtract

• Grouping symbols: If parentheses are inside one another, calculate the inside set first .

Parentheses ( ) are used in the inner most grouping .

Square brackets [ ] are used in the second higher level grouping .

Braces { } are used in the most outer grouping .


1. 4 × 32 + {[5 + (2+1)] - 3} = 4 × 32 + {[5 + 3] - 3} ( ), [ ]

= 4 × 32 + {8 - 3} { }

= 4 × 32 + 5 an

= 4 × 9 + 5 ×

= 36 + 5 +

= 41

2. |𝟒𝟒−𝟔𝟔|+𝟑𝟑∙𝟐𝟐𝟐𝟐𝟐𝟐+𝟔𝟔

= 2+3∙24+6

| | and an

= 2+64+6

×

= 810

= 𝟒𝟒𝟓𝟓

÷ and simplify

3. 𝟐𝟐𝟑𝟑+𝟒𝟒𝟐𝟐−𝟑𝟑∙𝟓𝟓𝟐𝟐|𝟑𝟑−𝟓𝟓|÷(−𝟒𝟒)

= 23+42−3∙52∙2÷(−4)

= 8+16−3∙52∙2÷(-4)

| | and an

= 8+16−15-1

= 9-1

= -𝟗𝟗 × and ÷

Page 1- 9





Page 1- 10

1-4 ALGEBRAIC EXPRESSIONS

Evaluating Expressions

• Review of basic algebraic terms

Algebraic Term Description Example

algebraic expression A mathematical phrase that contains numbers, variables, and arithmetic operations.

5x + 2, 3a – 4b + 6, 2𝑦𝑦𝑦𝑦3 + 4

constant A number. x + 2 constant: 2 variable A letter that can be assigned different values. 3 – x variable: x

coefficient The number that is in front of a variable. -6 x coefficient: -6 xz3 coefficient: 1

term A term can be a constant, variable, or the product of a number and variable(s). Terms are separated by addition or subtraction.

3x −25 + 13y2 + 73x

Terms: 3x, - 25 , 13y2 , 73x

like terms The terms that have the same variables and exponents.

2x – y2 −25 + 5x – 7 + 13y2

Like terms: 2x and 5x -y2 and 13y2 , - 2

5 and -7

Note: - In algebra we usually do not write the multiplication sign “×” (to avoid confusing it with the letter x).

- If there is no symbol or sign between a number and letter, it means multiplication, such as 5x = 5 ∙ x .

• To evaluate an expression: x - Replace the variable(s) with number(s).

- Calculate.

Example: Evaluate the following.

1. 𝒙𝒙𝒙𝒙𝒚𝒚𝒚𝒚 , given x = -3 and y = 5. Substitute x for -3 and y for 5.

𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 = -𝟑𝟑𝟑𝟑

𝟓𝟓𝟓𝟓

2. 3a – 4 + 2, given a = 5.

3a – 4 + 2 = 3 ∙ 5 – 4 + 2 Substitute a for 5.

= 15 – 4 + 2 Calculate.

= 13

3. 6𝑥𝑥𝑥𝑥2

𝑦𝑦𝑦𝑦−3+ 7𝑥𝑥𝑥𝑥 − 2, given x = 1 and y = 9.

6𝑥𝑥𝑥𝑥2

𝑦𝑦𝑦𝑦−3+ 7𝑥𝑥𝑥𝑥 − 2 = 6 ∙ 𝟏𝟏𝟏𝟏2

𝟗𝟗𝟗𝟗−3+ 7 ∙ 𝟏𝟏𝟏𝟏 − 2 Substitute x for 1 and y for 9.

= 6 6

+ 7 − 2 = 6 Calculate.





Translating Words Into Algebraic Expressions

• Key or clue words in word problemsAddition

(+)Subtraction

(-)Multiplication

(×)Division

(÷)Equals to

(=)add subtract times divided by equals

sum (of) difference product quotient isplus take away multiplied by over was

total (of) minus double split up arealtogether less (than) twice fit into were

increased by decreased by triple per amounts togain (of) loss (of) of each totalscombined balance how much (total) goes into results in

entire (amount) left how many as much as the same as in all savings out of gives

greater than withdraw ratio (of) yieldscomplete reduced by percenttogether fewer (than) share

more (than) how much more distributeand how many extra average

additional how farexceed

• Translate words into an algebraic expressionAlgebraic

Expression Word PhrasesAlgebraic

Expression Word Phrases

7 + y

the sum of 7 and y

t – 8

8 less than t7 more than y t decreased (or reduced) by 8y increased by 7 subtract 8 from t7 plus y the difference between t and 8

AlgebraicExpression Word Phrases


2x or2 • x

the product of 2 and xz ÷ 3 or 𝒛𝒛

𝟑𝟑

the quotient of z and 32 multiplied by x z divided by 3double (or twice) of x one third of z



y3

the third power of y 4y – 9 9 less than 4 times yy cubed 2(t – 5) twice the difference of t and 5

y raised to the third power 𝟔𝟔 +𝟐𝟐𝟐𝟐𝟑𝟑 6 more than the quotient of 2x by 3

Page 1- 11

Translating Words Into Algebraic Expressions

• Key or clue words in word problemsAddition

(+)Subtraction

(-)Multiplication

(×)Division

(÷)Equals to

(=)add subtract times divided by equals

sum (of) difference product quotient isplus take away multiplied by over was

total (of) minus double split up arealtogether less (than) twice fit into were

increased by decreased by triple per amounts togain (of) loss (of) of each totalscombined balance how much (total) goes into results in

entire (amount) left how many as much as the same as in all savings out of gives

greater than withdraw ratio (of) yieldscomplete reduced by percenttogether fewer (than) share

more (than) how much more distributeand how many extra average

additional how farexceed

• Translate words into an algebraic expressionAlgebraic

Expression Word PhrasesAlgebraic

Expression Word Phrases

7 + y

the sum of 7 and y

t – 8

8 less than t7 more than y t decreased (or reduced) by 8y increased by 7 subtract 8 from t7 plus y the difference between t and 8



2x or2 • x

the product of 2 and xz ÷ 3 or 𝒛𝒛

𝟑𝟑

the quotient of z and 32 multiplied by x z divided by 3double (or twice) of x one third of z



y3

the third power of y 4y – 9 9 less than 4 times yy cubed 2(t – 5) twice the difference of t and 5

y raised to the third power 𝟔𝟔 +𝟐𝟐𝟐𝟐𝟑𝟑 6 more than the quotient of 2x by 3

Page 1- 11





Properties of Addition and Multiplication

• Review properties of addition Additive Properties Example

commutative property a + b = b + a 4 + 7 = 7 + 4 associative property (a + b) + c = a + (b + c) (7 + 2) + 9 = 7 + (2 + 9)

identity property a + 0 = a 12 .7 + 0 = 12 .7 closure property If a and b are real numbers, If 9 and 11 are real numbers,

then a + b is a real number . then 9 + 11 = 20 is a real number .inverse property -a + a = 0 -100 + 100 = 0

Example: Name the properties . Answer

1. 7 x + 0 = 7x identity property of addition

2. (3 + x) + 11 = 3 + (x + 11) associative property of addition

3. 11y + 7 = 7 + 11y commutative property of addition

4. (4y + 3) + [- (4y + 3)] = 0 inverse property of addition

• Review properties of multiplication

Multiplicative Properties Examplecommutative property a b = b a 9 ∙ 5 = 5 ∙ 9associative property (a b) c = a (b c) (3 ∙ 7) 5 = 3 (7 ∙ 5)

identity property of 1 a ∙ 1 = a 100,000 ∙ 1 = 100,000

closure property If a and b are real numbers, If 3 and 5 are real numbers,then ab is a real number . then (3)(5) = 15 is a real number .

distributive property a (b + c) = ab + ac 2 (3 – 4) = 2 ∙ 3 – 2 ∙ 4zero product property a ∙ 0 = 0 -76 ∙ 0 = 0

inverse property 𝑎𝑎 ∙

𝟏𝟏𝒂𝒂

= 1 1)97(

1)97( =−

⋅− number its reciprocal

Example: Name the property which the given statement illustrates .Answer

1. 3(5x – 2) = 3 ∙ 5x – 3 ∙ 2 distributive property of multiplication= 15x – 6

2. xy = yx commutative property of multiplication

3. 1 ∙ -1(25+6𝑎𝑎)

= -1(25+6𝑎𝑎)

identity property of multiplication

4. (y x) 4z = y (x · 4z) associative property of multiplication

5. -(7 + 3𝑥𝑥) ∙ 1-(7+3𝑥𝑥)

= 1 inverse property of multiplication

6. 5a(2b – 3c) = 10ab – 15ac distributive property of multiplication

7. 14𝑥𝑥𝑥𝑥

· 0 = 0 zero product property of multiplication

(Switch the order)

(Switch the parentheses)

(Switch the order .)

(Switch the parentheses .)

Page 1- 12

Properties of Addition and Multiplication

• Review properties of addition Additive Properties Example


identity property a + 0 = a 12 .7 + 0 = 12 .7 closure property If a and b are real numbers, If 9 and 11 are real numbers,

then a + b is a real number . then 9 + 11 = 20 is a real number .inverse property -a + a = 0 -100 + 100 = 0

Example: Name the properties . Answer

1. 7 x + 0 = 7x identity property of addition

2. (3 + x) + 11 = 3 + (x + 11) associative property of addition

3. 11y + 7 = 7 + 11y commutative property of addition

4. (4y + 3) + [- (4y + 3)] = 0 inverse property of addition

• Review properties of multiplication

Multiplicative Properties Examplecommutative property a b = b a 9 ∙ 5 = 5 ∙ 9associative property (a b) c = a (b c) (3 ∙ 7) 5 = 3 (7 ∙ 5)


closure property If a and b are real numbers, If 3 and 5 are real numbers,then ab is a real number . then (3)(5) = 15 is a real number .

distributive property a (b + c) = ab + ac 2 (3 – 4) = 2 ∙ 3 – 2 ∙ 4zero product property a ∙ 0 = 0 -76 ∙ 0 = 0

inverse property 𝑎𝑎 ∙

𝟏𝟏𝒂𝒂

= 1 1)97(

1)97( =−

⋅− number its reciprocal

Example: Name the property which the given statement illustrates .Answer

1. 3(5x – 2) = 3 ∙ 5x – 3 ∙ 2 distributive property of multiplication= 15x – 6

2. xy = yx commutative property of multiplication

3. 1 ∙ -1(25+6𝑎𝑎)

= -1(25+6𝑎𝑎)

identity property of multiplication

4. (y x) 4z = y (x · 4z) associative property of multiplication

5. -(7 + 3𝑥𝑥) ∙ 1-(7+3𝑥𝑥)

= 1 inverse property of multiplication

6. 5a(2b – 3c) = 10ab – 15ac distributive property of multiplication

7. 14𝑥𝑥𝑥𝑥

· 0 = 0 zero product property of multiplication

(Switch the order)


(Switch the order .)

(Switch the parentheses .)

Page 1- 12





1-5 SIMPLIFYING ALGEBRAIC EXPRESSIONS

Equivalent Expressions

Equivalent expressions: Two expressions are equivalent if they have the same value for all

allowable replacements .

Example: Complete the table by evaluating the expression for the given values . Then

determine whether the expressions are equivalent .3x - x 2x

x = 1 2 2x = 2 4 4x = 0 0 0

Yes, 3x – x and 2x are equivalent.Examples

1. Multiply by 1 to find an equivalent expression with a given

denominator of 3x . Answer𝟐𝟐𝟑𝟑

= 23∙ 1 = 2

3∙ 𝑥𝑥𝑥𝑥

= 𝟐𝟐𝟐𝟐𝟑𝟑𝟐𝟐

2. Simplify: - 𝟖𝟖𝟖𝟖𝟏𝟏𝟐𝟐𝟖𝟖

- 8𝑡𝑡12𝑡𝑡

= - 𝟐𝟐𝟑𝟑

3. Factor: 4xyz – 2yz + 6z 4xyz – 2yz + 6z = 2z(2xy – y + 3)

4. Multiply: -3p(2q – r) -3p(2q – r) = -6pq + 3pr

5. List the terms: 3a – 2bc + d 3a, - 2bc, d

Example: Use properties of addition and multiplication to find an equivalent expression .

Answer

1. ac : ca commutative property of multiplication

2. wt + 6 : 6 + wt , 6 + tw , or tw+ 6 commutative property of addition/ multiplication

3. (a + 6) – b : a + (6 – b) associative property of addition

Example: Use properties of addition and multiplication to find three equivalent expressions .

Answer

1. (t + u) + 3 : (t + 3) + u, (3 + t) + u, t + (u + 3), …commutative/associative property of addition

2. (3 · y) ∙ z : (y · z) ∙ 3, (3 · z) ∙ y, 3 · (y ∙ z), …commutative/associative property of multiplication

Page 1- 13

1-5 SIMPLIFYING ALGEBRAIC EXPRESSIONS

Equivalent Expressions

Equivalent expressions: Two expressions are equivalent if they have the same value for all

allowable replacements .

Example: Complete the table by evaluating the expression for the given values . Then

determine whether the expressions are equivalent .3x - x 2x

x = 1 2 2x = 2 4 4x = 0 0 0

Yes, 3x – x and 2x are equivalent.Examples

1. Multiply by 1 to find an equivalent expression with a given

denominator of 3x . Answer𝟐𝟐𝟑𝟑

= 23∙ 1 = 2

3∙ 𝑥𝑥𝑥𝑥

= 𝟐𝟐𝟐𝟐𝟑𝟑𝟐𝟐

2. Simplify: - 𝟖𝟖𝟖𝟖𝟏𝟏𝟐𝟐𝟖𝟖

- 8𝑡𝑡12𝑡𝑡

= - 𝟐𝟐𝟑𝟑

3. Factor: 4xyz – 2yz + 6z 4xyz – 2yz + 6z = 2z(2xy – y + 3)

4. Multiply: -3p(2q – r) -3p(2q – r) = -6pq + 3pr

5. List the terms: 3a – 2bc + d 3a, - 2bc, d

Example: Use properties of addition and multiplication to find an equivalent expression .

Answer

1. ac : ca commutative property of multiplication

2. wt + 6 : 6 + wt , 6 + tw , or tw+ 6 commutative property of addition/ multiplication

3. (a + 6) – b : a + (6 – b) associative property of addition

Example: Use properties of addition and multiplication to find three equivalent expressions .

Answer

1. (t + u) + 3 : (t + 3) + u, (3 + t) + u, t + (u + 3), …commutative/associative property of addition

2. (3 · y) ∙ z : (y · z) ∙ 3, (3 · z) ∙ y, 3 · (y ∙ z), …commutative/associative property of multiplication

Page 1- 13





Combining Like Terms

• Like terms: terms that have the same variables and exponents (the numerical coefficients

can be different .)

Example Like or Unlike Terms3x and -14x like terms

-2x2, 43x2 , and -x2 like terms25𝑎𝑎2𝑏𝑏 and

-27𝑎𝑎2𝑏𝑏 like terms

-7t2w3 and 4t2w3 like terms4x and -35y unlike terms6x3 and -9x2 unlike terms

-7a2b3 and 4a3b2 unlike terms

• To combine (or collect) like terms, add or subtract their numerical coefficients and keep

the same variables and exponents .

Note: Unlike terms cannot be combined .

Example: Simplify the following expressions .

1. 7x + 2y – 3x + 11y = (7x – 3x) + (2y + 11y) Regroup like terms .

= 4x + 13y Combine like terms .

2. 2y² – 0.9x + 1.4x – 5y² = (2y² – 5y²) + (-0 .9x + 1 .4x) Regroup like terms .

= -3y² + 0.5x Combine like terms .

3. 2a²b + ab² – 7a²b - 8ab2

= 2a²b + ab² – 7a2b – 8ab² Mark or underline like terms and regroup .

= -5a²b – 7ab2Combine like terms .

4. 𝟏𝟏𝟑𝟑𝒙𝒙 + 𝟓𝟓

𝟐𝟐𝒚𝒚 − 𝟑𝟑

𝟒𝟒𝒙𝒙 − 𝟏𝟏

𝟑𝟑𝒚𝒚 = �1

3𝑥𝑥 − 3

4𝑥𝑥� + �5

2𝑦𝑦 − 1

3𝑦𝑦� Regroup like terms .

= � 412𝑥𝑥 − 9

12𝑥𝑥� + �15

6𝑦𝑦 − 2

6𝑦𝑦� Combine like terms .

= -𝟓𝟓𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟏𝟏𝟑𝟑

𝟔𝟔𝒚𝒚

Page 1- 14

Combining Like Terms

• Like terms: terms that have the same variables and exponents (the numerical coefficients

can be different .)

Example Like or Unlike Terms3x and -14x like terms

-2x2, 43x2 , and -x2 like terms25𝑎𝑎2𝑏𝑏 and

-27𝑎𝑎2𝑏𝑏 like terms

-7t2w3 and 4t2w3 like terms4x and -35y unlike terms6x3 and -9x2 unlike terms

-7a2b3 and 4a3b2 unlike terms

• To combine (or collect) like terms, add or subtract their numerical coefficients and keep

the same variables and exponents .

Note: Unlike terms cannot be combined .


1. 7x + 2y – 3x + 11y = (7x – 3x) + (2y + 11y) Regroup like terms .

= 4x + 13y Combine like terms .

2. 2y² – 0.9x + 1.4x – 5y² = (2y² – 5y²) + (-0 .9x + 1 .4x) Regroup like terms .

= -3y² + 0.5x Combine like terms .

3. 2a²b + ab² – 7a²b - 8ab2

= 2a²b + ab² – 7a2b – 8ab² Mark or underline like terms and regroup .

= -5a²b – 7ab2Combine like terms .

4. 𝟏𝟏𝟑𝟑𝒙𝒙 + 𝟓𝟓

𝟐𝟐𝒚𝒚 − 𝟑𝟑

𝟒𝟒𝒙𝒙 − 𝟏𝟏

𝟑𝟑𝒚𝒚 = �1

3𝑥𝑥 − 3

4𝑥𝑥� + �5

2𝑦𝑦 − 1

3𝑦𝑦� Regroup like terms .

= � 412𝑥𝑥 − 9

12𝑥𝑥� + �15

6𝑦𝑦 − 2

6𝑦𝑦� Combine like terms .

= -𝟓𝟓𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟏𝟏𝟑𝟑

𝟔𝟔𝒚𝒚

Page 1- 14





Removing Parentheses

• If the sign preceding the parentheses is positive (+), do not change the sign of terms

inside the parentheses, just remove the parentheses . Example: (a – 2) = a – 2

• If the sign preceding the parentheses is negative (-), remove the parentheses, omit the

negative (-) sign, and change the sign of terms inside the parentheses .

Example: - (a – 2) = -a + 2

• Remove parentheses

Algebraic Expression Remove Parentheses Example(Ax + B) Ax + B (3x + 4) = 3x + 4(Ax – B) Ax – B �5

6𝑥𝑥 − 3� = 5

6𝑥𝑥 − 3

- (Ax + B) -Ax – B - (2x + 7) = -2x - 7

- (Ax – B) -Ax + B - �13 𝑥𝑥 −25� = -1

3 𝑥𝑥 + 25


1. 2x² + 3 – (x² – 2) = 2x² + 3 – x² + 2 Remove parentheses .

= x² + 5 Combine like terms .

2. - (x² + 3x – 0.5) + 2(2x² – 7x + 𝟑𝟑𝟖𝟖)

= - x² – 3x + 0.5 + 4x² – 14x + 𝟑𝟑𝟒𝟒

Remove parentheses .

= 3x² – 17 x + 𝟓𝟓𝟒𝟒

Combine like terms .

3. -2(t ² – 4t) + 3(t – 4) – (6 + 2t – 3t ²)

= -2t ² + 8t + 3t – 12 – 6 – 2t + 3t ² Remove parentheses .

= t ² + 9t – 18 Combine like terms .

4. 𝟏𝟏𝟑𝟑

(a + 3) – 𝟏𝟏 𝟐𝟐

(a + 2) = 13

a + 1 – 1 2

a – 1 Remove parentheses .

= �13 𝑎𝑎 – 1

2 𝑎𝑎 � + (1 – 1) Combine like terms .

= -𝟏𝟏 𝟔𝟔

a

Page 1- 15

Removing Parentheses

• If the sign preceding the parentheses is positive (+), do not change the sign of terms

inside the parentheses, just remove the parentheses . Example: (a – 2) = a – 2

• If the sign preceding the parentheses is negative (-), remove the parentheses, omit the

negative (-) sign, and change the sign of terms inside the parentheses .

Example: - (a – 2) = -a + 2

• Remove parentheses

Algebraic Expression Remove Parentheses Example(Ax + B) Ax + B (3x + 4) = 3x + 4(Ax – B) Ax – B �5

6𝑥𝑥 − 3� = 5

6𝑥𝑥 − 3

- (Ax + B) -Ax – B - (2x + 7) = -2x - 7

- (Ax – B) -Ax + B - �13 𝑥𝑥 −25� = -1

3 𝑥𝑥 + 25


1. 2x² + 3 – (x² – 2) = 2x² + 3 – x² + 2 Remove parentheses .

= x² + 5 Combine like terms .

2. - (x² + 3x – 0.5) + 2(2x² – 7x + 𝟑𝟑𝟖𝟖)

= - x² – 3x + 0.5 + 4x² – 14x + 𝟑𝟑𝟒𝟒

Remove parentheses .

= 3x² – 17 x + 𝟓𝟓𝟒𝟒


3. -2(t ² – 4t) + 3(t – 4) – (6 + 2t – 3t ²)

= -2t ² + 8t + 3t – 12 – 6 – 2t + 3t ² Remove parentheses .

= t ² + 9t – 18 Combine like terms .

4. 𝟏𝟏𝟑𝟑

(a + 3) – 𝟏𝟏 𝟐𝟐

(a + 2) = 13

a + 1 – 1 2

a – 1 Remove parentheses .

= �13 𝑎𝑎 – 1

2 𝑎𝑎 � + (1 – 1) Combine like terms .

= -𝟏𝟏 𝟔𝟔

a

Page 1- 15





1-6 EXPONENTS & SCIENTIFIC NOTATION

Rules of Exponents

Example: Simplify the following .

1. 102 10-3 = 102 - 3 = 10-1 = 𝟏𝟏𝟏𝟏𝟏𝟏

am an = am + n , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

2. 𝒘𝒘-𝟓𝟓

𝒘𝒘𝟐𝟐= 𝑤𝑤-5−2 = 𝑤𝑤-7 = 𝟏𝟏

𝒘𝒘𝟕𝟕 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛 , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

3. (y -3) -2 = y -3(-2) = y 6 (an)m = an m

4. [(-3) ∙ (𝟏𝟏.𝟒𝟒)]2 = (-3)2 ∙ 0 .42 = (9) (0 .16) = 1.44 (a ∙b)n = an bn

5. (3x2 ∙ y -3 )3 = 33 ∙ x2∙3 ∙ y -3∙3 = 27x6 y -9 = 𝟐𝟐𝟕𝟕𝟐𝟐𝟔𝟔

𝒚𝒚𝟗𝟗(am∙bn)p = amp bnp , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

6. �𝟐𝟐𝒚𝒚�

-𝟑𝟑= 𝑥𝑥-3

𝑦𝑦-3 = 𝒚𝒚𝟑𝟑

𝟐𝟐𝟑𝟑 �𝑎𝑎

𝑏𝑏�𝑛𝑛

= 𝑎𝑎𝑛𝑛

𝑏𝑏𝑛𝑛 , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛 , 1


7. � 𝒕𝒕𝟓𝟓

𝒖𝒖-𝟐𝟐�𝟐𝟐

= 𝑡𝑡5∙2

𝑢𝑢�-2�(2) = 𝑡𝑡10

𝑢𝑢-4 = 𝒕𝒕𝟏𝟏𝟏𝟏𝒖𝒖𝟒𝟒 �𝑎𝑎𝑚𝑚

𝑏𝑏𝑛𝑛�𝑝𝑝

= 𝑎𝑎𝑚𝑚𝑝𝑝

𝑏𝑏𝑛𝑛𝑝𝑝, 1



1. (-𝟒𝟒𝒖𝒖𝟑𝟑)𝟐𝟐(𝟑𝟑𝒘𝒘𝟒𝟒)-𝟑𝟑(-𝟏𝟏𝟐𝟐𝟕𝟕𝟏𝟏-𝟏𝟏𝟓𝟓)𝟏𝟏

= (-4)2(𝑢𝑢3∙2)(3-3𝑤𝑤4(-3))(1) (am∙bn)p = amp bnp, a0 = 1

= 16(𝑢𝑢6)(3-3𝑤𝑤-12))

= 16(𝑢𝑢6)33𝑤𝑤12 = 𝟏𝟏𝟔𝟔𝒖𝒖𝟔𝟔

𝟐𝟐𝟕𝟕𝒘𝒘𝟏𝟏𝟐𝟐 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

2. �(𝟒𝟒𝟒𝟒𝟑𝟑)(𝒃𝒃𝟓𝟓)𝟐𝟐𝟒𝟒𝟐𝟐𝒃𝒃𝟒𝟒

�𝟐𝟐

= (4𝑎𝑎3)𝟐𝟐(𝑏𝑏5)𝟐𝟐

(2𝑎𝑎2𝑏𝑏4)𝟐𝟐= (4𝟐𝟐𝑎𝑎3∙𝟐𝟐)(𝑏𝑏5∙𝟐𝟐)

2𝟐𝟐𝑎𝑎2∙𝟐𝟐𝑏𝑏4∙𝟐𝟐�𝑎𝑎

𝑚𝑚


= 𝑎𝑎𝑚𝑚𝑚𝑚

𝑏𝑏𝑛𝑛𝑚𝑚, (a ∙ b)n = an bn

= (16𝑎𝑎6)(𝑏𝑏10)4𝑎𝑎4𝑏𝑏8

= 𝟒𝟒𝟒𝟒𝟐𝟐𝒃𝒃𝟐𝟐 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛

Name Rule Exampleproduct of like bases am an = am + n (a ≠ 0) 23 22 = 23 + 2 = 25 = 32quotient of like bases (the same base)

𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛 (a ≠ 0) 𝑦𝑦

3

𝑦𝑦2 = 𝑦𝑦3−2 = 𝑦𝑦1 = 𝑦𝑦

power of a power (am)n = amn (x3) 2 = x3 · 2 = x6

power of a product(different bases)

(a ∙ b)n = anbn (a, b ≠ 0) (2 ∙ 3)2 = 22 32 = 4 ∙ 9 = 36

(am ∙ bn)p = amp bnp (t 3 ∙ s 4)2 = t 3∙2 s 4∙2 = t6 s8

power of a quotient(different bases)

�𝑎𝑎𝑏𝑏�𝑛𝑛

= 𝑎𝑎𝑛𝑛

𝑏𝑏𝑛𝑛 (b ≠ 0) �

23�

2

=22

32 =49

�𝑎𝑎𝑚𝑚



𝑏𝑏𝑛𝑛𝑚𝑚 �

𝑞𝑞2

𝑝𝑝4�3

=𝑞𝑞2∙3

𝑝𝑝4∙3 =𝑞𝑞6

𝑝𝑝12

Page 1- 16





Page 1- 17

Scientific Notation

• Scientific notation is a special way of concisely expressing very large and small numbers.

Example: 300,000,000 = 3 × 108 m/sec the speed of light

0.00000000000000000016 = 1.6 × 10-19 C an electron

• Scientific notation: a product of a number between 1 and 10 and power of 10. N × 10±n

Scientific Notation Example N × 10±n

1 ≤ N < 10 n - integer

34,005.9 = 3.40059 × 104

standard form scientific notation

• Scientific vs. non-scientific notation

Scientific Notation Not Scientific Notation 3.5 × 103 35 × 102 35 > 10, 35 is not between 1 and 10 4.3 × 10-2 0.043 0.043 < 1, 0 is not between 1 and 10 5.3 × 1022 0.53 × 1023 0.53 < 1, 0 is not between 1 and 10 1.03 × 108 10.3 × 107 N should be < 10

• Writing a number in scientific notation x Step Example

- Move the decimal point after the first nonzero digit. 0.0035 43270000.

- Determine n (the power of 10) by counting the n = 3 n = 7 number of places you moved the decimal.

- If the decimal point is moved to the right: × 10-n 0.0035 = 3.5 × 10-3 3 places to the right

- If the decimal point is moved to the left: × 10n 43270000 = 4.327×107

7 places to the left Example: Write in scientific notation.

1. 135,000 =135,000. = 1.35× 105 5 places to the left, × 10n

2. 0.0000000548 = 5.48 × 10-8 8 places to the right,

× 10-n

Example: Simplify and write in scientific notation.

1. �3.4 × 10-4� (4.79 × 107) = (3.4 × 4.79) � 10-4+7� Multiply coefficients of 10±n, aman=am+n

= (16.286 × 103) 16.286 > 10, this is not in scientific notation.

= (𝟏𝟏𝟏𝟏.𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒) 1.6.286 < 10, this is in scientific notation.

2. (4×10-4)(1.5×103)5.2×106

= 4×1.55.2

× (10-4×103)106

Regroup coefficients of 10±n

≈ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏-𝟕𝟕𝟕𝟕 am an = am + n , 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚

𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚−𝑛𝑛𝑛𝑛

Page 1- 17

Scientific Notation

• Scientific notation is a special way of concisely expressing very large and small numbers.

Example: 300,000,000 = 3 × 108 m/sec the speed of light

0.00000000000000000016 = 1.6 × 10-19 C an electron

• Scientific notation: a product of a number between 1 and 10 and power of 10. N × 10±n

Scientific Notation Example N × 10±n


34,005.9 = 3.40059 × 104


• Scientific vs. non-scientific notation

Scientific Notation Not Scientific Notation 3.5 × 103 35 × 102 35 > 10, 35 is not between 1 and 10 4.3 × 10-2 0.043 0.043 < 1, 0 is not between 1 and 10 5.3 × 1022 0.53 × 1023 0.53 < 1, 0 is not between 1 and 10 1.03 × 108 10.3 × 107 N should be < 10

• Writing a number in scientific notation x Step Example

- Move the decimal point after the first nonzero digit. 0.0035 43270000.

- Determine n (the power of 10) by counting the n = 3 n = 7 number of places you moved the decimal.

- If the decimal point is moved to the right: × 10-n 0.0035 = 3.5 × 10-3 3 places to the right

- If the decimal point is moved to the left: × 10n 43270000 = 4.327×107

7 places to the left Example: Write in scientific notation.

1. 135,000 =135,000. = 1.35× 105 5 places to the left, × 10n

2. 0.0000000548 = 5.48 × 10-8 8 places to the right,

× 10-n

Example: Simplify and write in scientific notation.

1. �3.4 × 10-4� (4.79 × 107) = (3.4 × 4.79) � 10-4+7� Multiply coefficients of 10±n, aman=am+n

= (16.286 × 103) 16.286 > 10, this is not in scientific notation.

= (𝟏𝟏𝟏𝟏.𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒) 1.6.286 < 10, this is in scientific notation.

2. (4×10-4)(1.5×103)5.2×106

= 4×1.55.2

× (10-4×103)106

Regroup coefficients of 10±n

≈ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏-𝟕𝟕𝟕𝟕 am an = am + n , 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚

𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚−𝑛𝑛𝑛𝑛





Unit 1 Summary

• Roster Notation { } Roster Notation { } Example

List all the elements or numbers of the set .

The set of odd numbers between 3 and 17: { 5, 7, 9, 11, 13, 15}

• Set-builder notationSet-Builder Notation Example

{ x | x … }


{ x | x > 3 }


• The real number system x


4 , -2 .13

Integers: … -2, -1, 0, 1, … Irrational Numbers Whole Numbers: 0, 1, 2, 3 … √5 , π , …

Natural Numbers: 1,2, 3 …




orx < 3 3 x < 3 3


or -5 x ≥ -5 -5 x ≥ -5

• Order of operationsOrder of Operations

1 . brackets or parentheses and absolute values(innermost first)

( ) , [ ] , { } ,

2 . exponent an

3 . multiplication or division (from left to right) × and ÷ 4 . addition or subtraction (from left to right) + and –

Page 1- 18

Unit 1 Summary

• Roster Notation { } Roster Notation { } Example

List all the elements or numbers of the set .

The set of odd numbers between 3 and 17: { 5, 7, 9, 11, 13, 15}

• Set-builder notationSet-Builder Notation Example

{ x | x … }


{ x | x > 3 }


• The real number system x


4 , -2 .13

Integers: … -2, -1, 0, 1, … Irrational Numbers Whole Numbers: 0, 1, 2, 3 … √5 , π , …

Natural Numbers: 1,2, 3 …




orx < 3 3 x < 3 3


or -5 x ≥ -5 -5 x ≥ -5

• Order of operationsOrder of Operations

1 . brackets or parentheses and absolute values(innermost first)

( ) , [ ] , { } ,

2 . exponent an

3 . multiplication or division (from left to right) × and ÷ 4 . addition or subtraction (from left to right) + and –

Page 1- 18

Rational Numbers: 34, -2 .13Integers: … -2, -1, 0, 1, …Whole Numbers: 0, 1, 2, 3 …Natural Numbers: 1, 2, 3 …

)

[





Page 1- 19

• Signed numbers summary Operation Method


- Add two numbers with the same sign: add their values, and keep their common sign.

- Add two numbers with different signs: subtract their values, and keep the sign of the larger absolute value.

subtracting signed numbers Subtract a number by adding its opposite.

multiplying signed numbers (+)(+) = (+), (–)(–) = (+), (–)(+) = (–), (+)(–) = (–)

dividing signed numbers ++

= + , −−

= + , +−

= − , −+

= −

Note: 0𝐴𝐴𝐴𝐴

= 0 , 𝐴𝐴𝐴𝐴0

is undefined

• Review basic algebraic terms Algebraic Term Description Example

algebraic expression A mathematical phrase that contains numbers, variables, and arithmetic operations. 5x + 2, 3a – 4b + 6, 2𝑦𝑦𝑦𝑦3 + 4

constant A number. x + 2 constant: 2 variable A letter that can be assigned different values. 3 – x variable: x

coefficient The number that is in front of a variable. -6 x coefficient: -6 xz3 coefficient: 1

term A term can be a constant, variable, or the product of a number and variable(s). Terms are separated by addition or subtraction.

3x −25 + 13y2 + 73x

Terms: 3x, - 25 , 13y2 , 73x

like terms The terms that have the same variables and exponents.

2x – y2 −25 + 5x – 7 + 13y2

Like terms: 2x and 5x -y2 and 13y2 , - 2

5 and -7

• Properties of addition Additive Properties Example


identity property a + 0 = a 12.7 + 0 = 12.7 closure property If a and b are real numbers, If 9 and 11 are real numbers,

then a + b is a real number. then 9 + 11 = 20 is a real number. inverse property -a + a = 0 -100 + 100 = 0

• Properties of multiplication Multiplicative Properties Example

commutative property a b = b a 9 ∙ 5 = 5 ∙ 9 associative property (a b) c = a (b c) (3 ∙ 7) 5 = 3 (7 ∙ 5)


closure property If a and b are real numbers, If 3 and 5 are real numbers, then ab is a real number. then (3)(5) = 15 is a real number.

distributive property a (b + c) = ab + ac 2 (3 – 4) = 2 ∙ 3 – 2 ∙ 4 zero product property a ∙ 0 = 0 -76 ∙ 0 = 0

inverse property 1)97(

1)97( =−

⋅−

number its reciprocal

(Switch the order)


(Switch the order)


𝑎𝑎𝑎𝑎 ∙

𝟏𝟏𝟏𝟏𝒂𝒂𝒂𝒂

= 1





• Terms of operationsOperations

addition addend + addend = sumsubtraction subtrahend – minuend = difference

multiplication multiplicand × multiplier (factor) (factor)

= product

division dividend ÷ divisor = quotient

• Remove parenthesesAlgebraic Expression Remove Parentheses Example

(Ax + B) Ax + B (3x + 4) = 3x + 4(Ax – B) Ax – B �5

6𝑥𝑥 − 3� = 5

6𝑥𝑥 − 3

- (Ax + B) -Ax – B - (2x + 7) = -2x - 7- (Ax – B) -Ax + B - �1

3𝑥𝑥 − 2

5� = -1

3𝑥𝑥 + 2

5

• Rules of exponents


negative exponent 𝑎𝑎−𝑛𝑛 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛 (a ≠ 0) 4–2 = 1

42= 1

16 1

𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛 or 𝑎𝑎−𝑛𝑛 ∙ 𝑎𝑎𝑛𝑛 = 11

4-2 = 42 = 16

product of like bases am an = am + n (a ≠ 0) 23 22 = 23 + 2 = 25 = 32quotient of like bases (the same base)

𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛 (a ≠ 0) 𝑦𝑦

3

𝑦𝑦2 = 𝑦𝑦3−2 = 𝑦𝑦1 = 𝑦𝑦

power of a power (am)n = amn (x3) 2 = x3 · 2 = x6

power of a product(different bases)

(a ∙ b)n = anbn (a, b ≠ 0) (2 ∙ 3)2 = 22 32 = 4 ∙ 9 = 36

(am ∙ bn)p = amp bnp (t 3 ∙ s 4)2 = t 3∙2 s 4∙2 = t6 s8

power of a quotient(different bases)

�𝑎𝑎𝑏𝑏�𝑛𝑛

= 𝑎𝑎𝑛𝑛

𝑏𝑏𝑛𝑛 (b ≠ 0) �

23�

2

=22

32 =49

�𝑎𝑎𝑚𝑚



𝑏𝑏𝑛𝑛𝑚𝑚 �

𝑞𝑞2

𝑝𝑝4�3

=𝑞𝑞2∙3

𝑝𝑝4∙3 =𝑞𝑞6

𝑝𝑝12

• Absolute value |𝒙𝒙| : the distance of a number x from zero on the number lineNo negatives for absolute value: �-𝑏𝑏� = |𝑏𝑏|

• Opposite (or additive or negative inverse): the opposite of a number (two numbers whose sum is 0) .

• Equivalent expressions: two expressions are equivalent if they have the same value for all allowable replacements .

• Combine (or collect) like terms: add or subtract their numerical coefficients and keep the same variables and exponents .

• Scientific notation

Scientific Notation ExampleN × 10±n


34,005 .9 = 3 .40059 × 104


Page 1- 20





PRACTICE QUIZ

Unit 1 Fundamental Concepts

1 Write in set-builder notation and roster notation: .


. Given A = {-9, 7π, √5, 23

} . Specify the following sets .2

a. {x | x is a negative number}

b. {b | b is an irrational number}

. Sketch the graphs of the following inequalities .3

a . x < - 2

b. - 1 .5 ≤ x < 7

. Perform the indicated operations .4

a. (-1 .3) + (-2)

b . 12 − (-7)

c . (-3)(-0 .1)(-5)

d . �- 35� ÷ 1

5

. Evaluate the following .5

a . (- 0 .2)3

b . m ∙ m ∙ m

c . (-10855)0

𝐝𝐝. 32+ 52−2 ∙ 23|2−7|÷(-3)

. Evaluate 3𝑦𝑦2

𝑥𝑥−2+ 7𝑦𝑦 − 4 , given x = -1 and y = 3 .6

Page 1

PRACTICE QUIZ

Unit 1 Fundamental Concepts

1 Write in set-builder notation and roster notation: .


. Given A = {-9, 7π, √5, 23

} . Specify the following sets .2

a. {x | x is a negative number}

b. {b | b is an irrational number}

. Sketch the graphs of the following inequalities .3

a . x < - 2

b. - 1 .5 ≤ x < 7

. Perform the indicated operations .4

a. (-1 .3) + (-2)

b . 12 − (-7)

c . (-3)(-0 .1)(-5)

d . �- 35� ÷ 1

5

. Evaluate the following .5

a . (- 0 .2)3

b . m ∙ m ∙ m

c . (-10855)0

𝐝𝐝. 32+ 52−2 ∙ 23|2−7|÷(-3)

. Evaluate 3𝑦𝑦2

𝑥𝑥−2+ 7𝑦𝑦 − 4 , given x = -1 and y = 3 .6

Page 1





. Translate words into algebraic expression .7

a . 3 less than the product of 7 and y .

b . Twice the sum of t and 9 .

. Name the property . 8

a. (2x + 5) + [- (2x + 5)] = 0

b. (b a) 7c = b (a · 7c)

a . Factor: 5abc – 25bc + 35c9.b . Multiply: -3p(2q – r)

c . Combine like terms: 4x² – 0 .5y + 1 .5y – 2x²

d . Simplify: -3(x² – 2x) + 5(x – 3) – (4 + 3x – 2x²)

. Simplify the following .10

a . (-2𝑥𝑥2)3(𝑦𝑦3)-4(-2.357𝑧𝑧-178)0

b. �(2𝑥𝑥2) 𝑦𝑦4

4𝑥𝑥3𝑦𝑦2�3

. Simplify and write in scientific notation .11

a . (4.3 × 10-5)(3.25 × 109)

b .(3×10-5)(2.3×104)

1.2×107

Page 2




Algebra I & II Key Concepts, Practice, and Quizzes Unit 2 – Equations and Inequalities

UNIT 2 EQUATIONS AND INEQUALITIES

2-1 SOLVING EQUATIONS

Equations

• Equation: A mathematical statement that contains two expressions separated by an equal

sign (both sides of the equation have the same value) .

Example: 3 x + 2 = 5

• To solve an equation is the process of finding a particular value for the variable in the

equation that makes the equation true .

Example: For the equation 3x + 2 = 5 , only x = 1 can make it true, since 3 ∙ 1 + 2 = 5 .

• Solution, root, or zero of an equation: the particular value of the variable in the equation

that makes the equation true . This value is also called “root” or “zero” of the equation .

Example: For the equation 3x + 2 = 5, x = 1 is the solution .

More examples: Indicate whether each of the given number is a solution .? √

a. 5: 2x – 3 = 7 2∙5 – 3 = 7 7 = 7 Yes ? √

b. -3: 10 + 412

y = 9 10 + 412

(-3) = 9 9 = 9 Yes ?

c. 1: 3t + 2(t – 4) = 5t – 6 3∙1 + 2 (1 – 4) = 5∙1 – 6 ?

3 – 6 = -1 -3 ≠ -1 No

• Solution Set { }: the set of all values that makes the equation true .

Example: The solution set to x2 – 4 = 0 is {-2, 2}. √ √

Since (-2)2 – 4 = 0 and 22 – 4 = 0

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Linear Equations

• Linear equation (or first-degree equation) in one variable: an equation in which the

highest power (exponent) of the variable is one. (An equation whose graph is a straight line.)

• Standard form of a linear equation in one variable: Ax + B = 0 x = x1 A ≠ 0, A is a coefficient, B is a constant.

Examples of linear equations: 3x + 2 = 0

7y – 5 = 3 + 2y

9 + 5𝑥𝑥𝑥𝑥11

= 4 – 3x

• Second-degree equation: an equation in which the highest power of the variable is two.

Example: 5x2 + 6x + 7 = 0

In general, a first-degree equation contains an “x” term (or any variable), a second-degree equation contains an “x2” term, and a third-degree equation contains an “x3” term, etc.

• Equations of different degrees

Equation Standard Form Example Comments first-degree equation

(linear equation) A x + B = 0

(x = x1) 5x + 4 = 0 The highest power of x is 1.

second-degree equation (quadratic equation)

Ax2 + Bx + C = 0 2x2 + 7x – 3 = 0 The highest power of x is 2.

third-degree equation (cubic equation)

Ax3 + Bx2 + Cx + D = 0 3x3 + 4x2 – 8x + 1 = 0 The highest power of x is 3.

fourth-degree equation Ax4 + Bx3 + Cx2 +D x + E = 0 x4 – 9x3 + 3x2 + 2x – 5 = 0 The highest power of x is 4.

• Higher-degree equations are nonlinear equations.

• A linear equation in two variables: an equation with two variables in which the highest

power (exponent) of two variables is one.

Standard form: Ax + By = C A, and B are coefficients,

Example: 2x + y = 3 C is a constant.

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Linear Equations


highest power (exponent) of the variable is one. (An equation whose graph is a straight line.)

• Standard form of a linear equation in one variable: Ax + B = 0 x = x1 A ≠ 0, A is a coefficient, B is a constant.

Examples of linear equations: 3x + 2 = 0

7y – 5 = 3 + 2y

9 + 5𝑥𝑥𝑥𝑥11

= 4 – 3x

• Second-degree equation: an equation in which the highest power of the variable is two.

Example: 5x2 + 6x + 7 = 0

In general, a first-degree equation contains an “x” term (or any variable), a second-degree equation contains an “x2” term, and a third-degree equation contains an “x3” term, etc.


Equation Standard Form Example Comments first-degree equation

(linear equation) A x + B = 0

(x = x1) 5x + 4 = 0 The highest power of x is 1.

second-degree equation (quadratic equation)

Ax2 + Bx + C = 0 2x2 + 7x – 3 = 0 The highest power of x is 2.

third-degree equation (cubic equation)

Ax3 + Bx2 + Cx + D = 0 3x3 + 4x2 – 8x + 1 = 0 The highest power of x is 3.

fourth-degree equation Ax4 + Bx3 + Cx2 +D x + E = 0 x4 – 9x3 + 3x2 + 2x – 5 = 0 The highest power of x is 4.


• A linear equation in two variables: an equation with two variables in which the highest

power (exponent) of two variables is one.

Standard form: Ax + By = C A, and B are coefficients,

Example: 2x + y = 3 C is a constant.





Properties of Equality

Properties for solving equations

Properties Equality Example

property of addition A = B, A + C = B + CSolve y – 7 = 2y – 7 + 7 = 2 + 7, y = 9

property of subtraction A = B, A – C = B – CSolve x + 3 = -8x + 3 – 3 = -8 – 3, x = -11

property of multiplication A = B, A ∙ C = B ∙ C (C ≠ 0)

Solve -𝑡𝑡6

= 7-𝑡𝑡6

(-𝟔𝟔) = 7(- 6), t = -42

property of division A = B, 𝐴𝐴𝑪𝑪

= 𝐵𝐵𝑪𝑪

(C ≠ 0)

Solve 4a = -16

4𝑎𝑎𝟒𝟒

= -16𝟒𝟒

, a = -4

Example: Solve the following equations .

Solution

1. -7 + x = 3 -7 + x + 7 = 3 + 7 Property of addition

x = 10? √

Check: -7 + 10 = 3 3 = 3 Replace x with 10 .

2. y + 𝟏𝟏𝟑𝟑

= -1 y + 13 −

𝟏𝟏

𝟑𝟑 = -1 −

𝟏𝟏

𝟑𝟑Property of subtraction

y = - 𝟒𝟒𝟑𝟑

3. -𝟏𝟏𝟒𝟒

x = 5 -𝟒𝟒 ∙ -14 𝑥𝑥 = 5(-𝟒𝟒) Property of multiplication

x = -20

4. -2x = 14 -2𝑥𝑥-𝟐𝟐

= 14-𝟐𝟐

Property of division

x = -7

5. 0.3y = -0.96 0.3𝑦𝑦𝟎𝟎.𝟑𝟑

= -0.96𝟎𝟎.𝟑𝟑


y = -3.2

6. -2x – 3 = 5 -2x – 3 + 3 = 5 + 3 Property of addition

-2x = 8, -2𝑥𝑥-2

= 8-2


x = -4

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Properties of Equality

Properties for solving equations





Solve -𝑡𝑡6

= 7-𝑡𝑡6

(-𝟔𝟔) = 7(- 6), t = -42


= 𝐵𝐵𝑪𝑪

(C ≠ 0)

Solve 4a = -16

4𝑎𝑎𝟒𝟒

= -16𝟒𝟒

, a = -4

Example: Solve the following equations .

Solution

1. -7 + x = 3 -7 + x + 7 = 3 + 7 Property of addition

x = 10? √

Check: -7 + 10 = 3 3 = 3 Replace x with 10 .

2. y + 𝟏𝟏𝟑𝟑

= -1 y + 13 −

𝟏𝟏

𝟑𝟑 = -1 −

𝟏𝟏

𝟑𝟑Property of subtraction

y = - 𝟒𝟒𝟑𝟑

3. -𝟏𝟏𝟒𝟒

x = 5 -𝟒𝟒 ∙ -14 𝑥𝑥 = 5(-𝟒𝟒) Property of multiplication

x = -20

4. -2x = 14 -2𝑥𝑥-𝟐𝟐

= 14-𝟐𝟐


x = -7

5. 0.3y = -0.96 0.3𝑦𝑦𝟎𝟎.𝟑𝟑

= -0.96𝟎𝟎.𝟑𝟑


y = -3.2

6. -2x – 3 = 5 -2x – 3 + 3 = 5 + 3 Property of addition

-2x = 8, -2𝑥𝑥-2

= 8-2


x = -4

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Procedure for Solving Equations

Equation-Solving Strategy • Clear the fractions or decimal if necessary . • Remove parentheses . • Combine like terms on each side of the equation if necessary .• Collect the variable terms on one side of the equation and the numerical terms on

the other side . • Isolate the variable . • Check the solution with the original equation .

Procedure for solving linear equations

Steps Example: Solve 𝟏𝟏𝟐𝟐

(𝒙𝒙 + 𝟏𝟏) = 𝟑𝟑𝒙𝒙 − 𝟐𝟐𝒙𝒙 .

- Eliminate the denominators if the equation 2 ∙ 12

(𝑥𝑥 + 1) = 𝟐𝟐(3𝑥𝑥) − 𝟐𝟐(2𝑥𝑥)

has fractions . Multiply each term by 2 .

- Remove parentheses . x + 1 = 6𝑥𝑥 − 4𝑥𝑥

- Combine like terms . x + 1 = 2𝑥𝑥

- Collect variable terms on one side and the x + 1 – 2x – 1 = 2x – 2x – 1

constants on the other side . Subtract 2x and 1 from both sides .

- Isolate the variable . -x = -1 Divide both sides by -1 .

x = 1?

- Check . 12

(𝟏𝟏 + 1) = 3 ∙ 𝟏𝟏 − 2 ∙ 𝟏𝟏 √

1 = 1 Correct!

Example: Solve 4(y – 3) + 3y + 2 = 2(4 – y) . Remove parentheses.

4y – 12 + 3y + 2 = 8 – 2y Combine like terms .

7y – 10 = 8 – 2y Add 2y & 10 to both sides .

7y – 10 + 2y + 10 = 8 – 2y + 2y + 10

9y = 18 Isolate the variable .

y = 2 ?

Check: 4(2 – 3) + 3∙2 + 2 = 2(4 – 2) √

4 = 4 Correct!

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Procedure for Solving Equations

Equation-Solving Strategy • Clear the fractions or decimal if necessary . • Remove parentheses . • Combine like terms on each side of the equation if necessary .• Collect the variable terms on one side of the equation and the numerical terms on

the other side . • Isolate the variable . • Check the solution with the original equation .

Procedure for solving linear equations

Steps Example: Solve 𝟏𝟏𝟐𝟐

(𝒙𝒙 + 𝟏𝟏) = 𝟑𝟑𝒙𝒙 − 𝟐𝟐𝒙𝒙 .

- Eliminate the denominators if the equation 2 ∙ 12

(𝑥𝑥 + 1) = 𝟐𝟐(3𝑥𝑥) − 𝟐𝟐(2𝑥𝑥)

has fractions . Multiply each term by 2 .

- Remove parentheses . x + 1 = 6𝑥𝑥 − 4𝑥𝑥

- Combine like terms . x + 1 = 2𝑥𝑥

- Collect variable terms on one side and the x + 1 – 2x – 1 = 2x – 2x – 1

constants on the other side . Subtract 2x and 1 from both sides .

- Isolate the variable . -x = -1 Divide both sides by -1 .

x = 1?

- Check . 12

(𝟏𝟏 + 1) = 3 ∙ 𝟏𝟏 − 2 ∙ 𝟏𝟏 √

1 = 1 Correct!

Example: Solve 4(y – 3) + 3y + 2 = 2(4 – y) . Remove parentheses.

4y – 12 + 3y + 2 = 8 – 2y Combine like terms .

7y – 10 = 8 – 2y Add 2y & 10 to both sides .

7y – 10 + 2y + 10 = 8 – 2y + 2y + 10

9y = 18 Isolate the variable .

y = 2 ?

Check: 4(2 – 3) + 3∙2 + 2 = 2(4 – 2) √

4 = 4 Correct!

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Equations Involving Decimals/Fractions

• Equations involving decimals x

Steps Example: Solve 0.25x – 0.20 = -3.15x.

- Multiply each term by 100 to clear the decimal. 100(0.25x) – 100(0.20) = 100(-3.15x)

- Collect the variable terms on one side of the 25x – 20 = -315x

equation and the constants on the other side. 25x + 315x = 20

340x = 20

- Isolate the variable. x ≈ 0.06

Example: Solve 0.3y + 0.06 = 0.009. Multiply each term by 1,000.

1,000(0.3y) + 1,000(0.06) = 1,000(0.009)

300y + 60 = 9 Combine like terms.

300y = -51 Divide both sides by 300.

y = -0.17

Tip: Multiply every term of both sides of the equation by a power of 10 (10, 100, 1000, etc.) to clear the decimals.

• Equations involving fractions x

Steps Example: Solve 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

+ 𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒

= - 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐− 𝟐𝟐𝟐𝟐

𝟑𝟑𝟑𝟑 .

- Multiply each term by the LCD 12∙ 2𝑥𝑥𝑥𝑥3

+ 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 ∙ 14

= 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐�- 𝑥𝑥𝑥𝑥2� − 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 ∙ 2

3

(least common denominator).

- Collect the variable terms on one side of the 8x + 3 = -6x – 8

equation and the constants on the other side. 14x = -11

- Isolate the variable. 𝟐𝟐𝟐𝟐 = -𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒

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Equations Involving Decimals/Fractions

• Equations involving decimals x

Steps Example: Solve 0.25x – 0.20 = -3.15x.

- Multiply each term by 100 to clear the decimal. 100(0.25x) – 100(0.20) = 100(-3.15x)

- Collect the variable terms on one side of the 25x – 20 = -315x

equation and the constants on the other side. 25x + 315x = 20

340x = 20

- Isolate the variable. x ≈ 0.06

Example: Solve 0.3y + 0.06 = 0.009. Multiply each term by 1,000.

1,000(0.3y) + 1,000(0.06) = 1,000(0.009)

300y + 60 = 9 Combine like terms.

300y = -51 Divide both sides by 300.

y = -0.17

Tip: Multiply every term of both sides of the equation by a power of 10 (10, 100, 1000, etc.) to clear the decimals.

• Equations involving fractions x

Steps Example: Solve 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

+ 𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒

= - 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐− 𝟐𝟐𝟐𝟐

𝟑𝟑𝟑𝟑 .

- Multiply each term by the LCD 12∙ 2𝑥𝑥𝑥𝑥3

+ 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 ∙ 14

= 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐�- 𝑥𝑥𝑥𝑥2� − 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 ∙ 2

3

(least common denominator).

- Collect the variable terms on one side of the 8x + 3 = -6x – 8

equation and the constants on the other side. 14x = -11

- Isolate the variable. 𝟐𝟐𝟐𝟐 = -𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒





2-2 LINEAR EQUATIONS AND MODELING

Geometry Formulas

• Formula: an equation that contains more than one variable and is used to solve practical problems in everyday life .

• Recall some geometry formulas

P – perimeter, C – circumference, A – area, V – volume

Name of the Figure Formula Figure

rectangleP = 2l + 2w

A = lww

l

parallelogram P = 2a + 2bA = bh

h a b

circle 𝐴𝐴 = 𝜋𝜋𝜋𝜋2rdC π=π= 2

r d

triangle∠ X + ∠ Y + ∠ Z = 1800

bhA21

=

XhY b Z

trapezoid )(21 BbhA +=

b h B

cube V = s3

rectangular solid V = lwh h w

cylinder hrV 2π=r

h

sphere3

34 rV π=

r

cone hrV 2

31π= h

r

pyramid lwhV31

=

l wh

s

l

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Solving Formulas

Example: Solve the formula for the given letter .Solution

1. d = r t , for t 𝑑𝑑𝒓𝒓

= 𝑟𝑟𝑟𝑟𝒓𝒓

t = 𝒅𝒅𝒓𝒓

2. I = P r t , for r 𝐼𝐼𝑷𝑷𝑷𝑷

= 𝑃𝑃𝑟𝑟𝑟𝑟𝑷𝑷𝑷𝑷

r = 𝑰𝑰𝑷𝑷𝑷𝑷

3. P = 2 l + 2 w , for w P – 2 l = 2 l + 2 w – 2 l

P – 2 l = 2𝑤𝑤, 𝑃𝑃−2𝑙𝑙𝟐𝟐

= 2𝑤𝑤𝟐𝟐

𝒑𝒑−𝟐𝟐𝟐𝟐𝟐𝟐

= 𝒘𝒘

4. 𝐹𝐹 = 95𝐶𝐶 + 32 , for C 𝐹𝐹 − 𝟑𝟑𝟐𝟐 = 9

5𝐶𝐶 + 32 − 𝟑𝟑𝟐𝟐

𝐹𝐹 − 𝟑𝟑𝟐𝟐 = 95𝐶𝐶

(𝐹𝐹 − 32) 𝟓𝟓𝟗𝟗

= 95∙ 𝟓𝟓𝟗𝟗𝐶𝐶

𝑪𝑪 = 𝟓𝟓𝟗𝟗

(𝑭𝑭 − 𝟑𝟑𝟐𝟐)

Tip: Solve a formula for a given letter by isolating the given letter on one side of the equation .

• More formulas

Application Formula Component

distance d = rt, r = 𝑑𝑑𝑟𝑟

, t = 𝑑𝑑𝑟𝑟

d – distancer – speedt – time

simple interest I = P r t, P = 𝐼𝐼𝑟𝑟𝑟𝑟

, t = 𝐼𝐼𝑃𝑃𝑟𝑟

I – interestP – principler – interest rate (%)t – time (years)

compoundinterest B = P (100% + r) t

B – balanceP – principler – interest rate (%)t – time (years)

percent increase 𝑁𝑁 − 𝑂𝑂𝑂𝑂

N – new valueO – original value

percent decrease 𝑂𝑂 − 𝑁𝑁𝑂𝑂


sale price S = L – r L , 𝐿𝐿 = 𝑆𝑆1−𝑟𝑟

S – sale priceL – list pricer – discount rate

intelligence quotient (I.Q.) 𝐼𝐼 =

100𝑚𝑚𝑐𝑐

I – I .Q .m – mental agec – chronological age

temperature 𝐶𝐶 = 59

(𝐹𝐹 − 32) , 𝐹𝐹 = 95𝐶𝐶 + 32 C – Celsius

F – Fahrenheit

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Solving Formulas

Example: Solve the formula for the given letter .Solution

1. d = r t , for t 𝑑𝑑𝒓𝒓

= 𝑟𝑟𝑟𝑟𝒓𝒓

t = 𝒅𝒅𝒓𝒓

2. I = P r t , for r 𝐼𝐼𝑷𝑷𝑷𝑷

= 𝑃𝑃𝑟𝑟𝑟𝑟𝑷𝑷𝑷𝑷

r = 𝑰𝑰𝑷𝑷𝑷𝑷

3. P = 2 l + 2 w , for w P – 2 l = 2 l + 2 w – 2 l

P – 2 l = 2𝑤𝑤, 𝑃𝑃−2𝑙𝑙𝟐𝟐

= 2𝑤𝑤𝟐𝟐

𝒑𝒑−𝟐𝟐𝟐𝟐𝟐𝟐

= 𝒘𝒘

4. 𝐹𝐹 = 95𝐶𝐶 + 32 , for C 𝐹𝐹 − 𝟑𝟑𝟐𝟐 = 9

5𝐶𝐶 + 32 − 𝟑𝟑𝟐𝟐

𝐹𝐹 − 𝟑𝟑𝟐𝟐 = 95𝐶𝐶

(𝐹𝐹 − 32) 𝟓𝟓𝟗𝟗

= 95∙ 𝟓𝟓𝟗𝟗𝐶𝐶

𝑪𝑪 = 𝟓𝟓𝟗𝟗

(𝑭𝑭 − 𝟑𝟑𝟐𝟐)

Tip: Solve a formula for a given letter by isolating the given letter on one side of the equation .

• More formulas

Application Formula Component

distance d = rt, r = 𝑑𝑑𝑟𝑟



simple interest I = P r t, P = 𝐼𝐼𝑟𝑟𝑟𝑟












100𝑚𝑚𝑐𝑐



(𝐹𝐹 − 32) , 𝐹𝐹 = 95𝐶𝐶 + 32 C – Celsius

F – Fahrenheit

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PROBLEM SOLVING

Steps for solving word problems

Steps for Solving Word Problems• Organize the facts given from the problem .• Identify and label the unknown quantity (let x = unknown) .• Draw a diagram if it will make the problem clearer . • Convert words into a mathematical equation . • Solve the equation and find the solution(s) .• Check and state the answer .

Number Problems

English Phrase Algebraic Expression/Equation

3 more than the difference of a number and 9 . (x – 9) + 3

The quotient of 3 and the product of 7 and a number . 37𝑥𝑥

The product of seven and a number, decreased by five . 7x – 59 less than 4 times two numbers is 2 more than their sum. 4xy – 9 = 2 + x + yThe sum of the squares of two numbers is 4 less than their product . x2 + y2 = xy – 46 more than the quotient of 2x by 3 is 5 times that number . 6 + 2𝑥𝑥

3= 5x

(Let x = a number ; y = a number)

Example: Three more than two times a number is fifteen less than the number divided by five .

Find the number .

- Organize the facts . + 3 2x = - 15 + 𝑥𝑥5

Let x = number .

- Equation: 2x + 3 = 𝒙𝒙𝟓𝟓

– 15 Multiply each term by 5 .

5(2x) + 5∙3 = 5 �𝑥𝑥5 � – 5∙15 Remove parentheses .

10x + 15 = x – 75 Combine like terms .

9x = -90 Divide both sides by 9 .

- Solution: x = -10

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PROBLEM SOLVING

Steps for solving word problems


Number Problems

English Phrase Algebraic Expression/Equation

3 more than the difference of a number and 9 . (x – 9) + 3

The quotient of 3 and the product of 7 and a number . 37𝑥𝑥

The product of seven and a number, decreased by five . 7x – 59 less than 4 times two numbers is 2 more than their sum. 4xy – 9 = 2 + x + yThe sum of the squares of two numbers is 4 less than their product . x2 + y2 = xy – 46 more than the quotient of 2x by 3 is 5 times that number . 6 + 2𝑥𝑥

3= 5x

(Let x = a number ; y = a number)

Example: Three more than two times a number is fifteen less than the number divided by five .

Find the number .

- Organize the facts . + 3 2x = - 15 + 𝑥𝑥5

Let x = number .

- Equation: 2x + 3 = 𝒙𝒙𝟓𝟓

– 15 Multiply each term by 5 .

5(2x) + 5∙3 = 5 �𝑥𝑥5 � – 5∙15 Remove parentheses .

10x + 15 = x – 75 Combine like terms .

9x = -90 Divide both sides by 9 .

- Solution: x = -10

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?

- Check: 2(-10) + 3 = -𝟏𝟏𝟏𝟏5

– 15 ?

-20 + 3 = - 2 – 15 √-17 = -17 Correct!

- State the answer: the number is -10 .

Example: There are three numbers; the first is two less than four times the second, and the

third is five more than one-half of the first . The sum of these three numbers is

sixteen .

Find the second number .

- Organize the facts:

- Equation: (4x – 2) + x + �𝟏𝟏𝟐𝟐

(𝟒𝟒𝟒𝟒 − 𝟐𝟐) + 𝟓𝟓)� = 16 Remove parentheses .

4x – 2 + x + 2𝑥𝑥 − 1 + 5 = 16 Combine like terms .

7x + 2 = 16

7x = 14 Divide both sides by 7 .

- Solution: x = 2

1st Number 4x – 2 = 4∙2 – 2 = 62nd Number x = 23rd Number 1

2(4𝑥𝑥 − 2) + 5 = 1

2(4 ∙ 2 − 2) + 5 = 8

Number Wording Algebraic Expression2nd number let 2nd number = x x1st number 2 less than 4 times the 2nd number 4x – 2

3rd number 5 more than 12 of the 1st number 12

(4𝑥𝑥 − 2) + 5

sum the sum of three numbers is 16 1st # + 2nd # + 3rd # = 16

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Consecutive Integers

English Phrase Algebraic Expression Exampletwo consecutive integers x, x + 1 If x = 1, x + 1 = 2two consecutive odd integers x, x + 2 If x = 1, x + 2 = 3

two consecutive even integers x, x + 2 or 2x, 2x + 2

If x = 2, x + 2 = 4If x = 1, 2x = 2, 2x + 2 = 4

three consecutive odd integers x , x + 2, x + 4 If x = 1, x + 2 = 3, x + 4 = 5three consecutive even integers x, x + 2 , x + 4 If x = 2, x + 2 = 4, x + 4 = 6The product of two consecutive odd integers is 10 x (x + 2) = 10 Three consecutive even integers whose sum is 35 . x + (x + 2) + (x + 4) = 35

Example: The sum of four consecutive even integers is 20; find each number .- Organize the facts .

1st consecutive even number x 2nd consecutive even number x + 2 3rd consecutive even number x + 4 4th consecutive even number x + 6

- Equation: x + (x + 2) + (x + 4) + (x + 6) = 20 Combine like terms .

- Solution: 4x + 12 = 20 Solve for x.

x = 2- State the answer .

1st consecutive even number x = 2 2nd consecutive even number x + 2 = 2 + 2 = 4 3rd consecutive even number x + 4 = 2 + 4 = 6 4th consecutive even number x + 6 = 2 + 6 = 8

?- Check: 2, 4, 6, 8 = consecutive even integers Yes!

?2 + (2 + 2) + (2 + 4) + (2 + 6) = 20

√20 = 20 Correct!

Example: Find three consecutive odd integers such that three times the first integer is one less than the sum of the second and third integers.

- Organize the facts .Integer Consecutive Odd Integer

1st integer x2nd integer x + 23rd integer x + 4

- Equation: 3x = (x + 2) + (x + 4) −1- Solution: 3x = x + 2 + x + 4 - 1

x = 5- State the answer:

1st consecutive odd number x = 5 2nd consecutive odd number x + 2 = 5 + 2 = 7 3rd consecutive odd number x + 4 = 5 + 4 = 9

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Consecutive Integers

English Phrase Algebraic Expression Exampletwo consecutive integers x, x + 1 If x = 1, x + 1 = 2two consecutive odd integers x, x + 2 If x = 1, x + 2 = 3

two consecutive even integers x, x + 2 or 2x, 2x + 2

If x = 2, x + 2 = 4If x = 1, 2x = 2, 2x + 2 = 4

three consecutive odd integers x , x + 2, x + 4 If x = 1, x + 2 = 3, x + 4 = 5three consecutive even integers x, x + 2 , x + 4 If x = 2, x + 2 = 4, x + 4 = 6The product of two consecutive odd integers is 10 x (x + 2) = 10 Three consecutive even integers whose sum is 35 . x + (x + 2) + (x + 4) = 35

Example: The sum of four consecutive even integers is 20; find each number .- Organize the facts .

1st consecutive even number x 2nd consecutive even number x + 2 3rd consecutive even number x + 4 4th consecutive even number x + 6

- Equation: x + (x + 2) + (x + 4) + (x + 6) = 20 Combine like terms .

- Solution: 4x + 12 = 20 Solve for x.

x = 2- State the answer .

1st consecutive even number x = 2 2nd consecutive even number x + 2 = 2 + 2 = 4 3rd consecutive even number x + 4 = 2 + 4 = 6 4th consecutive even number x + 6 = 2 + 6 = 8

?- Check: 2, 4, 6, 8 = consecutive even integers Yes!

?2 + (2 + 2) + (2 + 4) + (2 + 6) = 20

√20 = 20 Correct!

Example: Find three consecutive odd integers such that three times the first integer is one less than the sum of the second and third integers.

- Organize the facts .Integer Consecutive Odd Integer

1st integer x2nd integer x + 23rd integer x + 4

- Equation: 3x = (x + 2) + (x + 4) −1- Solution: 3x = x + 2 + x + 4 - 1

x = 5- State the answer:

1st consecutive odd number x = 5 2nd consecutive odd number x + 2 = 5 + 2 = 7 3rd consecutive odd number x + 4 = 5 + 4 = 9

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Business Problems

Application Formula

Percent Increase valueOriginal valueOriginal valueNewincreasePercent −

= , O

ON−=x

Percent Decrease valueOriginal valueNew valueOriginaldecreasePercent −

= , O

NO−=x

Sales Tax sales tax = sales × tax rateCommission commission = sales × commission rate

Discount discount = original price × discount ratesale price = original price – discount

Markup markup = original price × markup rateoriginal price = selling price – markup

Simple Interest interest = principle × interest rate × time, I = P r tbalance = principle + interest

Compound Interest

balance = principle (100% + interest rate)t balance = P(100% + r)t

Example: A product increased production from 1,500 last month to 1,650 this month . Find the percent increase.

New value (N): 1,650 this month

Original value (O): 1,500 last month

Percent increase:

==−

=−

= 1 .0500,1

500,1650,1O

ONx 10% 10% increase

Example: A product was reduced from $33 to $29. What percent reduction is this?

Percent decrease:

=≈−

=−

= 12 .033

2933O

NOx 12% 12% decrease

Example: The tax rate is 7%, find the sales tax for a $1,050 laptop .

Sales tax = Sales × Tax rate

= ($1,050)(7%) = ($1,050)(0 .07) = $73.50

Example: The commission rate is 5%, find commission for a $550,000 house .

Commission = Sales × Commission rate

= ($550,000)(5%) = ($550,000)(0 .05) = $27,500

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Business Problems

Application Formula


= , O

ON−=x


= , O

NO−=x





Compound Interest


Example: A product increased production from 1,500 last month to 1,650 this month . Find the percent increase.

New value (N): 1,650 this month

Original value (O): 1,500 last month

Percent increase:

==−

=−

= 1 .0500,1

500,1650,1O

ONx 10% 10% increase

Example: A product was reduced from $33 to $29. What percent reduction is this?

Percent decrease:

=≈−

=−

= 12 .033

2933O

NOx 12% 12% decrease

Example: The tax rate is 7%, find the sales tax for a $1,050 laptop .

Sales tax = Sales × Tax rate

= ($1,050)(7%) = ($1,050)(0 .07) = $73.50

Example: The commission rate is 5%, find commission for a $550,000 house .

Commission = Sales × Commission rate

= ($550,000)(5%) = ($550,000)(0 .05) = $27,500

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Example: A women’s jacket originally priced at $47 is on sale at a 35% discount . Find the discount and sale price .

Discount = Original price × Discount rate

= ($47)(35%) = ($47)(0 .35) = $16.45

Sale price = Original price – Discount

= $47 – $16 .45 = $30.55

Example: A condo originally sold at $258,000, and the markup rate is 10%. What is the markup and the new selling price?

Markup = Original price × Markup rate

= ($258,000)(10%) = ($258,000)(0 .10) = $25,800

Selling price = Original price + Markup

= $258,000 + $25,800 = $283,800

Example: Tom borrowed $200,000 mortgage from a bank . Find the interest at 2.8% per year

for215 years, and the total amount that he paid the bank .

Interest = Principle × Interest rate × Time

I = 𝑃𝑃 𝑟𝑟 𝑡𝑡 = ($200,000)(2.8%)�215 �

= ($200,000)(0 .028)(5 .5) = $30,800

Balance = Principle + Interest

= $ 200,000 + $ 30,800 = $230,800

Example: Allan deposited $5,000 in an account at 3.5% interest compounded annually for 5 years . How much was in the account at the end of 5 years?

Balance = Principle (100% + Interest rate) t

= P(100% + r) t = $5,000 (100% + 3 .5%)5

= $5,000 (1+ 0 .035) 5 ≈ $ 5,938

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Application Formula

Sales

Discount: Original price – Discount rate × Original price = New price

x – % x = New pricePurchase: Original price + Tax rate × Original price = New price

x + % x = New price

Commission

Commission on the 1st $100,000 + Commission on the amount > $100,000 = Total commission

% ∙ 100,000 + % (S – 100,000) = Total commission

selling price

Example: Bob pays $1,050 for a laptop . The price includes a 12% sales tax . What is the price of the laptop itself?

original price (the price of the laptop itself) xtax rate 12%new price (price + tax) $1,050

- Equation: x + 12% x = $1,050 x + % x = New price

- Solution: x (1+ 0 .12) = $1,050

x = 1,0501.12

= $937.50 The price of the laptop itself is $937 .50 .

Example: The following is a real estate commission on the selling price of a house:

7% for the first $100,000

5% for the amount > $100,000

A realtor receives a commission of $20,000, what was the selling price?

- Formula: % ∙ 100,000 + % (S – 100,000) = Total commission

- Solution: 7% ∙ 100,000 + 5% (S – 100,000) = $20,000

0 .07(100,000) + 0 .05 S – 0 .05(100,000) = 20,000

7000 + 0 .05 S – 5,000 = 20,000

0 .05 S = 18,000

S = $ 360,000 Selling price

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Motion Problems

• FormulaDistance = Speed ∙ Time d = r t t = 𝑑𝑑

𝑟𝑟r = 𝑑𝑑

𝑡𝑡

Condition Speed (r) Time (t) Distance (d)A r t d = rtB r t d = rt

Total

Example: Two cyclists are 18 km apart and are travelling towards each other . Theirspeeds differ by 2 km per hour . What is the speed of each cyclist if they meet after 3 hours?

Condition Speed (r) Time (t) Distance (d = rt)bike A r (km/h) t = 3 h 3rbike B r – 2 (km/h) t = 3 h 3(r – 2)Total 18 km

Equation: 3r + 3 (r – 2) = 18 Distance of A + distance of B = 18km .

bike A: r = 4 km/h

bike B: r – 2 = 4 – 2 = 2 km/h

Example: John boats at a speed of 30 km per hour in still water . The river flows at a speed of 10 km per hour . How long will it take John to boat 2 km downstream?2 km upstream?

Condition Speed (r) Distance (d) Time ( t = 𝒅𝒅𝒓𝒓

)

downstream r = 30 + 10 = 40 km/h d = 2 km t = 𝑑𝑑𝑟𝑟

= 2𝑘𝑘𝑘𝑘40𝑘𝑘𝑘𝑘/ℎ

upstream r = 30 – 10 = 20 km/h d = 2 km t = 𝑑𝑑𝑟𝑟


downstream (fast): speed of boat + speed of riverupstream (slower): speed of boat - speed of river

downstream: t = 𝑑𝑑𝑟𝑟


= 0.05 h

upstream: t = 𝑑𝑑𝑟𝑟


= 0.1 h

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Motion Problems

• FormulaDistance = Speed ∙ Time d = r t t = 𝑑𝑑


𝑡𝑡

Condition Speed (r) Time (t) Distance (d)A r t d = rtB r t d = rt

Total

Example: Two cyclists are 18 km apart and are travelling towards each other . Theirspeeds differ by 2 km per hour . What is the speed of each cyclist if they meet after 3 hours?

Condition Speed (r) Time (t) Distance (d = rt)bike A r (km/h) t = 3 h 3rbike B r – 2 (km/h) t = 3 h 3(r – 2)Total 18 km

Equation: 3r + 3 (r – 2) = 18 Distance of A + distance of B = 18km .

bike A: r = 4 km/h

bike B: r – 2 = 4 – 2 = 2 km/h

Example: John boats at a speed of 30 km per hour in still water . The river flows at a speed of 10 km per hour . How long will it take John to boat 2 km downstream?2 km upstream?

Condition Speed (r) Distance (d) Time ( t = 𝒅𝒅𝒓𝒓

)

downstream r = 30 + 10 = 40 km/h d = 2 km t = 𝑑𝑑𝑟𝑟


upstream r = 30 – 10 = 20 km/h d = 2 km t = 𝑑𝑑𝑟𝑟


downstream (fast): speed of boat + speed of riverupstream (slower): speed of boat - speed of river

downstream: t = 𝑑𝑑𝑟𝑟


= 0.05 h

upstream: t = 𝑑𝑑𝑟𝑟


= 0.1 h

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Value Mixture Problems

Item Value of the Item Number of Items Total ValueA value of A # of item A (value of A)(# of item A)B value of B # of item B (value of B)(# of item B)C value of C # of item C (value of C)(# of item C)

total or mixture total value Let x = unknown .

Value of item A + Value of item B + Value of item C = Total value of the mixture

Example: Jack has $5.35 in nickels, dimes and quarters . If he has five less than two times quarters of dimes, and seven more nickels than quarters . How many of each coin does he have?

Coin Value of the Coin Number of Coins Total Value (in cents)quarter 25 ¢ x 25 xdime 10 ¢ 2x – 5 10(2x – 5)nickel 5 ¢ x + 7 5(x + 7)Total $5 .35 = 535 ¢

- Equation: 25x + 10(2x – 5) + 5(x + 7) = 535 value of quarters + value of dimes + value of nickels = 535 ¢

- Solution: 25x + 20x – 50 + 5x + 35 = 535 Remove parentheses .

50x – 15 = 535 Combine like terms .

x = 11 Solve for x.

- State the answer: number of quarters x = 11 number of dimes 2x – 5 = 2(11) – 5 = 17 number of nickels x + 7 = 11 + 7 = 18

Example: Evan purchased 46-cent, 66-cent, and 86-cent Canadian stamps with a total value of $16 .38 . If the number of 66-cent stamps is 5 more than the number of 46-cent stamps, and the number of 86-cent stamps is 8 more than one half the number of 46-cent stamps . How many of each did Evan receive?

Stamps Value of the Stamps Number of Stamps Total Value (in cents)46-cent 46 ¢ x 46 x66-cent 66 ¢ 5 + x 66(5 + x)

86-cent 86 ¢ 8 + 12

x 86(8 + 12

x)Total 1638 ¢

- Equation: 46x + 66 (5 + x) + 86 (𝟖𝟖 + 𝟏𝟏𝟐𝟐

x) = 1638- Solution: 46x + 330 + 66x + 688 + 43x = 1638 Remove parentheses & combine like terms .

155x = 620x = 4 46-cent: 4

5 + x = 5 + 4 = 9 66-cent: 9

8 + 𝟏𝟏𝟐𝟐

x = 8 + 12 ∙ 4 = 𝟏𝟏𝟏𝟏 86-cent: 10

Let x = numbers of quarters .

Let x = number of 46-cent stamps .

value of 46-cent + value of 66-cent + value of 86-cent = 1638 ¢

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Value Mixture Problems

Item Value of the Item Number of Items Total ValueA value of A # of item A (value of A)(# of item A)B value of B # of item B (value of B)(# of item B)C value of C # of item C (value of C)(# of item C)

total or mixture total value Let x = unknown .

Value of item A + Value of item B + Value of item C = Total value of the mixture

Example: Jack has $5.35 in nickels, dimes and quarters . If he has five less than two times quarters of dimes, and seven more nickels than quarters . How many of each coin does he have?

Coin Value of the Coin Number of Coins Total Value (in cents)quarter 25 ¢ x 25 xdime 10 ¢ 2x – 5 10(2x – 5)nickel 5 ¢ x + 7 5(x + 7)Total $5 .35 = 535 ¢

- Equation: 25x + 10(2x – 5) + 5(x + 7) = 535 value of quarters + value of dimes + value of nickels = 535 ¢

- Solution: 25x + 20x – 50 + 5x + 35 = 535 Remove parentheses .

50x – 15 = 535 Combine like terms .

x = 11 Solve for x.

- State the answer: number of quarters x = 11 number of dimes 2x – 5 = 2(11) – 5 = 17 number of nickels x + 7 = 11 + 7 = 18

Example: Evan purchased 46-cent, 66-cent, and 86-cent Canadian stamps with a total value of $16 .38 . If the number of 66-cent stamps is 5 more than the number of 46-cent stamps, and the number of 86-cent stamps is 8 more than one half the number of 46-cent stamps . How many of each did Evan receive?

Stamps Value of the Stamps Number of Stamps Total Value (in cents)46-cent 46 ¢ x 46 x66-cent 66 ¢ 5 + x 66(5 + x)

86-cent 86 ¢ 8 + 12

x 86(8 + 12

x)Total 1638 ¢

- Equation: 46x + 66 (5 + x) + 86 (𝟖𝟖 + 𝟏𝟏𝟐𝟐

x) = 1638- Solution: 46x + 330 + 66x + 688 + 43x = 1638 Remove parentheses & combine like terms .

155x = 620x = 4 46-cent: 4

5 + x = 5 + 4 = 9 66-cent: 9

8 + 𝟏𝟏𝟐𝟐

x = 8 + 12 ∙ 4 = 𝟏𝟏𝟏𝟏 86-cent: 10

Let x = numbers of quarters .

Let x = number of 46-cent stamps .

value of 46-cent + value of 66-cent + value of 86-cent = 1638 ¢

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Concentration/Mixture Problems

Item Concentration Volume AmountA concentration of A volume of A (concentration of A)(volume of A)B concentration of B volume of B (concentration of B)(volume of B)

Mixture concentration of mixture volume of mixture (concentration of mixture)(volume of the mixture)

Let x = unknown Amount of item A + Amount of item B = Amount of the mixture

Example: A chicken meal is 30% protein and a beef meal is 40% protein . Steve wants an 800

grams mixture that is 35% protein . How many grams of each meal should he have?

Meal Concentration Protein Volume Amountchicken meal 30% = 0 .3 x 0 .3 xbeef meal 40% = 0 .4 800 – x 0 .4 (800 – x)Mixture 35% = 0 .35 800 0 .35 (800)

Let x = Protein volume of the chicken meal

Equation: 0.3 x + 0.4 (800 – x) = (0.35)(800)Amount of chicken meal + Amount of beef meal = amount of the mixture

0 .3 x + 320 – 0 .4 x = 280

- 0 .1 x = -40

x = 400 g chicken meal

800 – x = 800 – 400 = 400 g beef meal

Example: How much 5% salt solution must be added to 20 liters of 25% solution to make a

10% solution?

Solution Concentration Volume Amount5% 0 .05 x 0 .05x

25% 0 .25 20 (0 .25)(20)10% 0 .1 20 + x 0 .1(20+ x)

Let x = Volume of 5% solution .

Equation: 0.05x + (0.25 )(20) = 0.1 (20 + x) Amount of 5% + Amount of 25% = Amount of 10%

0 .05x + 5 = 2 + 0 .1 x

0 .05 x = 3

x = 60 liters Volume of 5% solution .

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2-3 SETS AND INEQUALITIES

Inequalities

• An inequality is a mathematical statement that contains < , > , ≥ , or ≤ symbol .

• Inequality symbols

Symbols Meaning Example> is greater than 15x + 7 > 0

< is less than 4x – 3y < 1325

≥ is greater than or equal to x + 5y ≥ -12≤ Is less than or equal to 3x – 14 y ≤ 67

• Compound inequality is a statement that contains more than one inequality . a < x < b

• The solution of an inequality is the particular value of the variable in the inequality that

makes the inequality true .

Example: Indicate if x = 3, -5 and 12

are solutions of the inequality 6 – 2x < 5x .

?1. For x = 3 6 – 2 ∙ 3 < 5 ∙ 3 Substitute x for 3 .

√0 < 15 True x = 3 is a solution .

? 2. For x = -5 6 – 2 (-5) < 5 (-5) Substitute x for -5 .

? 6 + 10 < -25

× 16 < -25 False x = -5 is not a solution .

?

3. For x = 12

6 – 2 ∙ 12

< 5 ∙ 12

Substitute x for 12 .

?

6 - 1 < 52

×5 < 5

2False x = 1

2is not a solution .

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2-3 SETS AND INEQUALITIES

Inequalities

• An inequality is a mathematical statement that contains < , > , ≥ , or ≤ symbol .

• Inequality symbols

Symbols Meaning Example> is greater than 15x + 7 > 0

< is less than 4x – 3y < 1325

≥ is greater than or equal to x + 5y ≥ -12≤ Is less than or equal to 3x – 14 y ≤ 67

• Compound inequality is a statement that contains more than one inequality . a < x < b

• The solution of an inequality is the particular value of the variable in the inequality that

makes the inequality true .

Example: Indicate if x = 3, -5 and 12

are solutions of the inequality 6 – 2x < 5x .

?1. For x = 3 6 – 2 ∙ 3 < 5 ∙ 3 Substitute x for 3 .

√0 < 15 True x = 3 is a solution .

? 2. For x = -5 6 – 2 (-5) < 5 (-5) Substitute x for -5 .

? 6 + 10 < -25

× 16 < -25 False x = -5 is not a solution .

?

3. For x = 12

6 – 2 ∙ 12

< 5 ∙ 12

Substitute x for 12 .

?

6 - 1 < 52

×5 < 5

2False x = 1

2is not a solution .

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Intervals

• Interval: a set of numbers between or possibly including two given numbers .

• Interval notation:

Open interval ( ): the end points are not included .

Closed interval [ ]: the end points are included .

Half-open interval (a, b]: a is not included, but b is included .

[a, b): a is included, but b is not included .

Non-ending open interval (a, ∞): a is not included and infinity is always excluded .

(-∞, a): a is not included and infinity is always excluded .

Non-ending half-open interval (-∞, b]: b is included and infinity is always excluded .

[b, ∞): b is included and infinity is always excluded .

• Double inequality (a < x < b) indicates “betweenness”, meaning both a < x, x < b , and

a must be less than b .

• Strict inequalities: an inequality that uses the symbols < or > .

• Weak inequalities: an inequality that uses the symbol ≤ or ≥.

• Graphing real-number inequalities

The empty circle ○ or open interval ( ): the endpoints are excluded .

The filled in circle ● or closed interval [ ]: the endpoints are included .

Use a heavy line and an open or closed interval or an empty circle or filled-in circle to

graph intervals .

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Intervals

• Interval: a set of numbers between or possibly including two given numbers .

• Interval notation:

Open interval ( ): the end points are not included .

Closed interval [ ]: the end points are included .

Half-open interval (a, b]: a is not included, but b is included .

[a, b): a is included, but b is not included .

Non-ending open interval (a, ∞): a is not included and infinity is always excluded .

(-∞, a): a is not included and infinity is always excluded .

Non-ending half-open interval (-∞, b]: b is included and infinity is always excluded .

[b, ∞): b is included and infinity is always excluded .

• Double inequality (a < x < b) indicates “betweenness”, meaning both a < x, x < b , and

a must be less than b .

• Strict inequalities: an inequality that uses the symbols < or > .

• Weak inequalities: an inequality that uses the symbol ≤ or ≥.

• Graphing real-number inequalities

The empty circle ○ or open interval ( ): the endpoints are excluded .


Use a heavy line and an open or closed interval or an empty circle or filled-in circle to

graph intervals .

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• Interval summary

Inequality IntervalNotation

Set-Builder Notation

ExampleEmpty/filled in circle

GraphOpen/closed interval

a < x < b (a, b) { x | a < x < b } 2 < x < 5

a ≤ x ≤ b [a, b] { x | a ≤ x ≤ b } 2 ≤ x ≤ 5

a ≤ x < b [a, b) { x | a ≤ x < b } 2 ≤ x < 5

a < x ≤ b (a, b] { x | a < x ≤ b } 2 < x ≤ 5

a < x < ∞ (a, ∞) { x | x > a } x > 2

a ≤ x < ∞ [a, ∞) { x | x ≥ 𝑎𝑎 } x ≥ 2

- ∞ < x < b (-∞, b) { x | x < b } x < 5

- ∞ < x ≤ b (-∞, b] { x | x ≤ b } x ≤ 5

- ∞ < x < ∞ (-∞, ∞) { x | -∞ < x < ∞ } - ∞ < x < ∞

Example: Express the following in interval notation . Solution

1. { x | x < -3 } (-∞, -3)

2. { x | -5 ≤ x < 5 } [-5, 5)

3. ( (-2, 9] -2 9

4. ( (-1, ∞) -1

Example: Graph the following inequalities on a number line .

1. { x | x < 2 } or ) 2 2

2. { t | - 4 ≤ t < 3} or [ ) -4 3 -4 3

3. { z | - 2 < z ≤ 7 } or ( ] -2 7 -2 7

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2 5

2 5

2 5

2 5

2

2

5

5

]





`1 Page 2-20

Properties of Inequalities

• Addition property of inequality: add the same value on each side of an inequality and the inequality remains true.

If a > b, Example: x – 4 > 3 then a + c > b + c x – 4 + 4 > 3 + 4 x > 7

• Subtraction property of inequality: subtract the same value from each side of an inequality and the inequality remains true.

If a < b, Example: y + 2 < 3 then a – c < b – c y + 2 – 2 < 3 – 2 y < 1

• Multiplication property of inequality: multiply the same positive value on each side of an inequality and the inequality remains true.

If a ≥ b, Example: 23

𝑥𝑥𝑥𝑥 ≥ 5

then ac ≥ bc 23

𝑥𝑥𝑥𝑥 ∙ 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐 ≥ 5 ∙ 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐

x ≥ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

Note: When multiplying each side of the inequality by a negative number, reverse the inequality sign.

If a > b, Example: - 23𝑥𝑥𝑥𝑥 > 2

then a(-c) < b(-c) - 23𝑥𝑥𝑥𝑥 �- 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐 � < 2 �- 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐�

x < - 3

• Division property of equality: divide the same positive value on each side of an inequality and the inequality remains true.

If a ≤ b, Example: 4y ≤ 13

then 𝑎𝑎𝑎𝑎𝒄𝒄𝒄𝒄 ≤ 𝑏𝑏𝑏𝑏

𝒄𝒄𝒄𝒄 (c ≠ 0) 4𝑦𝑦𝑦𝑦

𝟒𝟒𝟒𝟒≤ 1

3 ∙ 𝟒𝟒𝟒𝟒 , y ≤ 𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

Note: When dividing each side of the inequality by a negative number, reverse the inequality sign.

If a < b, Example: -4y ≤ 13

then 𝑎𝑎𝑎𝑎-𝒄𝒄𝒄𝒄

> 𝑏𝑏𝑏𝑏-𝒄𝒄𝒄𝒄

- 4𝑦𝑦𝑦𝑦-𝟒𝟒𝟒𝟒≥ 1

3(-𝟒𝟒𝟒𝟒) , y ≥ - 𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

`1 Page 2-20

Properties of Inequalities

• Addition property of inequality: add the same value on each side of an inequality and the inequality remains true.

If a > b, Example: x – 4 > 3 then a + c > b + c x – 4 + 4 > 3 + 4 x > 7

• Subtraction property of inequality: subtract the same value from each side of an inequality and the inequality remains true.

If a < b, Example: y + 2 < 3 then a – c < b – c y + 2 – 2 < 3 – 2 y < 1

• Multiplication property of inequality: multiply the same positive value on each side of an inequality and the inequality remains true.

If a ≥ b, Example: 23

𝑥𝑥𝑥𝑥 ≥ 5

then ac ≥ bc 23

𝑥𝑥𝑥𝑥 ∙ 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐 ≥ 5 ∙ 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐

x ≥ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

Note: When multiplying each side of the inequality by a negative number, reverse the inequality sign.

If a > b, Example: - 23𝑥𝑥𝑥𝑥 > 2

then a(-c) < b(-c) - 23𝑥𝑥𝑥𝑥 �- 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐 � < 2 �- 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐�

x < - 3

• Division property of equality: divide the same positive value on each side of an inequality and the inequality remains true.

If a ≤ b, Example: 4y ≤ 13

then 𝑎𝑎𝑎𝑎𝒄𝒄𝒄𝒄 ≤ 𝑏𝑏𝑏𝑏

𝒄𝒄𝒄𝒄 (c ≠ 0) 4𝑦𝑦𝑦𝑦

𝟒𝟒𝟒𝟒≤ 1

3 ∙ 𝟒𝟒𝟒𝟒 , y ≤ 𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

Note: When dividing each side of the inequality by a negative number, reverse the inequality sign.

If a < b, Example: -4y ≤ 13

then 𝑎𝑎𝑎𝑎-𝒄𝒄𝒄𝒄

> 𝑏𝑏𝑏𝑏-𝒄𝒄𝒄𝒄

- 4𝑦𝑦𝑦𝑦-𝟒𝟒𝟒𝟒≥ 1

3(-𝟒𝟒𝟒𝟒) , y ≥ - 𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐





Solving Inequalities

• Solving an inequality is the process of finding a particular value for the variable in the

inequality that makes the inequality true .

• The procedure for solving linear inequalities is similar to solving basic equations .

Example: Solve the inequality 6x + 1 ≥ -17 and graph the solution set .

6x + 1 – 1 ≥ -17 – 1 Subtract 1 from both sides .

6x ≥ -18 Divide both sides by 6 .

x ≥ -3 [-3,∞) or { x | x ≥ -3 } [

Check: Method I Method II

Treat the inequality as an equation & check . Choose any number greater than -3 (say 0) .

6x + 1 = -17, x = -3 6x + 1 ≥ -17, x = 0 -? ?

6(-3) + 1 = -17 Replace x with -3 . 6(0) + 1 ≥ -17 Replace x with 0 .

? √ -18 + 1 = -17 1 ≥ -17 Correct!

√ -17 = -17 Correct!

Example: Solve the following inequality and graph the solution set .

4 – 2(x – 5) – 3x ≥ 2x + 11 Remove parenthesis .

4 – 2x + 10 – 3x ≥ 2x + 11 Combine like terms .

14 – 5x ≥ 2x + 11 Isolate x.

-7x ≥ -3 Divide both sides by -7 .

-7𝑥𝑥-7

≤ -3-7

, 𝑥𝑥 ≤ 37

Reverse the symbol .

{ x | 𝒙𝒙 ≤ 𝟑𝟑𝟕𝟕

} (-∞ , 𝟑𝟑𝟕𝟕 ] ]

Example: Solve the inequality 𝟏𝟏𝟐𝟐

(3 – x) – 𝟏𝟏𝟑𝟑 > 𝟏𝟏

𝟒𝟒 and graph the solution set .

122

(3 – x) – 13

∙ 12 > 14∙ 12 Multiply each term by the LCD .

6(3 – x) – 4 > 3, 18 – 6x – 4 > 3

14 – 6x > 3, - 6x > -11

𝒙𝒙 < 𝟏𝟏𝟏𝟏𝟔𝟔

or { x | 𝒙𝒙 < 𝟏𝟏𝟏𝟏𝟔𝟔

} )116

37

-3

`1 Page 2-21

Solving Inequalities

• Solving an inequality is the process of finding a particular value for the variable in the

inequality that makes the inequality true .

• The procedure for solving linear inequalities is similar to solving basic equations .

Example: Solve the inequality 6x + 1 ≥ -17 and graph the solution set .

6x + 1 – 1 ≥ -17 – 1 Subtract 1 from both sides .

6x ≥ -18 Divide both sides by 6 .

x ≥ -3 [-3,∞) or { x | x ≥ -3 } [

Check: Method I Method II

Treat the inequality as an equation & check . Choose any number greater than -3 (say 0) .

6x + 1 = -17, x = -3 6x + 1 ≥ -17, x = 0 -? ?

6(-3) + 1 = -17 Replace x with -3 . 6(0) + 1 ≥ -17 Replace x with 0 .

? √ -18 + 1 = -17 1 ≥ -17 Correct!

√ -17 = -17 Correct!

Example: Solve the following inequality and graph the solution set .

4 – 2(x – 5) – 3x ≥ 2x + 11 Remove parenthesis .

4 – 2x + 10 – 3x ≥ 2x + 11 Combine like terms .

14 – 5x ≥ 2x + 11 Isolate x.

-7x ≥ -3 Divide both sides by -7 .

-7𝑥𝑥-7

≤ -3-7

, 𝑥𝑥 ≤ 37

Reverse the symbol .

{ x | 𝒙𝒙 ≤ 𝟑𝟑𝟕𝟕

} (-∞ , 𝟑𝟑𝟕𝟕 ] ]

Example: Solve the inequality 𝟏𝟏𝟐𝟐

(3 – x) – 𝟏𝟏𝟑𝟑 > 𝟏𝟏

𝟒𝟒 and graph the solution set .

122

(3 – x) – 13

∙ 12 > 14∙ 12 Multiply each term by the LCD .

6(3 – x) – 4 > 3, 18 – 6x – 4 > 3

14 – 6x > 3, - 6x > -11

𝒙𝒙 < 𝟏𝟏𝟏𝟏𝟔𝟔

or { x | 𝒙𝒙 < 𝟏𝟏𝟏𝟏𝟔𝟔

} )116

37

-3

`1 Page 2-21

?

?

√





Writing and Solving Inequalities

Example: If seven more than twice a number is greater than five times the number plus three, how large is the number?

- Let x = the number .


7 more than twice a number is greater than 5 times the number plus 3+ 2x > 5x + 3

- Inequality: 7 + 2x > 5x + 3 Subtract 5x from both sides .

7 – 3x > 3 Subtract 7 from both sides .

- 3x > -4 Divide both sides by -3 .

x < 43

, { x | 𝒙𝒙 < 𝟒𝟒𝟑𝟑

} Reverse the symbol .

- Check: Choose any number less than 4 3

(say 0) ?

7 + 2 ∙ 0 > 5 ∙ 0 + 3 √

7 > 3 Correct!

Example: Neal got an 81% on the midterm exam in Math . To get an A, the average of his midterm and final exam must be between 85% and 90% . For what range of scores on the final exam will Neal need to get an A?

- Facts and unknown:

Facts 85 % < the average of midterm and final exam < 90%Unknown Let x = the final exam score .

- Inequality: 85 < 𝒙𝒙+𝟖𝟖𝟖𝟖𝟐𝟐

< 90 The average of midterm and final exam: x+812

.

2(85) < 2 �𝑥𝑥+812� < 2(90) Multiply 2 by each part .

170 < x + 81 < 180 Subtract 81 from each part .

170 – 81 < x + 81 – 81 < 180 – 81

- Solution: 89 < x < 99 Neal’s final exam score must be between 89% and 99% .

{ x | 𝟖𝟖𝟖𝟖 < 𝒙𝒙 < 𝟖𝟖𝟖𝟖 }

`1 Page 2-22

Writing and Solving Inequalities

Example: If seven more than twice a number is greater than five times the number plus three, how large is the number?

- Let x = the number .


7 more than twice a number is greater than 5 times the number plus 3+ 2x > 5x + 3

- Inequality: 7 + 2x > 5x + 3 Subtract 5x from both sides .

7 – 3x > 3 Subtract 7 from both sides .

- 3x > -4 Divide both sides by -3 .

x < 43

, { x | 𝒙𝒙 < 𝟒𝟒𝟑𝟑

} Reverse the symbol .

- Check: Choose any number less than 4 3

(say 0) ?

7 + 2 ∙ 0 > 5 ∙ 0 + 3 √

7 > 3 Correct!

Example: Neal got an 81% on the midterm exam in Math . To get an A, the average of his midterm and final exam must be between 85% and 90% . For what range of scores on the final exam will Neal need to get an A?

- Facts and unknown:

Facts 85 % < the average of midterm and final exam < 90%Unknown Let x = the final exam score .

- Inequality: 85 < 𝒙𝒙+𝟖𝟖𝟖𝟖𝟐𝟐

< 90 The average of midterm and final exam: x+812

.

2(85) < 2 �𝑥𝑥+812� < 2(90) Multiply 2 by each part .

170 < x + 81 < 180 Subtract 81 from each part .

170 – 81 < x + 81 – 81 < 180 – 81

- Solution: 89 < x < 99 Neal’s final exam score must be between 89% and 99% .

{ x | 𝟖𝟖𝟖𝟖 < 𝒙𝒙 < 𝟖𝟖𝟖𝟖 }

`1 Page 2-22





2-4 INTERSECTIONS AND UNIONS

Intersections, Unions, and Subsets

• Intersection of A and B (A ∩ B): the set of all elements contained in both A and B .

Example: 1. If A = {red, green, yellow, black} and B = {white, black, green},

then A ∩ 𝑩𝑩 = {black, green} .

2. If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8},

then A ∩ 𝐁𝐁 = {4, 5} . A intersect B

• Union of A and B (A ∪ B): the set of all elements contained in A or B, or both .

Example: 1. If A = {red, green, yellow} and B = {white, black},

then A ∪ 𝐁𝐁 = { red, green, yellow, white, black} .

2. If A = {1, 3, 5} and B = {2, 4, 6},

then A ∪ 𝐁𝐁 = {1, 2, 3, 4, 5, 6}. A union B

• Empty set (or null set) ∅: a set that contains no elements (disjoint) .

Example: If A = { x | x = Feb . 30}, then A = ∅

B = { x | x = Christmas day on Nov . 25} = ∅

• Subset (B ⊂ A): a set B is a subset of a set A if each element of B is an element of A . A subset B is a portion of another set A .

Example: If A = {4, 5, 8, 10, 17, 23}, B = {5, 10, 17},

then (B ⊂ A) .

• x ∈ A: x is an element of the set A (or x belongs to A) .

• x ∉ A: x is not an element of the set A (or x does not belong to A) .

Example: A = {1, 2, 3, 4, 5}3 ∈ A: 3 is an element of A . 6 ∉ A: 6 is not an element of A .

• Sets Summary Unions, Intersections, and Subsets Exampleunion of A and B

(A ∪ B) OR

The set of all elements contained in A orB, or both .

If A = {2, 5} and B = {1, 3, 4}then A ∪ 𝐵𝐵 = {1, 2, 3, 4, 5} .

intersection of A and B(A ∩ B) AND

The set of all elements contained in both A and B .

If A = {3, 6, 9} and B = {5, 6, 7, 8, 9}then A ∩ 𝐵𝐵 = {6, 9} .

empty set (or null set) ∅ A set that contains no elements . If A = { x | x = Feb . 30} , then A = ∅ .subset (B ⊂ A) The subset B is a portion of another set A. If A = {2, 5, 7, 11}, B = {5, 11}, then (B ⊂ A) .

x ∈ A x is an element of the set A . 23

∈ Rational numbersx ∉ A x is not an element of the set A. √5 ∉ Rational numbers

B A

A B

A B

A B

`1 Page 2-23

2-4 INTERSECTIONS AND UNIONS

Intersections, Unions, and Subsets

• Intersection of A and B (A ∩ B): the set of all elements contained in both A and B .

Example: 1. If A = {red, green, yellow, black} and B = {white, black, green},

then A ∩ 𝑩𝑩 = {black, green} .

2. If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8},

then A ∩ 𝐁𝐁 = {4, 5} . A intersect B

• Union of A and B (A ∪ B): the set of all elements contained in A or B, or both .

Example: 1. If A = {red, green, yellow} and B = {white, black},

then A ∪ 𝐁𝐁 = { red, green, yellow, white, black} .

2. If A = {1, 3, 5} and B = {2, 4, 6},

then A ∪ 𝐁𝐁 = {1, 2, 3, 4, 5, 6}. A union B

• Empty set (or null set) ∅: a set that contains no elements (disjoint) .

Example: If A = { x | x = Feb . 30}, then A = ∅

B = { x | x = Christmas day on Nov . 25} = ∅

• Subset (B ⊂ A): a set B is a subset of a set A if each element of B is an element of A . A subset B is a portion of another set A .

Example: If A = {4, 5, 8, 10, 17, 23}, B = {5, 10, 17},

then (B ⊂ A) .

• x ∈ A: x is an element of the set A (or x belongs to A) .

• x ∉ A: x is not an element of the set A (or x does not belong to A) .

Example: A = {1, 2, 3, 4, 5}3 ∈ A: 3 is an element of A . 6 ∉ A: 6 is not an element of A .


(A ∪ B) OR


If A = {2, 5} and B = {1, 3, 4}then A ∪ 𝐵𝐵 = {1, 2, 3, 4, 5} .



If A = {3, 6, 9} and B = {5, 6, 7, 8, 9}then A ∩ 𝐵𝐵 = {6, 9} .




B A

A B

A B

A B

`1 Page 2-23





Example: Given A = { a | a is a number between 6 and 10} . 7, 8, 9

B = { b | b is a prime number between 3 and 10} . 5, 7

Review: A prime number is a whole number that only has two factors, 1 and itself .

List the elements in A ∪ B and A ∩ B.

A ∪ B = {5, 7, 8, 9}, A ∩ B = {7}

Example: Let A = {2, 4, 6, 8}, B = {0, 1, 2, 3, 4}, and C = {-2, 3} . List the elements in

the following .

Solution

1. A ∪ 𝐵𝐵 {0, 1, 2, 3, 4, 6, 8}

2. A ∩ B { 2, 4}

3. A ∩ 𝐶𝐶 ∅

Example: Find the following sets .

1. {x, y, z} ∩ { u, v, w, z, x, y} {x, y, z}

2. {a, b, c} ∪ ∅ {a, b, c}

• Compound inequality review: a statement that contains more than one inequality . a < x < b

It means a < x and x < b.

Example: Graph and write using interval notation . Solution

1. -2 ≤ t and t < 3 [-2, 3) [ ) -2 3

2. 1 < b ≤ 5 (1, 5] ( ] 1 5

Example: Solve and sketch the graphs of the following inequalities .

1. -3 ≤ 4 + 3x < 5-3 - 4 ≤ 4 + 3x - 4 < 5 – 4 Subtract 4 from each term .

-7 ≤ 3x < 1 Divide each part by 3 .

-73

≤ 𝑥𝑥 < 13

�𝒙𝒙 | -𝟕𝟕𝟑𝟑

≤ 𝒙𝒙 < 𝟏𝟏𝟑𝟑� or �−𝟕𝟕

𝟑𝟑, 𝟏𝟏𝟑𝟑� [ )

-2 .33 0 .33

�−73

, 13� or [-2 .33, 0 .33)

2. -2 ≤ 𝟐𝟐𝒙𝒙+𝟐𝟐𝟑𝟑

< 6

-6 ≤ 2x + 2 < 18 Multiply 3 by each term .

-8 ≤ 2x < 16 Subtract 2 from each term .

-4 ≤ 𝑥𝑥 < 8 Divide each term by 2 .

{x | -𝟒𝟒 ≤ 𝐱𝐱 < 𝟖𝟖 } or [-𝟒𝟒,𝟖𝟖) [ ) -4 8

`1 Page 2-24





Inequality and Unions/ Intersections

Intersections and inequalities

Example: Solve and sketch the graphs of 2x – 1 ≤ 3 and 4 + x < 5 .Tip: “and” means all x values that satisfy both inequalities (use intersection ∩).

- Solve each inequality and graph . 2x – 1 ≤ 3 4 + x < 5

2x ≤ 4 x < 1 x ≤ 2 , (-∞, 2] (-∞, 1)

] ) 2 1

- Solution: { x | x ≤ 2} ∩ { x | x < 1} or { x | x < 1}The numbers common to both sets are those that are less than 1 .

- Graph the intersect of the two solution sets . ) (-∞, 2] ∩ (-∞, 1)

1

Unions and inequalities

Example: Solve and sketch the graphs of 2 – 3x > 5 or 4 + x ≥ 7 .Tip: “or” means x does not have to be in both solution sets to satisfy both inequalities (use union ∪ ) .

- Solve each inequality and graph .

2 - 3x > 5 4 + x ≥ 7-3x > 3 x ≥ 3x < -1, (-∞, -1) - [3, ∞)

) [ -1 3

- Graph the union of the two solution sets .

) [ - 1 3

- The solution set: { x | x < -1 or x ≥ 3 } or (-∞, -1) ∪ [3, ∞)

Example: Solve and sketch the graphs of 2x – 3 ≥ 1 or 4 + 3x < 16 . 2x ≥ 4 3x < 12

x ≥ 2 , [2, ∞) x < 4 , (-∞, 4)

[ ) 2 4

[2, ∞) ∪ (-∞, 4)

The solution set: (-∞, ∞), or all real numbers .

`1 Page 2-25

Inequality and Unions/ Intersections

Intersections and inequalities

Example: Solve and sketch the graphs of 2x – 1 ≤ 3 and 4 + x < 5 .Tip: “and” means all x values that satisfy both inequalities (use intersection ∩).

- Solve each inequality and graph . 2x – 1 ≤ 3 4 + x < 5

2x ≤ 4 x < 1 x ≤ 2 , (-∞, 2] (-∞, 1)

] ) 2 1

- Solution: { x | x ≤ 2} ∩ { x | x < 1} or { x | x < 1}The numbers common to both sets are those that are less than 1 .

- Graph the intersect of the two solution sets . ) (-∞, 2] ∩ (-∞, 1)

1

Unions and inequalities

Example: Solve and sketch the graphs of 2 – 3x > 5 or 4 + x ≥ 7 .Tip: “or” means x does not have to be in both solution sets to satisfy both inequalities (use union ∪ ) .

- Solve each inequality and graph .

2 - 3x > 5 4 + x ≥ 7-3x > 3 x ≥ 3x < -1, (-∞, -1) - [3, ∞)

) [ -1 3

- Graph the union of the two solution sets .

) [ - 1 3

- The solution set: { x | x < -1 or x ≥ 3 } or (-∞, -1) ∪ [3, ∞)

Example: Solve and sketch the graphs of 2x – 3 ≥ 1 or 4 + 3x < 16 . 2x ≥ 4 3x < 12

x ≥ 2 , [2, ∞) x < 4 , (-∞, 4)

[ ) 2 4

[2, ∞) ∪ (-∞, 4)

The solution set: (-∞, ∞), or all real numbers .

`1 Page 2-25





2-5 ABSOLUTE-VALUE EQUATIONS & INEQUALITIES

Absolute Value

• Absolute value review: geometrically, it is the distance of a number x from zero on the

number line . It is symbolized by vertical bars, as in “|𝑥𝑥|”, “|𝑦𝑦|” …


• No negatives for absolute value, �-𝒃𝒃� = |𝒃𝒃| : Distance is always positive, and the absolute

value is a distance, so the absolute value is never negative .


-2 0 2


-2 0 2

• Properties of absolute value

Absolute Value Example

absolute value |𝑥𝑥| = �𝑥𝑥, if 𝑥𝑥 ≥ 0 -𝑥𝑥, if 𝑥𝑥 < 0

If |2𝑥𝑥 – 3 | = 5, then 2𝑥𝑥 – 3 = 5 or 2𝑥𝑥 – 3 = -5

properties |𝑥𝑥𝑦𝑦| = |𝑥𝑥||𝑦𝑦| |-4𝑎𝑎| = |-4||𝑎𝑎| = 4|𝑎𝑎|

�𝑥𝑥𝑦𝑦� =

|𝑥𝑥||𝑦𝑦| (𝑦𝑦 ≠ 0) �3𝑥𝑥

3

5𝑦𝑦� = �3𝑥𝑥3�

|5𝑦𝑦|= 3�𝑥𝑥3�

5|𝑦𝑦|


1. �-𝟕𝟕𝟕𝟕� = �-7�|𝑥𝑥| = 7|𝟕𝟕| |𝑥𝑥𝑦𝑦| = |𝑥𝑥||𝑦𝑦| , �-𝑏𝑏� = |𝑏𝑏|

2. �𝟓𝟓𝟕𝟕𝟐𝟐

𝟏𝟏𝟓𝟓𝟕𝟕� = �𝑥𝑥

2

3𝑥𝑥� = �𝑥𝑥

3� = |𝑥𝑥|

|3|= |𝟕𝟕|

𝟑𝟑�𝑥𝑥𝑦𝑦� = |𝑥𝑥|

|𝑦𝑦|

3. �-𝟏𝟏𝟐𝟐𝟏𝟏𝟓𝟓

𝟐𝟐� = �-6𝑎𝑎5� = �-6� |𝑎𝑎5| = 𝟔𝟔�𝟏𝟏𝟓𝟓� |𝑥𝑥𝑦𝑦| = |𝑥𝑥||𝑦𝑦|, �-𝑏𝑏� = |𝑏𝑏|

`1 Page 2-26

2-5 ABSOLUTE-VALUE EQUATIONS & INEQUALITIES

Absolute Value

• Absolute value review: geometrically, it is the distance of a number x from zero on the

number line . It is symbolized by vertical bars, as in “|𝑥𝑥|”, “|𝑦𝑦|” …


• No negatives for absolute value, �-𝒃𝒃� = |𝒃𝒃| : Distance is always positive, and the absolute

value is a distance, so the absolute value is never negative .


-2 0 2


-2 0 2

• Properties of absolute value

Absolute Value Example


If |2𝑥𝑥 – 3 | = 5, then 2𝑥𝑥 – 3 = 5 or 2𝑥𝑥 – 3 = -5

properties |𝑥𝑥𝑦𝑦| = |𝑥𝑥||𝑦𝑦| |-4𝑎𝑎| = |-4||𝑎𝑎| = 4|𝑎𝑎|

�𝑥𝑥𝑦𝑦� =

|𝑥𝑥||𝑦𝑦| (𝑦𝑦 ≠ 0) �3𝑥𝑥

3

5𝑦𝑦� = �3𝑥𝑥3�

|5𝑦𝑦|= 3�𝑥𝑥3�

5|𝑦𝑦|


1. �-𝟕𝟕𝟕𝟕� = �-7�|𝑥𝑥| = 7|𝟕𝟕| |𝑥𝑥𝑦𝑦| = |𝑥𝑥||𝑦𝑦| , �-𝑏𝑏� = |𝑏𝑏|

2. �𝟓𝟓𝟕𝟕𝟐𝟐

𝟏𝟏𝟓𝟓𝟕𝟕� = �𝑥𝑥

2

3𝑥𝑥� = �𝑥𝑥

3� = |𝑥𝑥|

|3|= |𝟕𝟕|

𝟑𝟑�𝑥𝑥𝑦𝑦� = |𝑥𝑥|

|𝑦𝑦|

3. �-𝟏𝟏𝟐𝟐𝟏𝟏𝟓𝟓

𝟐𝟐� = �-6𝑎𝑎5� = �-6� |𝑎𝑎5| = 𝟔𝟔�𝟏𝟏𝟓𝟓� |𝑥𝑥𝑦𝑦| = |𝑥𝑥||𝑦𝑦|, �-𝑏𝑏� = |𝑏𝑏|

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Equations With Absolute Value

• Absolute value equation: an equation that includes absolute value(s) .

• |x| = A is equivalent to x = ±A: Example: |5𝑡𝑡 − 3| = 2 is equivalent to 5t – 3 = ±2 5t – 3 = 2, 5t – 3 = -2

• Procedure to solve an absolute value equation

Steps Example: |𝟐𝟐𝟐𝟐 − 𝟑𝟑| − 𝟓𝟓 = 𝟎𝟎

- Isolate the absolute value . |2𝑥𝑥 − 3| = 5 +5 to both sides .

+ – - Remove the absolute value symbol and set up 2x – 3 = 5 2x – 3 = -5

two equations (one positive and one negative) . |2x − 3| = 5 is equivalent to 2x – 3 = ±5 .

- Solve two equations . 2x = 8 2x = -2

x = 4 x = -1

? ?

- Check . |2 ∙ 4 − 3| = 5 �2(-1) − 3� = 5√ √

|5| = 5 �-5� = 5 Correct!

The solution set: {-1, 4}

Example: Solve for x . 2 |𝟐𝟐 + 𝟏𝟏| − 𝟑𝟑 = 𝟓𝟓

- Isolate |𝑥𝑥|. 2|𝑥𝑥 + 1| = 8 Add 3 to both sides .

|𝑥𝑥 + 1| = 4 Divide both sides by 2 .

- Remove the absolute value symbol and x + 1 = 4 x + 1 = -4set up two equations . |𝑥𝑥 + 1| = 4 is equivalent to x + 1 = ±4 .

- Solve two equations . x = 3 x = -5 ? ?

- Check . 2|3 + 1| − 3 = 5 2�-5 + 1� − 3 = 5 √ √

5 = 5 5 = 5 Correct!


Example: Solve for t . |𝟓𝟓𝟓𝟓 + 𝟐𝟐| = -𝟕𝟕

No solution The absolute value of an expression is never negative .

`1 Page 2-27

Equations With Absolute Value


• |x| = A is equivalent to x = ±A: Example: |5𝑡𝑡 − 3| = 2 is equivalent to 5t – 3 = ±2 5t – 3 = 2, 5t – 3 = -2

• Procedure to solve an absolute value equation

Steps Example: |𝟐𝟐𝟐𝟐 − 𝟑𝟑| − 𝟓𝟓 = 𝟎𝟎

- Isolate the absolute value . |2𝑥𝑥 − 3| = 5 +5 to both sides .

+ – - Remove the absolute value symbol and set up 2x – 3 = 5 2x – 3 = -5

two equations (one positive and one negative) . |2x − 3| = 5 is equivalent to 2x – 3 = ±5 .

- Solve two equations . 2x = 8 2x = -2

x = 4 x = -1

? ?

- Check . |2 ∙ 4 − 3| = 5 �2(-1) − 3� = 5√ √

|5| = 5 �-5� = 5 Correct!


Example: Solve for x . 2 |𝟐𝟐 + 𝟏𝟏| − 𝟑𝟑 = 𝟓𝟓

- Isolate |𝑥𝑥|. 2|𝑥𝑥 + 1| = 8 Add 3 to both sides .

|𝑥𝑥 + 1| = 4 Divide both sides by 2 .

- Remove the absolute value symbol and x + 1 = 4 x + 1 = -4set up two equations . |𝑥𝑥 + 1| = 4 is equivalent to x + 1 = ±4 .

- Solve two equations . x = 3 x = -5 ? ?

- Check . 2|3 + 1| − 3 = 5 2�-5 + 1� − 3 = 5 √ √

5 = 5 5 = 5 Correct!


Example: Solve for t . |𝟓𝟓𝟓𝟓 + 𝟐𝟐| = -𝟕𝟕

No solution The absolute value of an expression is never negative .

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• Equations containing two absolute-value expressions

Example: Solve for x . |𝟒𝟒𝟒𝟒 − 𝟓𝟓| = |𝟑𝟑𝟒𝟒 − 𝟐𝟐|

+ – - Remove the absolute value symbol and 4x – 5 = 3x – 2 4x – 5 = - (3x – 2)

set up two equations . 4x – 3x = - 2 + 5 4x – 5 = -3x + 2

- Solve two equations . x = 3 x = 1

? ?- Check . |4 ∙ 3 − 5| = |3 ∙ 3 − 2| |4 ∙ 1 − 5| = |3 ∙ 1 − 2|

? ? |12 − 5| = |9 − 2| �-1� = |1|

√ √7 = 7 1 = 1 Correct!

The solution set is {1, 3}

Example: Solve for x . �𝟔𝟔−𝟖𝟖𝟒𝟒𝟓𝟓� = �𝟕𝟕+𝟑𝟑𝟒𝟒

𝟐𝟐�

�6−8𝑥𝑥5� ∙ |10| = �7+3𝑥𝑥

2� ∙ |10| Multiply the LCD .

�6−8𝑥𝑥5

∙ 10� = �7+3𝑥𝑥2

∙ 10� |𝑥𝑥𝑥𝑥| = |𝑥𝑥||𝑥𝑥|

|12 − 16𝑥𝑥| = |35 + 15𝑥𝑥| |12 − 16𝑥𝑥| = |35 + 15𝑥𝑥| is equivalent to 12 – 16x = ±(35 + 15x) .

12 – 16x = 35 + 15x 12 – 16x = -(35 + 15x)

-31x = 23 12 – 16x = -35 – 15x

- x = -47

𝟒𝟒 = -𝟐𝟐𝟑𝟑𝟑𝟑𝟑𝟑

𝟒𝟒 = 47

The solution set: �-𝟐𝟐𝟑𝟑𝟑𝟑𝟑𝟑

, 𝟒𝟒𝟕𝟕�

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Absolute Value Inequalities

• |𝒙𝒙| < 𝑨𝑨: x is any value whose distance from zero is less than A units .

It can be written as { x | -A < x < A} or (-A, A) . ∵ ± x < A �𝑥𝑥 < 𝐴𝐴

-𝑥𝑥 < 𝐴𝐴, 𝑥𝑥 > -𝐴𝐴

Example: |𝑥𝑥| < 3, x is less than 3 units away from zero, i .e .

{ x | -3 < x < 3} or (-3, 3) -3 0 3

• |𝒙𝒙| ≤ 𝑨𝑨: x is any value whose distance from zero is less than or equal to A units .

It can be written as {x | -A ≤ x ≤ A } or [-A, A] . ∵ ± x ≤ A � 𝑥𝑥 ≤ 𝐴𝐴 -𝑥𝑥 ≤ 𝐴𝐴, 𝑥𝑥 ≥ -𝐴𝐴

Example: |𝑥𝑥| ≤ 3, x is less than or equal to 3 units away from zero, i .e .{ x | -3 ≤ x ≤ 3} or [-3, 3] ∙ ∙

-3 0 3

• |𝒙𝒙| > 𝑨𝑨: x is any value whose distance from zero is greater than A units .

It can be written as {x | x < -A or x > A } or (-∞, -A) ∪ (A, ∞) .

∵ ± x > A � 𝑥𝑥 > 𝐴𝐴 -𝑥𝑥 > 𝐴𝐴, 𝑥𝑥 < -𝐴𝐴

Example: |𝑥𝑥| > 4, x is any value whose distance from zero is greater than 4 units, i .e .

{x | x < -4 or x > 4} or (-∞, -4) ∪ (4, ∞) ) ( -4 0 4

• |𝒙𝒙| ≥ 𝑨𝑨: x is any value whose distance from zero is at least A units . ∵ ± x ≥ A � 𝑥𝑥 ≥ 𝐴𝐴 -𝑥𝑥 ≥ 𝐴𝐴, 𝑥𝑥 ≤ -𝐴𝐴

It can be written as {x | x ≤ -A or x ≥ A } or (-∞, -A] ∪ [A, ∞) .

Example: |𝑥𝑥| ≥ 4, x is at least 4 units away from zero, i .e .{x | x ≤ -4 or x ≥ 4} or (-∞, -4] ∪ [4, ∞) ] [

-4 0 4

Absolute value inequalities summaryAbsolute Value Inequality Example

|𝒙𝒙| < 𝑨𝑨 -A < x < A or (-A, A) |𝑥𝑥| < 2: { x | -2 < x < 2} or (-2, 2)|𝒙𝒙| ≤ 𝑨𝑨 -A ≤ x ≤ A or [-A, A] |𝑥𝑥| ≤ 2: {x | -2 ≤ x ≤ 2 } or [-2, 2]|𝒙𝒙| > 𝑨𝑨 x < -A or x > A or (-∞, -A) ∪ (A, ∞) |𝑥𝑥| > 2 : {x | x < -2 or x > 2} or (-∞, -2) ∪ (2, ∞)|𝒙𝒙| ≥ 𝑨𝑨 x ≤ -A or x ≥ A or (-∞, -A] ∪ [A, ∞) |𝑥𝑥| ≥ 2: {x | x ≤ -2 or x ≥ 2} or (-∞, -2] ∪ [2, ∞)

`1 Page 2-29

Absolute Value Inequalities

• |𝒙𝒙| < 𝑨𝑨: x is any value whose distance from zero is less than A units .

It can be written as { x | -A < x < A} or (-A, A) . ∵ ± x < A �𝑥𝑥 < 𝐴𝐴

-𝑥𝑥 < 𝐴𝐴, 𝑥𝑥 > -𝐴𝐴

Example: |𝑥𝑥| < 3, x is less than 3 units away from zero, i .e .

{ x | -3 < x < 3} or (-3, 3) -3 0 3

• |𝒙𝒙| ≤ 𝑨𝑨: x is any value whose distance from zero is less than or equal to A units .

It can be written as {x | -A ≤ x ≤ A } or [-A, A] . ∵ ± x ≤ A � 𝑥𝑥 ≤ 𝐴𝐴 -𝑥𝑥 ≤ 𝐴𝐴, 𝑥𝑥 ≥ -𝐴𝐴

Example: |𝑥𝑥| ≤ 3, x is less than or equal to 3 units away from zero, i .e .{ x | -3 ≤ x ≤ 3} or [-3, 3] ∙ ∙

-3 0 3

• |𝒙𝒙| > 𝑨𝑨: x is any value whose distance from zero is greater than A units .

It can be written as {x | x < -A or x > A } or (-∞, -A) ∪ (A, ∞) .

∵ ± x > A � 𝑥𝑥 > 𝐴𝐴 -𝑥𝑥 > 𝐴𝐴, 𝑥𝑥 < -𝐴𝐴

Example: |𝑥𝑥| > 4, x is any value whose distance from zero is greater than 4 units, i .e .

{x | x < -4 or x > 4} or (-∞, -4) ∪ (4, ∞) ) ( -4 0 4

• |𝒙𝒙| ≥ 𝑨𝑨: x is any value whose distance from zero is at least A units . ∵ ± x ≥ A � 𝑥𝑥 ≥ 𝐴𝐴 -𝑥𝑥 ≥ 𝐴𝐴, 𝑥𝑥 ≤ -𝐴𝐴

It can be written as {x | x ≤ -A or x ≥ A } or (-∞, -A] ∪ [A, ∞) .

Example: |𝑥𝑥| ≥ 4, x is at least 4 units away from zero, i .e .{x | x ≤ -4 or x ≥ 4} or (-∞, -4] ∪ [4, ∞) ] [

-4 0 4

Absolute value inequalities summaryAbsolute Value Inequality Example

|𝒙𝒙| < 𝑨𝑨 -A < x < A or (-A, A) |𝑥𝑥| < 2: { x | -2 < x < 2} or (-2, 2)|𝒙𝒙| ≤ 𝑨𝑨 -A ≤ x ≤ A or [-A, A] |𝑥𝑥| ≤ 2: {x | -2 ≤ x ≤ 2 } or [-2, 2]|𝒙𝒙| > 𝑨𝑨 x < -A or x > A or (-∞, -A) ∪ (A, ∞) |𝑥𝑥| > 2 : {x | x < -2 or x > 2} or (-∞, -2) ∪ (2, ∞)|𝒙𝒙| ≥ 𝑨𝑨 x ≤ -A or x ≥ A or (-∞, -A] ∪ [A, ∞) |𝑥𝑥| ≥ 2: {x | x ≤ -2 or x ≥ 2} or (-∞, -2] ∪ [2, ∞)

`1 Page 2-29





`1 Page 2-30

Example 1. Solve for x: |𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑| ≤ 𝟓𝟓𝟓𝟓 |𝑥𝑥𝑥𝑥| ≤ 𝐴𝐴𝐴𝐴 : -A ≤ x ≤ A

-5 ≤ 2𝑥𝑥𝑥𝑥 − 3 ≤ 5 Remove the absolute value symbol.

-2 ≤ 2𝑥𝑥𝑥𝑥 ≤ 8 Isolate x term (add 3 to each term).

{ x | -𝟏𝟏𝟏𝟏 ≤ 𝟐𝟐𝟐𝟐 ≤ 𝟒𝟒𝟒𝟒 } or [-1, 4] Divide each term by 2.

2. Solve for x: |𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐 − 𝟕𝟕𝟕𝟕| > 𝟐𝟐𝟐𝟐 |𝑥𝑥𝑥𝑥| > 𝐴𝐴𝐴𝐴: x < -A or x > A

3𝑥𝑥𝑥𝑥 − 7 < -2 3𝑥𝑥𝑥𝑥 − 7 > 2 Add 7 to both sides.

3𝑥𝑥𝑥𝑥 < 5 3𝑥𝑥𝑥𝑥 > 9 Divide 3 by both sides.

{x | 𝟐𝟐𝟐𝟐 < 𝟓𝟓𝟓𝟓𝟑𝟑𝟑𝟑 or x > 3 } or (-∞, 𝟓𝟓𝟓𝟓

𝟑𝟑𝟑𝟑� ∪ (3, ∞)





Unit 2 Summary

• Equation: a mathematical statement that contains two expressions separated by an equal

sign .

• Solution, root or zero of an equation: a solution is the particular value of the variable in the




highest power of the variable is one . (An equation whose graph is a straight line .)


Equation Standard Form Example Commentsfirst-degree equation

(linear equation)A x + B = 0

(x = x1) 5x + 4 = 0 The highest power of x is 1 .

second-degree equation(quadratic equation)

Ax2 + Bx + C = 0 2x2 + 7x - 3 = 0 The highest power of x is 2 .

third-degree equation(cubic equation)

Ax3 + Bx2 + Cx + D = 0 3x3 + 4x2 - 8x + 1 = 0 The highest power of x is 3 .

fourth-degree equation Ax4 + Bx3 + Cx2 +D x + E = 0 x4 – 9x3 + 3x2 + 2x – 5 = 0 The highest power of x is 4 .


• A linear equation in two variables: an equation that contains two variables in which the

highest power (exponent) of two variables is one .

• Formula: an equation that contains more than one variable and is used to solve practical

problems in everyday life .

• An inequality: a mathematical statement that contains < , > , ≥ , or ≤ symbol .

• Solution of an inequality: the particular value(s) of the variable in the inequality that makes

the inequality true .

• Compound inequality: a statement that contains more than one inequality . a < x < b


• |x| = A is equivalent to x = ±A. Example: |5𝑡𝑡 − 3| = 2 is equivalent to 5t – 3 = ±2 .

`1 Page 2-31

Unit 2 Summary

• Equation: a mathematical statement that contains two expressions separated by an equal

sign .

• Solution, root or zero of an equation: a solution is the particular value of the variable in the




highest power of the variable is one . (An equation whose graph is a straight line .)


Equation Standard Form Example Commentsfirst-degree equation

(linear equation)A x + B = 0

(x = x1) 5x + 4 = 0 The highest power of x is 1 .

second-degree equation(quadratic equation)

Ax2 + Bx + C = 0 2x2 + 7x - 3 = 0 The highest power of x is 2 .

third-degree equation(cubic equation)

Ax3 + Bx2 + Cx + D = 0 3x3 + 4x2 - 8x + 1 = 0 The highest power of x is 3 .

fourth-degree equation Ax4 + Bx3 + Cx2 +D x + E = 0 x4 – 9x3 + 3x2 + 2x – 5 = 0 The highest power of x is 4 .


• A linear equation in two variables: an equation that contains two variables in which the

highest power (exponent) of two variables is one .

• Formula: an equation that contains more than one variable and is used to solve practical

problems in everyday life .

• An inequality: a mathematical statement that contains < , > , ≥ , or ≤ symbol .

• Solution of an inequality: the particular value(s) of the variable in the inequality that makes

the inequality true .

• Compound inequality: a statement that contains more than one inequality . a < x < b


• |x| = A is equivalent to x = ±A. Example: |5𝑡𝑡 − 3| = 2 is equivalent to 5t – 3 = ±2 .

`1 Page 2-31





• Equations involving decimals

- Multiply each term by a power of 10 (10, 100, 1,000, etc .) to clear the decimals .

- Collect the variable terms on one side of the equation and the constants on the other side .

- Isolate the variable .

• Equations involving fractions

- Multiply each term by the LCD .

- Collect the variable terms on one side of the equation and the constants on the other side .

- Isolate the variable .

• Properties for solving equations





Solve -𝑡𝑡6

= 7-𝑡𝑡6

(-𝟔𝟔) = 7(-6), t = -42


= 𝐵𝐵𝑪𝑪

(C ≠ 0)

Solve 4a = -16

4𝑎𝑎𝟒𝟒

= −16𝟒𝟒

, a = -4

• Equation-solving strategy

Equation-Solving Strategy • Clear the fractions or decimals if necessary . • Remove parentheses . • Combine like terms on each side of the equation if necessary .• Collect the variable terms on one side of the equation and the numerical

terms on the other side . • Isolate the variable . • Check the solution with the original equation .

• Steps for solving word problems


`1 Page 2-32






(A ∪ B) OR


If A = {2, 5} and B = {1, 3, 4}then A ∪ 𝐵𝐵 = {1, 2, 3, 4, 5} .



If A = {3, 6, 9} and B = {5, 6, 7, 8, 9}then A ∩ 𝐵𝐵 = {6, 9} .




• Properties of absolute valueAbsolute Value Example


If |2𝑥𝑥 – 3 | = 5 Then 2𝑥𝑥 – 3 = 5 or 2𝑥𝑥 – 3 = -5

properties |𝑥𝑥𝑥𝑥| = |𝑥𝑥||𝑥𝑥| |-4𝑎𝑎| = |-4||𝑎𝑎| = 4|𝑎𝑎|

�𝑥𝑥𝑥𝑥� =

|𝑥𝑥||𝑥𝑥| (𝑥𝑥 ≠ 0) �3𝑥𝑥

3

5𝑦𝑦� = �3𝑥𝑥3�

|5𝑦𝑦|= 3�𝑥𝑥3�

5|𝑦𝑦|

• Procedure to solve an absolute value equation:

- Isolate the absolute value .- Remove the absolute value symbol and set up two equations (one positive and one

negative .) - Solve two equations .- Check .

• Absolute value inequalities summaryAbsolute Value Inequality Example

|𝒙𝒙| < 𝑨𝑨 -A < x < A or (-A, A) |𝑥𝑥| < 2: { x | -2 < x < 2} or (-2, 2)|𝒙𝒙| ≤ 𝑨𝑨 -A ≤ x ≤ A or [-A, A] |𝑥𝑥| ≤ 2: {x | -2 ≤ x ≤ 2 } or [-2, 2]|𝒙𝒙| > 𝑨𝑨 x < -A or x > A or (-∞, -A) ∪ (A, ∞) |𝑥𝑥| > 2 : {x | x < -2 or x > 2} or (-∞, -2) ∪ (2, ∞) |𝒙𝒙| ≥ 𝑨𝑨 x ≤ -A or x ≥ A or (-∞, -A] ∪ [A, ∞) |𝑥𝑥| ≥ 2: {x | x ≤ -2 or x ≥ 2} or (-∞, -2] ∪ [2, ∞)

• Business formulasApplication Formula


= , O

ON−=x


= , O

NO−=x





Compound Interest


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• Recall some geometry formulas

• More formulasApplication Formula Component

distance d = rt, r = 𝑑𝑑𝑡𝑡



simple interest I = P r t, P = 𝐼𝐼𝑟𝑟𝑡𝑡












100𝑚𝑚𝑐𝑐



(𝐹𝐹 − 32) , 𝐹𝐹 = 95𝐶𝐶 + 32 C – Celsius

F – Fahrenheit

Name of the Figure Formula Figure

rectangleP = 2l + 2w

A = lww

l

parallelogram P = 2a + 2bA = bh

h a b

circle 𝐴𝐴 = 𝜋𝜋𝜋𝜋2rdC π=π= 2

r d

triangle∠ X + ∠ Y + ∠ Z = 1800

bhA21

=

XhY b Z

trapezoid )(21 BbhA +=

b h B

cube V = s3

rectangular solid V = lwh h w

cylinder hrV 2π=r

h

sphere3

34 rV π=

r

cone hrV 2

31π= h

r

pyramid lwhV31

=

l wh

s

l

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PRACTICE QUIZ

Unit 2 Equations and Inequalities

1. Solve the following equations .

a. 3(x – 2) + 4x –7 = 3(5 – x)

b. 0 .3y – 0 .27 = -4 .36y

c. 3𝑥𝑥4− 2

3= 𝑥𝑥

2+ 1

4

2 . Five less than four times a number is nine more than the number divided by two . Find the number .

3 . Find three consecutive even integers such that four times the first integer is two less than the sum of the second and third integers.

4. Two vehicles are 340 km apart and are traveling towards each other . Their

speeds differ by 10 km per hour . What is the speed of each vehicle if they meet after 2 hours?

5. Alice boats at a speed of 26 km per hour in still water . The river flows at a speed of 12 km per hour . How long will it take Alice to boat 4 km downstream? 3 km upstream?

6 . Tom purchased 46-cent, 66-cent, and 86-cent Canadian stamps with a total value of $6 .80 . If the number of 66-cent stamps is 3 more than the number of 46-cent stamps, and the number of 86-cent stamps is 2 more than one half the number of 46-cent stamps . How many stamps of each did Tom receive?

7. Solve the following inequalities and graph the solution sets .

a . -7x – 3 ≥ 11 -

b. 3 – 2(4x – 5) + 7x > 2x + 10

c. 34

(5 – y) – 52 ≤ 1

3

Page 3

PRACTICE QUIZ

Unit 2 Equations and Inequalities


a. 3(x – 2) + 4x –7 = 3(5 – x)

b. 0 .3y – 0 .27 = -4 .36y

c. 3𝑥𝑥4− 2

3= 𝑥𝑥

2+ 1

4

2 . Five less than four times a number is nine more than the number divided by two . Find the number .

3 . Find three consecutive even integers such that four times the first integer is two less than the sum of the second and third integers.

4. Two vehicles are 340 km apart and are traveling towards each other . Their

speeds differ by 10 km per hour . What is the speed of each vehicle if they meet after 2 hours?

5. Alice boats at a speed of 26 km per hour in still water . The river flows at a speed of 12 km per hour . How long will it take Alice to boat 4 km downstream? 3 km upstream?

6 . Tom purchased 46-cent, 66-cent, and 86-cent Canadian stamps with a total value of $6 .80 . If the number of 66-cent stamps is 3 more than the number of 46-cent stamps, and the number of 86-cent stamps is 2 more than one half the number of 46-cent stamps . How many stamps of each did Tom receive?

7. Solve the following inequalities and graph the solution sets .

a . -7x – 3 ≥ 11 -

b. 3 – 2(4x – 5) + 7x > 2x + 10

c. 34

(5 – y) – 52 ≤ 1

3

Page 3





8 . Amanda got a 78% on the midterm exam in English . To get a B+, the average of her midterm and final exam must be between 76% and 80% . For what range of scores on the final exam will Amanda need to get a B+?

9 . Indicate whether each of the following is true or false .

a . - 516

∈ rational numbers

b . √13 ∈ rational numbers

10. a . Given A = { a | a is a prime number between 10 and 18}

B = { b | b is a number between 12 and 16}

List the numbers in A ∪ B and A ∩ B

b. Given A = {3, 5, 7}, B = {1, 2, 3, 4, 5} and C = {-3, -2} . List the

elements in the following:

A ∪ 𝐵𝐵A ∩ BA ∩ 𝐶𝐶

11 . Solve the following and graph the solution set .

-3 < 1+2𝑥𝑥3

≤ 1


a. 2|𝑥𝑥 + 3| − 4 = 6

b. |3𝑥𝑥 − 4| = |5𝑥𝑥 − 2|

13 . Solve the following inequalities .

a. |3𝑥𝑥 − 4| ≤ 7

b . |5𝑥𝑥 − 3| > 4

Page 4




Algebra I & II Key Concepts, Practice, and Quizzes Unit 3 – Functions and Graphs

UNIT 3 FUNCTIONS AND GRAPHS

3-1 GRAPHING EQUATIONS

The Coordinate Plane

• The coordinate plane (or Cartesian / rectangular coordinate system): a powerful tool to

mark a point and the solution of a linear equation on a graph .

Coordinate axes:

x axis – the horizontal line .

y axis – the vertical line .

The origin: the intersection of the x and y axes where both lines are 0 .

• Ordered pair: (x, y): Each point on the plane corresponds to an ordered pair .

(x , y) Example: (2, 1) 1st coordinate (abscissa) 2nd coordinate (ordinate)

Example: (soda, $0 .90) , (juice, $1 .25)

• Coordinate: the numbers in an ordered pair .

• Four quadrants

Quadrant (x, y) ExampleThe 1st quadrant I (+x, +y) (+2, +3)The 2nd quadrant II (-x, +y) (-2, +3) The 3rd quadrant III (-x, -y) (-2, -3) The 4th quadrant IV (+x, -y) (+2, -3)

Example: Plot the points and name the quadrant .

(1, 3) (-3, 2) (-2, -2) (2, -1)

(1, 3): I , (-3, 2): II , (-2, -2): III , (2, -1): IV

• x-intercept: the point at which the graph crosses the x-axis .

Example: (x, y) = (3, 0) (0, 2)

• y-intercept: the point at which the graph crosses the y-axis .

Example: (x, y) = (0, 2)

Points are on axes

II

III IV

I y

x

x(3, 0)

y

∙ (-3, 2)

∙ (2, -1)

y

x

∙ (1, 3)

∙ (-2,- 2)

∙ (0, 0) ∙ (2, 1)

y

x

y

x ∙ (0, 0)

∙

∙

Page 1

UNIT 3 FUNCTIONS AND GRAPHS

3-1 GRAPHING EQUATIONS

The Coordinate Plane

• The coordinate plane (or Cartesian / rectangular coordinate system): a powerful tool to

mark a point and the solution of a linear equation on a graph .

Coordinate axes:

x axis – the horizontal line .

y axis – the vertical line .

The origin: the intersection of the x and y axes where both lines are 0 .

• Ordered pair: (x, y): Each point on the plane corresponds to an ordered pair .

(x , y) Example: (2, 1) 1st coordinate (abscissa) 2nd coordinate (ordinate)

Example: (soda, $0 .90) , (juice, $1 .25)

• Coordinate: the numbers in an ordered pair .

• Four quadrants

Quadrant (x, y) ExampleThe 1st quadrant I (+x, +y) (+2, +3)The 2nd quadrant II (-x, +y) (-2, +3) The 3rd quadrant III (-x, -y) (-2, -3) The 4th quadrant IV (+x, -y) (+2, -3)

Example: Plot the points and name the quadrant .

(1, 3) (-3, 2) (-2, -2) (2, -1)

(1, 3): I , (-3, 2): II , (-2, -2): III , (2, -1): IV

• x-intercept: the point at which the graph crosses the x-axis .

Example: (x, y) = (3, 0) (0, 2)

• y-intercept: the point at which the graph crosses the y-axis .

Example: (x, y) = (0, 2)

Points are on axes

II

III IV

I y

x

x(3, 0)

y

∙ (-3, 2)

∙ (2, -1)

y

x

∙ (1, 3)

∙ (-2,- 2)

∙ (0, 0) ∙ (2, 1)

y

x

y

x ∙ (0, 0)

∙

∙

Page 1





Page 2

Graphs of Linear Equations

• A linear (first-degree) equation: an equation whose graph is a straight line.

• A linear (first-degree) equation in two variables: a linear equation that contains two

variables, such as 2x + y = 3.

• The standard form of linear equation in two variables: Ax + By = C

Standard Form Example Ax + By = C 5x – 7y = 4

• Solutions of equations: Solutions for a linear equation in two variables are an ordered pair.

They are the particular values of the variables in the equation that makes the equation true.

Example: Determine whether the given point is a solution.

? √ 1. (2, -1): 2x – 3y = 7 2 ∙ 2 – 3(-1) = 7 7 = 7 Yes

? 2. (0, 3): 10p + 3q = -4 10 ∙ 0 + 3 ∙ 3 = -4 9 ≠ -4 No

• The graph of an equation is the diagram obtained by plotting the set of points where the

equation is true (or satisfies the equation).

• Procedure to graph a linear equation

Steps Example: Graph 2x – y = 3

- Choose two values of x, calculate the x y = 2x – 3 (x, y) corresponding y, and make a table. 0 2∙0 – 3 = -3 (0, -3) y-intercept

- Plot these two points on the coordinate plane. 1 2∙1 – 3 = -1 (1, -1) Select x Calculate y Ordered pair

- Connect the points with a straight line. (Any two points determine a straight line.)

- Check with the third point.

Is third point (2, 1) on the line? Yes. Correct! x y = 2x – 3 (x, y) 2 2∙2 – 3 = 1 (2, 1)

Example: Graph y = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 − 𝟑𝟑𝟑𝟑 and determine another point.

x y (x, y) 0 - 3 (0, -3) 2 - 2 (2, -2)

y

x ∙ (2, 1)

∙ (1, -1)

∙ (0, -3)

∙ (8, 1)

(0, -3) ∙ ∙ (2, -2)

y

x

Another solution

Solutions

3rd point

Page 2

Graphs of Linear Equations

• A linear (first-degree) equation: an equation whose graph is a straight line.

• A linear (first-degree) equation in two variables: a linear equation that contains two

variables, such as 2x + y = 3.

• The standard form of linear equation in two variables: Ax + By = C

Standard Form Example Ax + By = C 5x – 7y = 4

• Solutions of equations: Solutions for a linear equation in two variables are an ordered pair.

They are the particular values of the variables in the equation that makes the equation true.

Example: Determine whether the given point is a solution.

? √ 1. (2, -1): 2x – 3y = 7 2 ∙ 2 – 3(-1) = 7 7 = 7 Yes

? 2. (0, 3): 10p + 3q = -4 10 ∙ 0 + 3 ∙ 3 = -4 9 ≠ -4 No

• The graph of an equation is the diagram obtained by plotting the set of points where the

equation is true (or satisfies the equation).

• Procedure to graph a linear equation

Steps Example: Graph 2x – y = 3

- Choose two values of x, calculate the x y = 2x – 3 (x, y) corresponding y, and make a table. 0 2∙0 – 3 = -3 (0, -3) y-intercept

- Plot these two points on the coordinate plane. 1 2∙1 – 3 = -1 (1, -1) Select x Calculate y Ordered pair

- Connect the points with a straight line. (Any two points determine a straight line.)

- Check with the third point.

Is third point (2, 1) on the line? Yes. Correct! x y = 2x – 3 (x, y) 2 2∙2 – 3 = 1 (2, 1)

Example: Graph y = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 − 𝟑𝟑𝟑𝟑 and determine another point.

x y (x, y) 0 - 3 (0, -3) 2 - 2 (2, -2)

y

x ∙ (2, 1)

∙ (1, -1)

∙ (0, -3)

∙ (8, 1)

(0, -3) ∙ ∙ (2, -2)

y

x

Another solution

Solutions

3rd point





Graphing Nonlinear Equation With Two Variables

• Nonlinear equation: an equation whose graph is not a straight line .

Example: 2x2 – 5y = 3 , 3y3 + 7x2 – 3xy + 8 = 0Recall: Higher-degree equations are nonlinear equations .

• Procedure to graph a nonlinear equation with two variables

- Choose a few values of x, calculate the corresponding y, and make a table .

- Plot these points on the coordinate plane (plot more points to get a cleaner shape of

the graph) .

- Connect the points with a smooth curve .

Example: Graph the equation y = 5 + x2 .x y = 5 + x2 Ordered Pair0 5 + 02 = 5 (0, 5)1 5 + 12 = 6 (1, 6)

-1 5 + (-1)2 = 6 (-1, 6)2 5 + 22 = 9 (2, 9)

-2 5 + (-2)2 = 9 (-2, 9)

Example: Graph the equation y = 𝟒𝟒𝒙𝒙

.

x y = 𝟒𝟒𝒙𝒙 Ordered Pair

1 41

= 4 (1, 4)

-14-1

= -4 (-1, -4)

2 42

= 2 (2, 2)

-24-2

= -2 (-2, -2)

4 44

= 1 (4, 1)

-44-4

= -1 (-4, -1)

Example: Graph the equation y = |𝒙𝒙 − 𝟏𝟏| .x y = |𝑥𝑥 − 1| Ordered Pair0 |0 − 1|= 1 (0, 1) 1 |1 − 1|= 0 (1, 0)

-1 �-1 − 1�= 2 (-1, 2) 2 |2 − 1|= 1 (2, 1) 3 |3 − 1|= 2 (3, 2)

∙ (2, 9) (-2, 9) ∙

y y

∙ ∙

y

∙ (1, 4)y

y

∙ (0, 1)

x

∙ (2, 1)(-1, 2) ∙ ∙ (3, 2)

∙ (2, 2) ∙ (4, 1)

∙ (-1,- 4)

∙ (-2, -2) ∙ (-4, -1)

∙ (1, 6)(-1, 6) ∙

x

∙∙∙

(0, 5) ∙

x

x ∙ (1, 0)

Page 3

Graphing Nonlinear Equation With Two Variables


Example: 2x2 – 5y = 3 , 3y3 + 7x2 – 3xy + 8 = 0Recall: Higher-degree equations are nonlinear equations .

• Procedure to graph a nonlinear equation with two variables

- Choose a few values of x, calculate the corresponding y, and make a table .

- Plot these points on the coordinate plane (plot more points to get a cleaner shape of

the graph) .


Example: Graph the equation y = 5 + x2 .x y = 5 + x2 Ordered Pair0 5 + 02 = 5 (0, 5)1 5 + 12 = 6 (1, 6)

-1 5 + (-1)2 = 6 (-1, 6)2 5 + 22 = 9 (2, 9)

-2 5 + (-2)2 = 9 (-2, 9)

Example: Graph the equation y = 𝟒𝟒𝒙𝒙

.

x y = 𝟒𝟒𝒙𝒙 Ordered Pair

1 41

= 4 (1, 4)

-14-1

= -4 (-1, -4)

2 42

= 2 (2, 2)

-24-2

= -2 (-2, -2)

4 44

= 1 (4, 1)

-44-4

= -1 (-4, -1)

Example: Graph the equation y = |𝒙𝒙 − 𝟏𝟏| .x y = |𝑥𝑥 − 1| Ordered Pair0 |0 − 1|= 1 (0, 1) 1 |1 − 1|= 0 (1, 0)

-1 �-1 − 1�= 2 (-1, 2) 2 |2 − 1|= 1 (2, 1) 3 |3 − 1|= 2 (3, 2)

∙ (2, 9) (-2, 9) ∙

y y

∙ ∙

y

∙ (1, 4)y

y

∙ (0, 1)

x

∙ (2, 1)(-1, 2) ∙ ∙ (3, 2)

∙ (2, 2) ∙ (4, 1)

∙ (-1,- 4)

∙ (-2, -2) ∙ (-4, -1)

∙ (1, 6)(-1, 6) ∙

x

∙∙∙

(0, 5) ∙

x

x ∙ (1, 0)

Page 3





3-2 FUNCTIONS

Function

• Function: a special type of relation (or correspondence) which matches each element of

the domain (x-value or first set) with exactly one element of the range (y-value or second

set) .

• All functions are relations (correspondence), but not all relations are functions.

Example: 1. Name (x) Social Security Number (y)Tom 618-31-4123Steve 312-15-7432

This is a function (Since each person is assigned only one SSN number .)

2. Name (x) Address (y)

Adam 123 First Ave .Shawn 234 Second Ave . (Home)

456 Univ . Way (Student residency)

This is a relation but not a function. (Shawn has two addresses .)

Example: Determine if the following relation (correspondence) is a function .

Domain (x) Range (y) Function Comments abc

123

Yes Each value of x is assigned onlyone value of y .

3-2

4-3 No -2 is assigned more than one

value of the range (4 and -3) . -540

2

3Yes -5 and 4 are assigned only one

value of the range (2) .


Domain (x) Range (y) Correspondence Function Comments

famous writers a set of book titles

a book that the writer has published No Some writers have publised

more than one book .

a set of numbers

a set of positive numbers

square each number and then divide by 3 Yes

𝑦𝑦 = 𝑥𝑥2

3The result will be a unique positive number .

Page 3-4

3-2 FUNCTIONS

Function


the domain (x-value or first set) with exactly one element of the range (y-value or second

set) .


Example: 1. Name (x) Social Security Number (y)Tom 618-31-4123Steve 312-15-7432

This is a function (Since each person is assigned only one SSN number .)

2. Name (x) Address (y)

Adam 123 First Ave .Shawn 234 Second Ave . (Home)

456 Univ . Way (Student residency)

This is a relation but not a function. (Shawn has two addresses .)


Domain (x) Range (y) Function Comments abc

123

Yes Each value of x is assigned onlyone value of y .

3-2

4-3 No -2 is assigned more than one

value of the range (4 and -3) . -540

2

3Yes -5 and 4 are assigned only one

value of the range (2) .


Domain (x) Range (y) Correspondence Function Comments

famous writers a set of book titles

a book that the writer has published No Some writers have publised

more than one book .

a set of numbers

a set of positive numbers

square each number and then divide by 3 Yes

𝑦𝑦 = 𝑥𝑥2

3The result will be a unique positive number .

Page 3-4





Finding Function Values

• Function notation: The notation for a function is f (x), P(x), g(x), h(x) , … Read f (x) as “ f of x” . f (x) does not mean f times x .

• Function values: The value of a function at “x = a” is denoted as “f (a)” . f (a) is the value

of f (x) when a is replaced by x in f (x) .

f (x) | x = a = f (a), �𝑎𝑎 is a constant replace 𝑥𝑥 by 𝑎𝑎

Example: f (x) = 3x + 1 If x = 2 If x = -4

f ( ) = 3( ) + 1 f ( ) = 3( ) + 1

f (2) = 3(2) + 1 = 7 f (-4) = 3(-4) + 1 = -11(Substitute x for 2 .) (Substitute x for -4 .)

Example: Evaluate the functions and simplify at the indicated values .

Solution

1. f (-2) for f (x) = 3 – 5x f (-2) = 3 – 5(-2) = 3 + 10 = 13

2. p(3) for p(t) = 2t2 – 7 p(3) = 2(3)2 – 7 = 2 ∙ 9 – 7 = 11

3. f (a – 2) for f (x) = 8 + 3x3 f (a – 2) = 8 + 3(a – 2)3 Replace (a – 2) by x in f (x) .

4. g (0) for g (r) = r2 – 3r + 2 g(0) = 02 – 3∙0 + 2 = 2


f (x) = 3x + 1 Solution

1. f (-2) f (-2) = 3(-2) + 1 = -6 + 1 = -5

2. f (b + 2) f (b + 2) = 3(b + 2) + 1 = 3b + 6 + 1 = 3b + 7

3. f (a) + f (3) f (a) + f (3) = (3a + 1) + (3∙3 + 1) = 3a + 1 + 10 = 3a + 11

4. f �𝟏𝟏𝟔𝟔� f �𝟏𝟏

𝟔𝟔� = 3�𝟏𝟏

𝟔𝟔� + 1 = 1

2+ 1 = 3

2= 𝟏𝟏 𝟏𝟏

𝟐𝟐

h(r) = 3r2 – 2

1. h (0) h (0) = 3∙02 – 2 = -2

2. 𝒉𝒉�- 𝟐𝟐𝟑𝟑� ℎ �- 𝟐𝟐

𝟑𝟑� = 3 �- 𝟐𝟐

𝟑𝟑�2− 2 = 3 �4

9� − 2 = 4

3− 2 = 4

3− 6

3= - 𝟐𝟐

𝟑𝟑

g(x) = |𝒙𝒙 + 𝟑𝟑| + 𝟐𝟐𝒙𝒙

1. g (-4) g (-4) = �-𝟒𝟒 + 3� + 2�-𝟒𝟒� = �-1� − 8 = 1 − 8 = -𝟕𝟕

2. g (a - b) g (a − b) = |𝒂𝒂 − 𝐛𝐛 + 𝟑𝟑| + 𝟐𝟐(𝒂𝒂 − 𝒃𝒃)

Page 3-5

Finding Function Values

• Function notation: The notation for a function is f (x), P(x), g(x), h(x) , … Read f (x) as “ f of x” . f (x) does not mean f times x .

• Function values: The value of a function at “x = a” is denoted as “f (a)” . f (a) is the value

of f (x) when a is replaced by x in f (x) .

f (x) | x = a = f (a), �𝑎𝑎 is a constant replace 𝑥𝑥 by 𝑎𝑎

Example: f (x) = 3x + 1 If x = 2 If x = -4

f ( ) = 3( ) + 1 f ( ) = 3( ) + 1

f (2) = 3(2) + 1 = 7 f (-4) = 3(-4) + 1 = -11(Substitute x for 2 .) (Substitute x for -4 .)


Solution

1. f (-2) for f (x) = 3 – 5x f (-2) = 3 – 5(-2) = 3 + 10 = 13

2. p(3) for p(t) = 2t2 – 7 p(3) = 2(3)2 – 7 = 2 ∙ 9 – 7 = 11

3. f (a – 2) for f (x) = 8 + 3x3 f (a – 2) = 8 + 3(a – 2)3 Replace (a – 2) by x in f (x) .

4. g (0) for g (r) = r2 – 3r + 2 g(0) = 02 – 3∙0 + 2 = 2


f (x) = 3x + 1 Solution

1. f (-2) f (-2) = 3(-2) + 1 = -6 + 1 = -5

2. f (b + 2) f (b + 2) = 3(b + 2) + 1 = 3b + 6 + 1 = 3b + 7

3. f (a) + f (3) f (a) + f (3) = (3a + 1) + (3∙3 + 1) = 3a + 1 + 10 = 3a + 11

4. f �𝟏𝟏𝟔𝟔� f �𝟏𝟏

𝟔𝟔� = 3�𝟏𝟏

𝟔𝟔� + 1 = 1

2+ 1 = 3

2= 𝟏𝟏 𝟏𝟏

𝟐𝟐

h(r) = 3r2 – 2

1. h (0) h (0) = 3∙02 – 2 = -2

2. 𝒉𝒉�- 𝟐𝟐𝟑𝟑� ℎ �- 𝟐𝟐

𝟑𝟑� = 3 �- 𝟐𝟐

𝟑𝟑�2− 2 = 3 �4

9� − 2 = 4

3− 2 = 4

3− 6

3= - 𝟐𝟐

𝟑𝟑

g(x) = |𝒙𝒙 + 𝟑𝟑| + 𝟐𝟐𝒙𝒙

1. g (-4) g (-4) = �-𝟒𝟒 + 3� + 2�-𝟒𝟒� = �-1� − 8 = 1 − 8 = -𝟕𝟕

2. g (a - b) g (a − b) = |𝒂𝒂 − 𝐛𝐛 + 𝟑𝟑| + 𝟐𝟐(𝒂𝒂 − 𝒃𝒃)

Page 3-5





Graphing a Function

• The graph of a function: the diagram obtained by plotting the set of all points where the

function y = f (x) is true .

• Procedure to graph a function: (similar to graph an equation)

- Choose a few values of x, calculate the corresponding values of functions y = f (x) and

make a table .

- Plot these points on the coordinate plane .


Example: Graph the function f (x) = 2x2 + x – 1 .

x y = f (x) = 2x2 + x – 1 (x, y) 0 2∙02 + 0 – 1 = -1 (0, -1)1 2∙12 + 1 – 1 = 2 (1, 2)

-1 2∙(-1)2 + (-1) – 1 = 0 (-1, 0)2 2∙22 + 2 – 1 = 9 (2, 9)

-2 2∙(-2)2 + (-2) – 1 = 5 (-2, 5)

• Identify the f (x) in a graph

- Locate the x-value(s) on the x-axis, and plot vertical line(s) to the curve and draw a

solid point . f(x)

- Plot horizontal line(s) from the point to the y-axis to determine y = f (x) value(s) . f (x) is the y-value that is corresponded with x .

Example: The following graph shows the number of car rentals as a function of time in a

vehicle rental store . How many cars were rented in 2011?

(-1, 0) ∙

(-2, 5) ∙

∙ (2, 9)y

x

∙ (1, 2)

∙ (0, -1)

x

∙

x (year)

f(x)# of car rentals100 ∙

50 ∙

∙ 2000

∙ 2011

∙ 2005

f (2011) = 100

(There were 100 cars rented in 2011 .)

∙

Page 3-6

Graphing a Function

• The graph of a function: the diagram obtained by plotting the set of all points where the

function y = f (x) is true .

• Procedure to graph a function: (similar to graph an equation)

- Choose a few values of x, calculate the corresponding values of functions y = f (x) and

make a table .

- Plot these points on the coordinate plane .


Example: Graph the function f (x) = 2x2 + x – 1 .

x y = f (x) = 2x2 + x – 1 (x, y) 0 2∙02 + 0 – 1 = -1 (0, -1)1 2∙12 + 1 – 1 = 2 (1, 2)

-1 2∙(-1)2 + (-1) – 1 = 0 (-1, 0)2 2∙22 + 2 – 1 = 9 (2, 9)

-2 2∙(-2)2 + (-2) – 1 = 5 (-2, 5)

• Identify the f (x) in a graph

- Locate the x-value(s) on the x-axis, and plot vertical line(s) to the curve and draw a

solid point . f(x)

- Plot horizontal line(s) from the point to the y-axis to determine y = f (x) value(s) . f (x) is the y-value that is corresponded with x .

Example: The following graph shows the number of car rentals as a function of time in a

vehicle rental store . How many cars were rented in 2011?

(-1, 0) ∙

(-2, 5) ∙

∙ (2, 9)y

x

∙ (1, 2)

∙ (0, -1)

x

∙

x (year)

f(x)# of car rentals100 ∙

50 ∙

∙ 2000

∙ 2011

∙ 2005

f (2011) = 100

(There were 100 cars rented in 2011 .)

∙

Page 3-6





The Vertical Line Test

• The vertical line test can determine whether a relation is a function .

• The vertical line test: If a vertical line cuts the relation’s graph more than once, then the

relation is not a function .

• Recall: A function is a special type of relation which matches each element of the

domain (x-value) with exactly one element of the range (y-value) .

Example: Determine if the following graphs are functions .

Yes, cuts once . No, cuts twice .


Yes, cuts once .

∙

∙

∙

∙

∙

∙

∙

Page 3-7

The Vertical Line Test

• The vertical line test can determine whether a relation is a function .



• Recall: A function is a special type of relation which matches each element of the

domain (x-value) with exactly one element of the range (y-value) .

Example: Determine if the following graphs are functions .



Yes, cuts once .

∙

∙

∙

∙

∙

∙

∙

Page 3-7





3-3 DOMAIN, RANGE, AND RELATION

Relation

• Relation: a set of ordered pairs (x, y) .

• Domain: the set of the values of the independent variable (x-value) Example

for which a function is defined . (3, 2)

• Range: the set of the values of the dependent variable (y-value) domain range

for which a function is defined .

Example Domain Range Ordered Pair

2 - can of Coke $1 .50 (2-Coke, $1 .50)1 - can of juice $1 .25 (1-juice, $1 .25)3 - can of soup $3 .00 (3-soup, $3 .00)

domain range (The range depends on the domain .)

• Correspondence diagram: an arrow points from each domain to the range .

Example: Name (x) Age (y)

Diane 25Susan 23Ann 18TomDomain Range

Example: Express the relation R = {(2, 4) (-1, 3) (5, -2) (-4, -3)} as a table, correspondence diagram, and domain/range set .

x y2 4-1 35 -2-4 -3

x y 2 4

-1 35 -2-4 -3

Domain Range

- Domain: {2, -1, 5, -4}Range: {4, 3, -2, -3}

- Table:

- Correspondence diagram:

Page 3-8

3-3 DOMAIN, RANGE, AND RELATION

Relation


• Domain: the set of the values of the independent variable (x-value) Example

for which a function is defined . (3, 2)

• Range: the set of the values of the dependent variable (y-value) domain range

for which a function is defined .

Example Domain Range Ordered Pair

2 - can of Coke $1 .50 (2-Coke, $1 .50)1 - can of juice $1 .25 (1-juice, $1 .25)3 - can of soup $3 .00 (3-soup, $3 .00)

domain range (The range depends on the domain .)

• Correspondence diagram: an arrow points from each domain to the range .

Example: Name (x) Age (y)

Diane 25Susan 23Ann 18TomDomain Range

Example: Express the relation R = {(2, 4) (-1, 3) (5, -2) (-4, -3)} as a table, correspondence diagram, and domain/range set .

x y2 4-1 35 -2-4 -3

x y 2 4

-1 35 -2-4 -3

Domain Range

- Domain: {2, -1, 5, -4}Range: {4, 3, -2, -3}

- Table:

- Correspondence diagram:

Page 3-8





Finding Domain and Range

Example: Answer the following questions regarding each graph (a function) below .

a. f (-2), b. the domain, c. all x-values such that f (x) = 1, and d. the range .

1.

f (-2) Domain (x-values) All x-values such that f (x) = 1 Range (y-values)1 {-2, -1, 1} When y = 1 , x = -2 and 1 {1, 3}

2.

f (-2) Domain (x-values) All x-values such that f (x) = 1 Range (y-values)-1 {-2, 3} When y = 1 , x = 0 {-1, 2}

Example: Identify the domain of the following functions .

f (x) Domain (x-values) Comments

5 – 3x all real numbers

Since any real number can be used to calculate y . x y0 51 2

… …5

3 − 𝑥𝑥

{x | x is a real number and x ≠ 3}or (-∞, 3) ∪ (3, ∞)

Find out what values of x “do not work”If x = 3, 5

3−3= 5

0 is undefined .

3|3𝑥𝑥 − 2|

{x | x is a real number and x ≠ 23}

or (-∞, 23 ) ∪ ( 2

3 , ∞)

If x = 23 , 3

�3�23�−2�= 3

0 is undefined .

Note: The domain x is the set of real numbers that will yield a real number for the range y .(The set of the x-values for which a function is defined .)

(3, 2)(-1, 3) ∙

f (x) = y

x ∙ (1, 1) ∙ (-2, 1)

(-2, -1) ∙

y

x

∙ (3, 2)

Open dot ○ : the point is not included .

Page 3-9

Finding Domain and Range

Example: Answer the following questions regarding each graph (a function) below .

a. f (-2), b. the domain, c. all x-values such that f (x) = 1, and d. the range .

1.

f (-2) Domain (x-values) All x-values such that f (x) = 1 Range (y-values)1 {-2, -1, 1} When y = 1 , x = -2 and 1 {1, 3}

2.

f (-2) Domain (x-values) All x-values such that f (x) = 1 Range (y-values)-1 {-2, 3} When y = 1 , x = 0 {-1, 2}

Example: Identify the domain of the following functions .

f (x) Domain (x-values) Comments

5 – 3x all real numbers

Since any real number can be used to calculate y . x y0 51 2

… …5

3 − 𝑥𝑥

{x | x is a real number and x ≠ 3}or (-∞, 3) ∪ (3, ∞)

Find out what values of x “do not work”If x = 3, 5

3−3= 5

0 is undefined .

3|3𝑥𝑥 − 2|

{x | x is a real number and x ≠ 23}

or (-∞, 23 ) ∪ ( 2

3 , ∞)

If x = 23 , 3

�3�23�−2�= 3

0 is undefined .

Note: The domain x is the set of real numbers that will yield a real number for the range y .(The set of the x-values for which a function is defined .)

(3, 2)(-1, 3) ∙

f (x) = y

x ∙ (1, 1) ∙ (-2, 1)

(-2, -1) ∙

y

x

∙ (3, 2)

Open dot ○ : the point is not included .

Page 3-9

1





3-4 LINEAR FUNCTIONS

Slope-Intercept Function of a Line

• Slope-intercept form of a linear function Slope-Intercept Function of a Line

f (x) = m x + b 𝑚𝑚: the slope of the line 𝑏𝑏: y-intercept

b

• The y intercept: the point at which the line crosses the y axis . b = (0, y)

Example: Identify the slope and y-intercept of the following equations .

1. f (x) = -0.3x – 5 f (x) = m x + b

The slope: m = -0.3

y-intercept: b = -5 or (0, -5)

2. 2x + 3y = 4 – x – 4y Combine like terms .

7y = -3x + 4 Divide both sides by 7 .

𝑦𝑦 = - 37𝑥𝑥 + 4

7f (x) = m x + b

The slope: 𝒎𝒎 = -𝟑𝟑𝟕𝟕

y-intercept: b = 𝟒𝟒𝟕𝟕

or (0, 𝟒𝟒𝟕𝟕�

3. 𝟑𝟑𝟑𝟑 + 𝟏𝟏𝟐𝟐𝒚𝒚 = 𝟖𝟖

3𝑥𝑥 ∙ 2 + 12

y ∙ 2 = 8 ∙ 2 Multiply 2 for each term .

6x + y = 16

y = -6x + 16 f (x) = m x + b

The slope: m = -6

y-intercept: b = 16 or (0, 16)

∙

y = f (x)

x

Page 3-10

3-4 LINEAR FUNCTIONS

Slope-Intercept Function of a Line


f (x) = m x + b 𝑚𝑚: the slope of the line 𝑏𝑏: y-intercept

b

• The y intercept: the point at which the line crosses the y axis . b = (0, y)

Example: Identify the slope and y-intercept of the following equations .

1. f (x) = -0.3x – 5 f (x) = m x + b

The slope: m = -0.3

y-intercept: b = -5 or (0, -5)

2. 2x + 3y = 4 – x – 4y Combine like terms .

7y = -3x + 4 Divide both sides by 7 .

𝑦𝑦 = - 37𝑥𝑥 + 4

7f (x) = m x + b

The slope: 𝒎𝒎 = -𝟑𝟑𝟕𝟕

y-intercept: b = 𝟒𝟒𝟕𝟕

or (0, 𝟒𝟒𝟕𝟕�

3. 𝟑𝟑𝟑𝟑 + 𝟏𝟏𝟐𝟐𝒚𝒚 = 𝟖𝟖

3𝑥𝑥 ∙ 2 + 12

y ∙ 2 = 8 ∙ 2 Multiply 2 for each term .

6x + y = 16

y = -6x + 16 f (x) = m x + b

The slope: m = -6

y-intercept: b = 16 or (0, 16)

∙

y = f (x)

x

Page 3-10





Slope

• Recall: The graph of a linear equation is a straight line .

• Slope (m): The slope of a straight line is the rate of change . It is a measure of the

“steepness” or incline of the line and indicates whether the line rises or falls .

• A line with a positive slope rises from left to right and a line with a negative slope falls.

• The slope formula

The Slope Formula

slope =the change in 𝑦𝑦the change in 𝑥𝑥

=riserun

The slope of the straight line that passes through two points (x1, y1) and (x2, y2):

m = 𝑦𝑦2− 𝑦𝑦1𝑥𝑥2− 𝑥𝑥1

or m = 𝑦𝑦1− 𝑦𝑦2𝑥𝑥1− 𝑥𝑥2

x1 ≠ x2

Example: Determine the slope containing points (2, -1) and (1, 3) .

m = 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

= 3 − (-1)1−2

= 4- 1

= - 4

or m = 𝑦𝑦1−𝑦𝑦2𝑥𝑥1−𝑥𝑥2

= -1−32−1

= - 41

= - 4

Example: Determine the slope of 5x – y – 7 = 0 .

x y = 5x – 7 (x, y) 0 -7 (x1, y1) = (0, -7) 1 -2 (x2, y2) = (1, -2)

Choose Calculate


= -2 − (-7)1− 0

= 51

= 5 or m = 𝑦𝑦1− 𝑦𝑦2𝑥𝑥1− 𝑥𝑥2

= -7− (-2)0−1

= - 5- 1

= 5

Other points on the line will obtain the same slope m .

x y = 5x – 7 (x, y)2 3 (2, 3)3 8 (3, 8)

Choose Calculate

m = 8−33−2

= 51

= 5

x

y

Change in x

Change in yrun

rise ∙

∙(x2, y2)

(x1, y1)

Page 3-11

Slope

• Recall: The graph of a linear equation is a straight line .

• Slope (m): The slope of a straight line is the rate of change . It is a measure of the


• A line with a positive slope rises from left to right and a line with a negative slope falls.


The Slope Formula


=riserun



or m = 𝑦𝑦1− 𝑦𝑦2𝑥𝑥1− 𝑥𝑥2

x1 ≠ x2

Example: Determine the slope containing points (2, -1) and (1, 3) .


= 3 − (-1)1−2

= 4- 1

= - 4


= -1−32−1

= - 41

= - 4

Example: Determine the slope of 5x – y – 7 = 0 .

x y = 5x – 7 (x, y) 0 -7 (x1, y1) = (0, -7) 1 -2 (x2, y2) = (1, -2)

Choose Calculate


= -2 − (-7)1− 0

= 51

= 5 or m = 𝑦𝑦1− 𝑦𝑦2𝑥𝑥1− 𝑥𝑥2

= -7− (-2)0−1

= - 5- 1

= 5

Other points on the line will obtain the same slope m .

x y = 5x – 7 (x, y)2 3 (2, 3)3 8 (3, 8)

Choose Calculate

m = 8−33−2

= 51

= 5

x

y

Change in x

Change in yrun

rise ∙

∙(x2, y2)

(x1, y1)

Page 3-11





Example: Identify the slope (or rate of change) .


= 2 − �-1�3 − 0

= 33

= 1

Example: Identify the slope .


= 34

= 𝟎𝟎.𝟕𝟕𝟕𝟕

Example: Tom purchased a car for $23,000 in 2007 . The car was worth $19,000 in 2011 .

Find the average annual rate of change .

Year x y (x, y) Assuming2007 7 $23,000 (7, 23,000) 2000 = 0 , 2001 = 1, … 2007 = 72011 11 $19,000 (11, 19,000) 2011 = 11


= 19,000−23,00011−7

= - 4,0004

= - 1,000

The result means that the value of Tom’s car decreased by $1,000 per year .

x (year)

y (amount $1,000)

∙∙

(11, 19,000)19

(7, 23,000)

7 11

23

x

y

∙ (3, 2)

∙ (0, -1)

3m

4m

Page 3-12





3-5 GRAPHING LINEAR EQUATIONS

Graphing Linear Equations Using the Intercept Method

• Recall: The x-intercept is the point at which the line crosses the x-axis . (x, 0)

The y-intercept is the point at which the line crosses the y-axis . (0, y)

• Procedures to graph a linear equation using the intercept method

Steps Example: 2x – y = 3

- Choose x = 0 and calculate the corresponding y . x y = 2x – 3 (x, y) Intercept j 0 -3 (0, -3) y-intercept

- Choose y = 0 and calculate the corresponding x. 1 .5 0 (1 .5, 0) x-intercept

- Plot these two points on the coordinate plane .

- Connect the points with a straight line .

- Check with the third point .

Is third point (-1, -5) on the line? Yes . Correct!

x y = 2x – 3 (x, y) j -1 -5 (-1, -5)

Example: Graph the equation x = 3y .

- Choose x = 0 and calculate the x y = 𝐱𝐱𝟑𝟑 (x, y) j

corresponding y . 0 03

= 0 (0, 0)

- Choose x = 3 and calculate the 3 33

= 1 (3, 1)corresponding y .

- Plot (0, 0) and (3, 1) .

- Connect two points with a straight line .

- Check (use the third point) . x y = 𝐱𝐱𝟑𝟑 (x, y) j

Is third point (6, 2) on the line? Yes . Correct! 6 2 (6, 2)

∙ (0, -3)

∙ (1 .5, 0)

∙ (-1, -5)

∙ (6, 2) ∙ (3, 1)

x

y

∙ (0, 0)

Page 3-13

3-5 GRAPHING LINEAR EQUATIONS

Graphing Linear Equations Using the Intercept Method

• Recall: The x-intercept is the point at which the line crosses the x-axis . (x, 0)

The y-intercept is the point at which the line crosses the y-axis . (0, y)

• Procedures to graph a linear equation using the intercept method

Steps Example: 2x – y = 3

- Choose x = 0 and calculate the corresponding y . x y = 2x – 3 (x, y) Intercept j 0 -3 (0, -3) y-intercept

- Choose y = 0 and calculate the corresponding x. 1 .5 0 (1 .5, 0) x-intercept

- Plot these two points on the coordinate plane .

- Connect the points with a straight line .

- Check with the third point .

Is third point (-1, -5) on the line? Yes . Correct!

x y = 2x – 3 (x, y) j -1 -5 (-1, -5)

Example: Graph the equation x = 3y .

- Choose x = 0 and calculate the x y = 𝐱𝐱𝟑𝟑 (x, y) j

corresponding y . 0 03

= 0 (0, 0)

- Choose x = 3 and calculate the 3 33

= 1 (3, 1)corresponding y .

- Plot (0, 0) and (3, 1) .

- Connect two points with a straight line .

- Check (use the third point) . x y = 𝐱𝐱𝟑𝟑 (x, y) j

Is third point (6, 2) on the line? Yes . Correct! 6 2 (6, 2)

∙ (0, -3)

∙ (1 .5, 0)

∙ (-1, -5)

∙ (6, 2) ∙ (3, 1)

x

y

∙ (0, 0)

Page 3-13





Graphing Using the Slope and the y - Intercept

• Recall: Slope-intercept function: f (x) = 𝑚𝑚x + b

� 𝑚𝑚 = slope

b = 𝑦𝑦-intercept

• The slope and a point can determine a straight line .

Example: Graph the function using the slope and the y-intercept . f (x) = - 𝟑𝟑𝟓𝟓

x + 4

- Plot the y-intercept (0, 4) . The change in y: the rise (move 3 units down, ∵ y is negative) .

- Determine the rise and run: m = - 35

The change in x: the run (move 5 units to the right, ∵ x is positive) .

- Plot another point by moving 3 units down and 5 units to the right .

- Connect the two points with a line . Starting point: y-intercept

Example: Graph the function using the slope and the y-intercept . -6x + 2 ∙ f (x) = -10

- Convert to the slope-intercept form . 2 ∙ f (x) = 6x – 10 Add 6x to both sides .

f (x) = 3x – 5 Divide both sides by 2 .

- y-intercept: (0, -5) f (x) = 𝑚𝑚x + b

- Slope: 𝑚𝑚 = 3 = 31

Move 3 units up and 1 unit to the right (both x & y are positive) .

• Tip: m = riserun

= change in 𝑦𝑦change in 𝑥𝑥

⎩⎪⎨

⎪⎧+𝑦𝑦: move up

- 𝑦𝑦: move down +𝑥𝑥: move to the right- 𝑥𝑥: move to the left

x

f (x)

∙ (5, 1)

∙ (0, 4) Starting point

Ending point

4

∙

∙ - 5

x

f (x)

0

1

5

+x

+y

-x

- y

0

Starting point

Ending point

Page 3-14

Graphing Using the Slope and the y - Intercept

• Recall: Slope-intercept function: f (x) = 𝑚𝑚x + b

� 𝑚𝑚 = slope

b = 𝑦𝑦-intercept

• The slope and a point can determine a straight line .

Example: Graph the function using the slope and the y-intercept . f (x) = - 𝟑𝟑𝟓𝟓

x + 4

- Plot the y-intercept (0, 4) . The change in y: the rise (move 3 units down, ∵ y is negative) .

- Determine the rise and run: m = - 35

The change in x: the run (move 5 units to the right, ∵ x is positive) .

- Plot another point by moving 3 units down and 5 units to the right .

- Connect the two points with a line . Starting point: y-intercept

Example: Graph the function using the slope and the y-intercept . -6x + 2 ∙ f (x) = -10

- Convert to the slope-intercept form . 2 ∙ f (x) = 6x – 10 Add 6x to both sides .

f (x) = 3x – 5 Divide both sides by 2 .

- y-intercept: (0, -5) f (x) = 𝑚𝑚x + b

- Slope: 𝑚𝑚 = 3 = 31

Move 3 units up and 1 unit to the right (both x & y are positive) .

• Tip: m = riserun

= change in 𝑦𝑦change in 𝑥𝑥

⎩⎪⎨

⎪⎧+𝑦𝑦: move up

- 𝑦𝑦: move down +𝑥𝑥: move to the right- 𝑥𝑥: move to the left

x

f (x)

∙ (5, 1)

∙ (0, 4) Starting point

Ending point

4

∙

∙ - 5

x

f (x)

0

1

5

+x

+y

-x

- y

0

Starting point

Ending point

Page 3-14





Vertical and Horizontal Lines

• Horizontal line: a line that is parallel to the x-axis . It has a slope of 0 and a

y-intercept (0, b), or y = b .

Example: y = -3x y (x, y)1 -3 (1, -3)4 -3 (4, -3)


= -3 − (- 3)4−1

= 03

= 0

• Vertical line: a line that is parallel to the y-axis . It has an infinite slope with an

x-intercept (a, 0), or x = a .

Example: x = -1

x y (x, y)-1 3 (-1, 3)-1 -1 (-1, -1)


= -1−3-1− (-1)

= - 40

= ∞ Undefined

• Horizontal and vertical lineLine Equation m Example Graph

horizontal line y = b 0 y = 3

vertical line x = a ∞ x = -2

Example: Graph the function 2 + 5 f (x) = 12 and determine the slope .

5 f (x) = 10 Isolate f (x) .

f (x) = 2 or y = 2

Slope: m = 0 Horizontal line

x

y

∙ (4, -3)

y = -3

y

x = -1

∙(-1, 3)

∙(1,- 3)

3

-2

x

f(x)

y = 2

0

0

0

0

x ∙(-1, -1) 0

Page 3-15

Vertical and Horizontal Lines

• Horizontal line: a line that is parallel to the x-axis . It has a slope of 0 and a

y-intercept (0, b), or y = b .

Example: y = -3x y (x, y)1 -3 (1, -3)4 -3 (4, -3)


= -3 − (- 3)4−1

= 03

= 0

• Vertical line: a line that is parallel to the y-axis . It has an infinite slope with an

x-intercept (a, 0), or x = a .

Example: x = -1

x y (x, y)-1 3 (-1, 3)-1 -1 (-1, -1)


= -1−3-1− (-1)

= - 40

= ∞ Undefined

• Horizontal and vertical lineLine Equation m Example Graph



Example: Graph the function 2 + 5 f (x) = 12 and determine the slope .

5 f (x) = 10 Isolate f (x) .

f (x) = 2 or y = 2

Slope: m = 0 Horizontal line

x

y

∙ (4, -3)

y = -3

y

x = -1

∙(-1, 3)

∙(1,- 3)

3

-2

x

f(x)

y = 2

0

0

0

0

x ∙(-1, -1) 0

Page 3-15





Page 3-16

Perpendicular and Parallel Lines

• Parallel lines are always the same distance from each other and they will never intersect.

Two parallel lines have the same slope m1 = m2 .

L1 ∥ L2 L1 L2

• Perpendicular lines intersect to form a 90-degree angle, and they have negative reciprocal

slopes.

m1 = - 1𝑚𝑚𝑚𝑚2

L1 ⊥ L2 900

• Parallel and perpendicular lines

Line Slope Two parallel lines ( ∥ ) m1 = m2

Two perpendicular lines (⊥) m1 = - 1𝑚𝑚𝑚𝑚2

Example: Determine if the graphs of two straight line equations are parallel or

perpendicular.

1. 5y + 2x = 1 and 3 – 4x = 10y

5y = -2x + 1 10y = -4x + 3 Convert to f (x) = mx + b.

y = - 𝟐𝟐𝟐𝟐𝟓𝟓𝟓𝟓 x + 1

5 y = - 𝟐𝟐𝟐𝟐

𝟓𝟓𝟓𝟓 x + 3

10

m1 = - 25 m2 = - 2

5

m1 = m2 =-25

, L1 ∥ L2

2. 3x = 8 + y and 3y + x + 4 = 0

y = 3x – 8 3y = - x – 4 Convert to f (x) = mx + b.

y = 𝟑𝟑𝟑𝟑 ∙ x – 8 y = - 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑 x – 4

3

m1 = 3 m2 = - 13

m1 = - 1𝑚𝑚𝑚𝑚2

, L1 ⊥ L2

Perpendicular symbol

Parallel symbol

∟

Page 3-16

Perpendicular and Parallel Lines

• Parallel lines are always the same distance from each other and they will never intersect.

Two parallel lines have the same slope m1 = m2 .

L1 ∥ L2 L1 L2

• Perpendicular lines intersect to form a 90-degree angle, and they have negative reciprocal

slopes.

m1 = - 1𝑚𝑚𝑚𝑚2

L1 ⊥ L2 900


Line Slope Two parallel lines ( ∥ ) m1 = m2

Two perpendicular lines (⊥) m1 = - 1𝑚𝑚𝑚𝑚2

Example: Determine if the graphs of two straight line equations are parallel or

perpendicular.

1. 5y + 2x = 1 and 3 – 4x = 10y

5y = -2x + 1 10y = -4x + 3 Convert to f (x) = mx + b.

y = - 𝟐𝟐𝟐𝟐𝟓𝟓𝟓𝟓 x + 1

5 y = - 𝟐𝟐𝟐𝟐

𝟓𝟓𝟓𝟓 x + 3

10

m1 = - 25 m2 = - 2

5

m1 = m2 =-25

, L1 ∥ L2

2. 3x = 8 + y and 3y + x + 4 = 0

y = 3x – 8 3y = - x – 4 Convert to f (x) = mx + b.

y = 𝟑𝟑𝟑𝟑 ∙ x – 8 y = - 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑 x – 4

3

m1 = 3 m2 = - 13

m1 = - 1𝑚𝑚𝑚𝑚2

, L1 ⊥ L2

Perpendicular symbol

Parallel symbol

∟





3-6 STRAIGHT LINE EQUATIONS

Point-Slope Equation of a Line

• Point-slope equation of a straight line

Point-Slope Equationy – y1 = m (x – x1) 𝑚𝑚− the slope of the line

(𝑥𝑥1, 𝑦𝑦1) − the given point on the line (𝑥𝑥, 𝑦𝑦) − any other point on the line

(𝑥𝑥, 𝑦𝑦)

(𝑥𝑥1,𝑦𝑦1)

• Derive: from the slope formula m = 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

, let x2 = x and y2 = y .

then m = 𝑦𝑦−𝑦𝑦1𝑥𝑥−𝑥𝑥1

Replace (x2 , y2) by (x, y) .

m (x – x1) = 𝑦𝑦−𝑦𝑦1𝑥𝑥−𝑥𝑥1

(𝑥𝑥 − 𝑥𝑥1) Multiply both sides by x – x1 .

point-slope equation: y – y1 = m (x – x1)

Example: Graph the line with slope 13

that passes through the point (2, 3) . Write an

equation in point-slope form .

- Slope and point: 𝑚𝑚 = 13

, (x1, y1) = (2, 3)

- Equation: y – 3 = 13

(x – 2) Point-slope equation: y – y1 = m (x – x1) .

Substitute y1 = 3, x1 = 2 and m = 13 .

- Graph: 𝑚𝑚 = 13

The change in y (move 1 unit up) .

The change in x (move 3 units to the right) .

∙

∙

x

y

(5, 4)(2, 3)

∙ ∙

x

y

4

2 5

3

y – 3 = 13

(x – 2)

Page 3-17

3-6 STRAIGHT LINE EQUATIONS

Point-Slope Equation of a Line

• Point-slope equation of a straight line

Point-Slope Equationy – y1 = m (x – x1) 𝑚𝑚− the slope of the line

(𝑥𝑥1, 𝑦𝑦1) − the given point on the line (𝑥𝑥, 𝑦𝑦) − any other point on the line

(𝑥𝑥, 𝑦𝑦)

(𝑥𝑥1,𝑦𝑦1)

• Derive: from the slope formula m = 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

, let x2 = x and y2 = y .

then m = 𝑦𝑦−𝑦𝑦1𝑥𝑥−𝑥𝑥1

Replace (x2 , y2) by (x, y) .

m (x – x1) = 𝑦𝑦−𝑦𝑦1𝑥𝑥−𝑥𝑥1

(𝑥𝑥 − 𝑥𝑥1) Multiply both sides by x – x1 .

point-slope equation: y – y1 = m (x – x1)

Example: Graph the line with slope 13

that passes through the point (2, 3) . Write an

equation in point-slope form .

- Slope and point: 𝑚𝑚 = 13

, (x1, y1) = (2, 3)

- Equation: y – 3 = 13

(x – 2) Point-slope equation: y – y1 = m (x – x1) .

Substitute y1 = 3, x1 = 2 and m = 13 .

- Graph: 𝑚𝑚 = 13

The change in y (move 1 unit up) .

The change in x (move 3 units to the right) .

∙

∙

x

y

(5, 4)(2, 3)

∙ ∙

x

y

4

2 5

3

y – 3 = 13

(x – 2)

Page 3-17





Finding an Equation of a Line

Straight-Line Equation Equation Examplegeneral form Ax + By = C 2x + y = 3 A = 2, B = 1, C = 3point-slope form y – y1 = m (x – x1) y – 1 = -2 (x + 5) m = -2 y1 = 1, x1 = -5

slope-intercept form y = mx + b y = 8x – 27

m = 8 , b = - 27

• Finding an equation of a line when the slope and the y-intercept are given

Example: Graph the line with slope -3 and y-intercept 5 and write the slope intercept

equation .

y = m x + b m = -3, b = 5

y = -3x + 5 𝑚𝑚 = -3 = - 31

Move 3 units down and 1 unit to the right .

• Finding an equation of a line when the slope and a point are given

Example: Write an equation of the line passing the point (3, 2) with slope m = -2 .

Start with: y = mx + b Slope-intercept equation

Solve for b: 2 = -2 ∙ 3 + b Replace (x , y) by (3, 2) & m by -2 .

b = 8 Equation of the line: y = -2x + 8 m = -2 , b = 8

• Finding an equation of a line when two points are given

Example: Write an equation of the line that passes through the points (1, 1) and (5, -7) .

The slope: m = 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

= -7−15−1

= - 84

= -2 Substitute (x1 , y1) = (1, 1)

(x2 , y2) = (5, -7) .


Solve for b: 1 = -2 ∙ 1 + b Replace (x , y) by (1, 1) & m by – 2 .

b = 3 Use (x , y) = (5, -7) will get the same result .

Equation of the line: y = -2x + 3 m = -2 , b = 3

2

∙

∙

5y

Page 3-18

Finding an Equation of a Line



m = 8 , b = - 27

• Finding an equation of a line when the slope and the y-intercept are given

Example: Graph the line with slope -3 and y-intercept 5 and write the slope intercept

equation .

y = m x + b m = -3, b = 5

y = -3x + 5 𝑚𝑚 = -3 = - 31

Move 3 units down and 1 unit to the right .

• Finding an equation of a line when the slope and a point are given

Example: Write an equation of the line passing the point (3, 2) with slope m = -2 .


Solve for b: 2 = -2 ∙ 3 + b Replace (x , y) by (3, 2) & m by -2 .

b = 8 Equation of the line: y = -2x + 8 m = -2 , b = 8

• Finding an equation of a line when two points are given

Example: Write an equation of the line that passes through the points (1, 1) and (5, -7) .

The slope: m = 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

= -7−15−1

= - 84

= -2 Substitute (x1 , y1) = (1, 1)

(x2 , y2) = (5, -7) .


Solve for b: 1 = -2 ∙ 1 + b Replace (x , y) by (1, 1) & m by – 2 .

b = 3 Use (x , y) = (5, -7) will get the same result .

Equation of the line: y = -2x + 3 m = -2 , b = 3

2

∙

∙

5y

Page 3-18

0





Example: Write an equation of the line passing through the points .

1. (-5, 3) (4, 3) and 2. (2, -1) (2, 3) . Solution:

1. m = 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

= 3−34−(-5)

= 09

= 0 Let (x1, y1) = (-5, 3) , (x2 , y2) = (4, 3) .

y = m𝑥𝑥 + b y = 0 · 𝑥𝑥 + b y = b y = 3 (horizontal line)

2. m = 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

= 3−(-1)2−2

= 40

= ∞ undefined Let (x1, y1) = (2, -1) , (x2 , y2) = (2, 3) .

x = 2 (Vertical line)

(2, -1) (2, 3)

• Find an equation of the line passing through a point and is parallel or perpendicular to a given line

Example: Write an equation of the line passing through the point (1, -4) and is:

1. parallel to and 2. perpendicular to the line 2x + 4y – 8 = 0 .

Solution: 2x + 4y = 8 Add 8 on both sides .

- Convert to slope-intercept form . 4y = -2x + 8 Subtract 2x from both sides .

y = - 1 2

x + 2 Divide 4 on both sides .

- Determine the slope for the line 1 . m1 = - 1 2

y = mx + b

1. A line L2 is parallel to the line L1 (2x + 4y – 8 = 0) and has a slope of m2 = - 1 2

.Parallel：m1 = m2

- Start with: y = mx + b

- Solve for b: -4 = - 1 2∙ 1 + b Replace (x , y) by (1, -4) & m by - 1

2 .

-8 = - 1 + 2b Multiply 2 for each term .

b = - 7 2

- Equation of the line: y = - 𝟏𝟏𝟐𝟐x – 𝟕𝟕

𝟐𝟐L2 ∥ L1

y = 3x

y

y

x

x = 2

0

0

Page 3-19





Page 3-20

2. A line L2 is perpendicular to the line L1 (2x + 4y – 8 = 0) and has a slope of m2 = - 1𝑚𝑚𝑚𝑚1

,

or m2 = -1𝑚𝑚𝑚𝑚1

= �-1� -12

= 2 𝑚𝑚𝑚𝑚1 = −12

- Start with: y = mx + b Replace (x , y) by (1, -4) & m by 2.

- Solve for b: -4 = 2 ∙ 1 + b , b = -6

- Equation of the line: y = 2x – 𝟔𝟔𝟔𝟔 m = 2 , b = -6 , L2 ⊥ L1

• Applications

Example: Tom bought a laptop for $1,000. The value of the laptop decreases at a rate of

$100 per year. Write an equation for the value f (t) of the laptop after t years.

Graph the equation and determine the value of the laptop after 4 years.

- Equation: f (t) = 1,000 – 100t y = mx + b , f (t) = -100t + 1,000

- Solve algebraically: f (4) = 1,000 – 100t t = 4 years

= 1,000 – 100 (4) = $ 600

- Solve graphically: t f (t) =1,000 – 100t 0 $ 1,000 2 $ 800

The laptop will be worth $600 after 4 years.

t (year)

f (t) cost $

2

1,000

0

∙ 800

4

∙

500 600





Unit 3 Summary

• Ordered pair (x, y): each point on the plane corresponds to an ordered pair .(x , y)

1st coordinate (abscissa) 2nd coordinate (ordinate)

• Four quadrantsQuadrant (x, y) Example

The 1st quadrant I (+x, +y) (+2, +3)The 2nd quadrant II (-x, +y) (-2, +3) The 3rd quadrant III (-x, -y) (-2, -3) The 4th quadrant IV (+x, -y) (+2, -3)

• x-intercept (x, 0): the point at which the graph crosses the x-axis . ∙ (0, y)

• y-intercept (0, y): the point at which the graph crosses the y-axis .

• A linear (first-degree) equation in two variables: a linear equation that contains two variables . (A linear equation whose graph is a straight line .)

Standard Form ExampleAx + By = C 5x – 7y = 4


• Procedure to graph a nonlinear equation (or function) with two variables:

- Choose a few values of x, calculate the corresponding y, and make a table . - Plot these points on the coordinate plane (plot more points to get the cleaner shape of

the graph) .- Connect the points with a smooth curve .


the domain with exactly one element of the range .


• Domain: the set of the values of the independent variable (x-value) for which a function is

defined .

• Range: the set of the values of the dependent variable (y-value) for which a function is defined .


• Function notation: the notation for a function is f (x), P(x), g(x), h(x) , …

• Function values:

f (x) | x = a = f (a) , �𝑎𝑎 is a constant replace 𝑥𝑥 by 𝑎𝑎

II

III IV

I x

x∙ (x, 0)

y

y

Page 3-21

Unit 3 Summary

• Ordered pair (x, y): each point on the plane corresponds to an ordered pair .(x , y)

1st coordinate (abscissa) 2nd coordinate (ordinate)

• Four quadrantsQuadrant (x, y) Example

The 1st quadrant I (+x, +y) (+2, +3)The 2nd quadrant II (-x, +y) (-2, +3) The 3rd quadrant III (-x, -y) (-2, -3) The 4th quadrant IV (+x, -y) (+2, -3)

• x-intercept (x, 0): the point at which the graph crosses the x-axis . ∙ (0, y)

• y-intercept (0, y): the point at which the graph crosses the y-axis .

• A linear (first-degree) equation in two variables: a linear equation that contains two variables . (A linear equation whose graph is a straight line .)

Standard Form ExampleAx + By = C 5x – 7y = 4


• Procedure to graph a nonlinear equation (or function) with two variables:

- Choose a few values of x, calculate the corresponding y, and make a table . - Plot these points on the coordinate plane (plot more points to get the cleaner shape of

the graph) .- Connect the points with a smooth curve .


the domain with exactly one element of the range .


• Domain: the set of the values of the independent variable (x-value) for which a function is

defined .

• Range: the set of the values of the dependent variable (y-value) for which a function is defined .


• Function notation: the notation for a function is f (x), P(x), g(x), h(x) , …

• Function values:

f (x) | x = a = f (a) , �𝑎𝑎 is a constant replace 𝑥𝑥 by 𝑎𝑎

II

III IV

I x

x∙ (x, 0)

y

y

Page 3-21








f (x) = m x + b 𝑚𝑚: 𝑡𝑡he slope of the line 𝑏𝑏: y-intercept

• Slope (m): the slope of a straight line is the rate of change . It is a measure of the



The Slope Formula


=riserun




x1 ≠ x2

• Horizontal and vertical linesLine Equation m Example Graph




Line Slopetwo lines are parallel (∥) m1 = m2

two lines are perpendicular ( ⊥ ) m1 = - 1𝑚𝑚2

• Equations of the straight lines



m = 8 , b = - 27

3

-2

0

0

Page 3-22





PRACTICE QUIZ

Unit 3 Functions and Graphs

1. Graph the following equations .

a. 3x – y = 2

b. y = |𝑥𝑥 + 2|

2. Evaluate the functions at the indicated values .

a. f (-3) for f (x) = 7 + 5x2

b. q (0) for q (r) = 5r2 + 2r – 1

3 . Evaluate the functions and simplify at the indicated values .

a. f (x) = 5x – 3 , f (a – 2) = ?

b. h (x) = |𝑥𝑥 − 2| + 3𝑥𝑥, h (5) = ?

4. The following graph shows the number of bicycle rentals as a function of

time in a rental store . How many bicycle rentals were there in 2012?

5. Given the relation: (3, 4), (-1, 6), (6, 3), (-4, 3)

a. Identify the domain .

b . Identify the range .

6. For the following functions, identify their domains .

a. 𝑓𝑓(𝑥𝑥) = 13

5−𝑥𝑥

b. 𝑓𝑓(𝑥𝑥) = 6

|7𝑥𝑥−5|

100 ∙

∙ 2000

∙ 2010

50 ∙x (year)

# of bicycles rentals f(x)

Page 5

PRACTICE QUIZ

Unit 3 Functions and Graphs

1. Graph the following equations .

a. 3x – y = 2

b. y = |𝑥𝑥 + 2|

2. Evaluate the functions at the indicated values .

a. f (-3) for f (x) = 7 + 5x2

b. q (0) for q (r) = 5r2 + 2r – 1

3 . Evaluate the functions and simplify at the indicated values .

a. f (x) = 5x – 3 , f (a – 2) = ?

b. h (x) = |𝑥𝑥 − 2| + 3𝑥𝑥, h (5) = ?

4. The following graph shows the number of bicycle rentals as a function of

time in a rental store . How many bicycle rentals were there in 2012?

5. Given the relation: (3, 4), (-1, 6), (6, 3), (-4, 3)

a. Identify the domain .

b . Identify the range .

6. For the following functions, identify their domains .

a. 𝑓𝑓(𝑥𝑥) = 13

5−𝑥𝑥

b. 𝑓𝑓(𝑥𝑥) = 6

|7𝑥𝑥−5|

100 ∙

∙ 2000

∙ 2010

50 ∙x (year)

# of bicycles rentals f(x)

Page 5





7. a . Identify the slope and y-intercept of 2𝑥𝑥 − 13𝑦𝑦 = 5 .

b. Identify the slope of the line .

8. Mary purchased a laptop for $1,000 in 2008 . The laptop was worth $500 in

2012 . Find the average annual rate of change .

9. Graph using the slope and the y-intercept . -4x + 2 f (x) = -14

10 . Determine if the graphs of two straight line equations are parallel or

perpendicular .

2𝑦𝑦 + 7x = 1 and 3 – 14x = 4y

11 . Write an equation of the line that passes through the points (2, 3) and (3, -4) .

12. Write an equation of the line passing through the point (2, -3) and

a. parallel to and b. perpendicular to the line 9y = -3x + 1 .

13. Sam bought a car for $20,000 . The value of the car decreases at a rate of

$1,000 per year . Write an equation for the value f (t) of the car after t years .

Graph the equation and determine the value of the car after 5 years .

y

∙ (4, 1) x

∙ (0, -1)

Page 6




Algebra I & II Key Concepts, Practice, and Quizzes Unit 4 – Systems of Equations and Inequalities

UNIT 4 SYSTEMS OF EQUATIONS & INEQUALITIES

4-1 SYSTEMS OF EQUATIONS

A System of Equations

• A system of equations is a group of two or more equations with the same variables

(unknowns) .

Example: � 2𝑥𝑥2 + 3𝑦𝑦 = 24𝑥𝑥2 – 5𝑦𝑦 = 7

• A system of linear equations is a group of two or more first-degree equations .

(First-degree equation: The highest power of the variable is one .)

• A system of two linear equations in two variables: two linear equations in two unknowns .

Example: � 𝑥𝑥 − 𝑦𝑦 = 2 2𝑥𝑥 + 𝑦𝑦 = 13

• A system of linear equations

Standard Form Example 2×2 system

2 equations 2 unknowns�𝐴𝐴1𝑥𝑥 + 𝐵𝐵1𝑦𝑦 = 𝐶𝐶1𝐴𝐴2𝑥𝑥 + 𝐵𝐵2𝑦𝑦 = 𝐶𝐶1

�3𝑥𝑥 − 4𝑦𝑦 = 5 5𝑥𝑥 + 7𝑦𝑦 = -2

3×3 system3 equations 3 unknowns

�𝐴𝐴1𝑥𝑥 + 𝐵𝐵1𝑦𝑦 + 𝐶𝐶1𝑧𝑧 = 𝐷𝐷1𝐴𝐴2𝑥𝑥 + 𝐵𝐵2𝑦𝑦 + 𝐶𝐶2𝑧𝑧 = 𝐷𝐷2𝐴𝐴3𝑥𝑥 + 𝐵𝐵3𝑦𝑦 + 𝐶𝐶3𝑧𝑧 = 𝐷𝐷3

�2𝑥𝑥 − 3𝑦𝑦 + 4𝑧𝑧 = 7 4𝑥𝑥 − 2𝑦𝑦 − 𝑧𝑧 = 3 5𝑥𝑥 − 6𝑦𝑦 + 2𝑧𝑧 = 2

• The solutions for a system of equations: the values for variables that make all equations in

the system true .

Example: Verify that the ordered pair (5, 3) is a solution of the system .

� 𝑥𝑥 − 𝑦𝑦 = 2 2𝑥𝑥 + 𝑦𝑦 = 13

x − y = 2 2x + y = 13? ?

5 − 3 = 2 2(5) + 3 = 13 Replace (x, y) by (5, 3) in both equations . √ √

2 = 2 Yes! 13 = 13

(5, 3) makes both equations true, it is the solution of the system .

Page 4-1

UNIT 4 SYSTEMS OF EQUATIONS & INEQUALITIES

4-1 SYSTEMS OF EQUATIONS

A System of Equations

• A system of equations is a group of two or more equations with the same variables

(unknowns) .

Example: � 2𝑥𝑥2 + 3𝑦𝑦 = 24𝑥𝑥2 – 5𝑦𝑦 = 7

• A system of linear equations is a group of two or more first-degree equations .

(First-degree equation: The highest power of the variable is one .)

• A system of two linear equations in two variables: two linear equations in two unknowns .

Example: � 𝑥𝑥 − 𝑦𝑦 = 2 2𝑥𝑥 + 𝑦𝑦 = 13

• A system of linear equations

Standard Form Example 2×2 system

2 equations 2 unknowns�𝐴𝐴1𝑥𝑥 + 𝐵𝐵1𝑦𝑦 = 𝐶𝐶1𝐴𝐴2𝑥𝑥 + 𝐵𝐵2𝑦𝑦 = 𝐶𝐶1

�3𝑥𝑥 − 4𝑦𝑦 = 5 5𝑥𝑥 + 7𝑦𝑦 = -2



�2𝑥𝑥 − 3𝑦𝑦 + 4𝑧𝑧 = 7 4𝑥𝑥 − 2𝑦𝑦 − 𝑧𝑧 = 3 5𝑥𝑥 − 6𝑦𝑦 + 2𝑧𝑧 = 2


the system true .

Example: Verify that the ordered pair (5, 3) is a solution of the system .

� 𝑥𝑥 − 𝑦𝑦 = 2 2𝑥𝑥 + 𝑦𝑦 = 13

x − y = 2 2x + y = 13? ?

5 − 3 = 2 2(5) + 3 = 13 Replace (x, y) by (5, 3) in both equations . √ √

2 = 2 Yes! 13 = 13

(5, 3) makes both equations true, it is the solution of the system .

Page 4-1





Solving Linear Systems by Graphing

• Solving systems of equations – graphing method: Graph both equations in the system on

the coordinate plane . The point(s) at which the lines intersect will be the solution(s) to the

system .

• Procedure for graphing

Example: Solve the following system graphically .

�𝑥𝑥 − 𝑦𝑦 = 2 (1)𝑥𝑥 + 𝑦𝑦 = 4 (2)

- Graph x – y = 2 (1)

x y = x – 2 (x, y)0 -2 (0, -2)2 0 (2, 0)

- Graph x + y = 4 (2)

x y = 4 – x (x, y)0 4 (0, 4)4 0 (4, 0)

- Find the intersection of two lines (x, y) . (x, y) = (3, 1) Solution

- Check . x − y = 2 x + y = 4

? ?3 − 1 = 2 3 + 1 = 4 √ √ 2 = 2 4 = 4 Correct!

∙ (0, 4)

y

x – y = 2

∙(4, 0)

∙ (2, 0)

x

∙ (0, -2) x + y = 4

∙ (3, 1)

Page 4-2

Solving Linear Systems by Graphing



system .

• Procedure for graphing

Example: Solve the following system graphically .

�𝑥𝑥 − 𝑦𝑦 = 2 (1)𝑥𝑥 + 𝑦𝑦 = 4 (2)

- Graph x – y = 2 (1)

x y = x – 2 (x, y)0 -2 (0, -2)2 0 (2, 0)

- Graph x + y = 4 (2)

x y = 4 – x (x, y)0 4 (0, 4)4 0 (4, 0)

- Find the intersection of two lines (x, y) . (x, y) = (3, 1) Solution

- Check . x − y = 2 x + y = 4

? ?3 − 1 = 2 3 + 1 = 4 √ √ 2 = 2 4 = 4 Correct!

∙ (0, 4)

y

x – y = 2

∙(4, 0)

∙ (2, 0)

x

∙ (0, -2) x + y = 4

∙ (3, 1)

Page 4-2





Properties of a Linear System

• Consistent and independent: The system has one solution (the lines of equations intersect

at one point) . The equations in the system are independent .

Example: (last example)

� 𝑥𝑥 − 𝑦𝑦 = 2 𝑥𝑥 + 𝑦𝑦 = 4 Solution: (x, y) = (3, 1)

• Consistent and dependent: The system has infinite number of solutions (the lines of the

equations coincide) .

Example: �𝑥𝑥 − 𝑦𝑦 = 2 (1)2𝑥𝑥 − 2𝑦𝑦 = 4 (2)

𝒙𝒙 − 𝒚𝒚 = 𝟐𝟐 (𝟏𝟏) 𝟐𝟐𝒙𝒙 − 𝟐𝟐𝒚𝒚 = 𝟒𝟒 (𝟐𝟐)x y = x − 2 x y = x − 20 -2 0 -2 2 0 2 0

• Inconsistent system: the system has no solution, and the lines of the equations are parallel .

(The solution set to the system is an empty set ∅ .)

Example: �2𝑥𝑥 + 𝑦𝑦 = 6 (1)2𝑥𝑥 + 𝑦𝑦 = - 8 (2)

𝟐𝟐𝒙𝒙 + 𝒚𝒚 = 𝟔𝟔 (𝟏𝟏) 𝟐𝟐𝒙𝒙 + 𝒚𝒚 = −𝟖𝟖 (𝟐𝟐)x y = 6 − 2x x y = -8 − 2x0 6 0 -83 0 -4 0

• Properties of linear equations

Property Numbers of Solution Lines Graphconsistent& independent one solution lines intersect

consistent& dependent infinitely number of solutions lines coincide

(the same line)

inconsistent no solution ∅ lines are parallel

∙ Solution

y

x

∙ (0, 6)

y

2x + y = -8

∙ (0, -8) 2x + y = 6

∙(-4, 0)

∙ (3, 0)

∙ (0, -2)

x

y

∙ (2, 0)

x

Page 4-3

Properties of a Linear System

• Consistent and independent: The system has one solution (the lines of equations intersect

at one point) . The equations in the system are independent .

Example: (last example)

� 𝑥𝑥 − 𝑦𝑦 = 2 𝑥𝑥 + 𝑦𝑦 = 4 Solution: (x, y) = (3, 1)

• Consistent and dependent: The system has infinite number of solutions (the lines of the

equations coincide) .

Example: �𝑥𝑥 − 𝑦𝑦 = 2 (1)2𝑥𝑥 − 2𝑦𝑦 = 4 (2)

𝒙𝒙 − 𝒚𝒚 = 𝟐𝟐 (𝟏𝟏) 𝟐𝟐𝒙𝒙 − 𝟐𝟐𝒚𝒚 = 𝟒𝟒 (𝟐𝟐)x y = x − 2 x y = x − 20 -2 0 -2 2 0 2 0

• Inconsistent system: the system has no solution, and the lines of the equations are parallel .

(The solution set to the system is an empty set ∅ .)

Example: �2𝑥𝑥 + 𝑦𝑦 = 6 (1)2𝑥𝑥 + 𝑦𝑦 = - 8 (2)

𝟐𝟐𝒙𝒙 + 𝒚𝒚 = 𝟔𝟔 (𝟏𝟏) 𝟐𝟐𝒙𝒙 + 𝒚𝒚 = −𝟖𝟖 (𝟐𝟐)x y = 6 − 2x x y = -8 − 2x0 6 0 -83 0 -4 0

• Properties of linear equations

Property Numbers of Solution Lines Graphconsistent& independent one solution lines intersect

consistent& dependent infinitely number of solutions lines coincide

(the same line)


∙ Solution

y

x

∙ (0, 6)

y

2x + y = -8

∙ (0, -8) 2x + y = 6

∙(-4, 0)

∙ (3, 0)

∙ (0, -2)

x

y

∙ (2, 0)

x

Page 4-3





4-2 SOLVING SYSTEMS BY SUBSTITUTION OR ELIMINATION

Solving Systems by Substitution

• Substitution method: Solve one variable in one equation and substitute the result into the

other equation to solve another variable . (Objective is to eliminate one of the unknown

variables) .

• Using the substitution method to solve systems:

Steps Example: � 𝒙𝒙 − 𝟐𝟐𝒚𝒚 = 𝟒𝟒 𝟐𝟐𝒙𝒙 + 𝒚𝒚 = 𝟑𝟑

- Label the equation as (1) & (2) . � 𝑥𝑥 − 2𝑦𝑦 = 4 (1) 2𝑥𝑥 + 𝑦𝑦 = 3 (2)

- Choose one equation and isolate one Choose (1) and isolate x .

variable (x or y), and name the equation (3) . x = 2y + 4 (3)

- Substitute the isolated variable into the Substitute x into (2)

other equation . 2(2y + 4) + y = 3 Replace x by 2y + 4 .

- Solve for the other variable . 4y + 8 + y = 3 Solve for y.

5y = -5 y = -1

- Substitute the solved value into x = 2 (-1) + 4 y = -1 (3)

equation (3) and solve for y or x . x = 2

The solution is (2, -1)

- Check . x – 2y = 4 2x + y = 3? ?

2 – 2(-1) = 4 2(2) + (-1) = 3 √ √2 + 2 = 4 4 – 1 = 3

x – 2y = 4 2x + y = 3

Consistent

x y = 𝒙𝒙

𝟐𝟐− 𝟐𝟐 x y = 3 - 2x

0 -2 0 34 0 1 1

x

∙ (0, 3)

y

∙ (2, -1) ∙ (0, -2)

∙ (4, 0) ∙ (1, 1)

solution

Page 4-4

4-2 SOLVING SYSTEMS BY SUBSTITUTION OR ELIMINATION

Solving Systems by Substitution

• Substitution method: Solve one variable in one equation and substitute the result into the

other equation to solve another variable . (Objective is to eliminate one of the unknown

variables) .

• Using the substitution method to solve systems:

Steps Example: � 𝒙𝒙 − 𝟐𝟐𝒚𝒚 = 𝟒𝟒 𝟐𝟐𝒙𝒙 + 𝒚𝒚 = 𝟑𝟑

- Label the equation as (1) & (2) . � 𝑥𝑥 − 2𝑦𝑦 = 4 (1) 2𝑥𝑥 + 𝑦𝑦 = 3 (2)

- Choose one equation and isolate one Choose (1) and isolate x .

variable (x or y), and name the equation (3) . x = 2y + 4 (3)

- Substitute the isolated variable into the Substitute x into (2)

other equation . 2(2y + 4) + y = 3 Replace x by 2y + 4 .

- Solve for the other variable . 4y + 8 + y = 3 Solve for y.

5y = -5 y = -1

- Substitute the solved value into x = 2 (-1) + 4 y = -1 (3)

equation (3) and solve for y or x . x = 2

The solution is (2, -1)

- Check . x – 2y = 4 2x + y = 3? ?

2 – 2(-1) = 4 2(2) + (-1) = 3 √ √2 + 2 = 4 4 – 1 = 3

x – 2y = 4 2x + y = 3

Consistent

x y = 𝒙𝒙

𝟐𝟐− 𝟐𝟐 x y = 3 - 2x

0 -2 0 34 0 1 1

x

∙ (0, 3)

y

∙ (2, -1) ∙ (0, -2)

∙ (4, 0) ∙ (1, 1)

solution

Page 4-4





Solving Systems by Elimination

• Elimination method: Add or subtract the equations to eliminate one of the variables(unknowns), and then solve the resulting equation in one variable .Objective: eliminate one of the variables .

• Using the elimination method to solve systems:

Steps Example: Solve � 𝟐𝟐𝟐𝟐 − 𝟏𝟏 = - 𝒚𝒚 𝟑𝟑𝟐𝟐 = 𝒚𝒚 + 𝟒𝟒

. - Write the system of equations in standard

form and label them as (1) and (2) . �2𝑥𝑥 + 𝑦𝑦 = 1 (1) 3𝑥𝑥 − 𝑦𝑦 = 4 (2)

Standard form: �𝐴𝐴1𝑥𝑥 + 𝐵𝐵1𝑦𝑦 = 𝐶𝐶1𝐴𝐴2𝑥𝑥 + 𝐵𝐵2𝑦𝑦 = 𝐶𝐶2

- Add equations (1) and (2) . 2x + y = 1 + 3x – y = 4

5x = 5 Solve for x. x = 1- Substitute the isolated variable into (1) or (2) .

2(1) + y = 1 x = 1 (1)

- Solve for the other variable . y = -1 Solve for y.

Solution: (1, -1)

- Check . 2x – 1 = - y 3x = y + 4? ?

2(1) – 1 = - (-1) 3(1) = -1 + 4 √ √2 – 1 = 1 3 = 3 Correct!

Example: Solve �𝟐𝟐𝟐𝟐 − 𝟓𝟓 = -𝒚𝒚 - 𝟐𝟐 + 𝟑𝟑𝒚𝒚 = 𝟏𝟏

.

Steps Solution

- Rewrite in standard form and label them as (1) and (2) . �2𝑥𝑥 + 𝑦𝑦 = 5 (1)-𝑥𝑥 + 3𝑦𝑦 = 1 (2)

- Multiply one or both equations by the appropriate numbers to eliminate one variable (x or y) . -2x + 6y = 2 (3) Multiply (2) by 2.

Note: if add equations (1) and (2), nothing cancels out: 2x + y = 5 + -x + 3y = 1

x + 4y = 6

- Add equations (1) and (3) and solve for y . 2x + y = 5 (1)+ -2x + 6y = 2 (3)

7y = 7 Solve for y.

y = 1- Substitute y = 1 into equation (1) & solve for x . 2x + 1 = 5 Replace y by 1 .

x = 2 Solve for x. Solution: (2, 1)

Page 4-5

Solving Systems by Elimination

• Elimination method: Add or subtract the equations to eliminate one of the variables(unknowns), and then solve the resulting equation in one variable .Objective: eliminate one of the variables .

• Using the elimination method to solve systems:

Steps Example: Solve � 𝟐𝟐𝟐𝟐 − 𝟏𝟏 = - 𝒚𝒚 𝟑𝟑𝟐𝟐 = 𝒚𝒚 + 𝟒𝟒

. - Write the system of equations in standard

form and label them as (1) and (2) . �2𝑥𝑥 + 𝑦𝑦 = 1 (1) 3𝑥𝑥 − 𝑦𝑦 = 4 (2)

Standard form: �𝐴𝐴1𝑥𝑥 + 𝐵𝐵1𝑦𝑦 = 𝐶𝐶1𝐴𝐴2𝑥𝑥 + 𝐵𝐵2𝑦𝑦 = 𝐶𝐶2

- Add equations (1) and (2) . 2x + y = 1 + 3x – y = 4

5x = 5 Solve for x. x = 1- Substitute the isolated variable into (1) or (2) .

2(1) + y = 1 x = 1 (1)

- Solve for the other variable . y = -1 Solve for y.

Solution: (1, -1)

- Check . 2x – 1 = - y 3x = y + 4? ?

2(1) – 1 = - (-1) 3(1) = -1 + 4 √ √2 – 1 = 1 3 = 3 Correct!

Example: Solve �𝟐𝟐𝟐𝟐 − 𝟓𝟓 = -𝒚𝒚 - 𝟐𝟐 + 𝟑𝟑𝒚𝒚 = 𝟏𝟏

.

Steps Solution

- Rewrite in standard form and label them as (1) and (2) . �2𝑥𝑥 + 𝑦𝑦 = 5 (1)-𝑥𝑥 + 3𝑦𝑦 = 1 (2)

- Multiply one or both equations by the appropriate numbers to eliminate one variable (x or y) . -2x + 6y = 2 (3) Multiply (2) by 2.

Note: if add equations (1) and (2), nothing cancels out: 2x + y = 5 + -x + 3y = 1

x + 4y = 6

- Add equations (1) and (3) and solve for y . 2x + y = 5 (1)+ -2x + 6y = 2 (3)

7y = 7 Solve for y.

y = 1- Substitute y = 1 into equation (1) & solve for x . 2x + 1 = 5 Replace y by 1 .

x = 2 Solve for x. Solution: (2, 1)

Page 4-5





Page 4-6

Systems Involving Decimals or Fractions

• System involving decimals:

Example: Solve the system of equations by elimination. �𝒙𝒙𝒙𝒙 − 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 = 𝟑𝟑𝟑𝟑 𝟎𝟎𝟎𝟎.𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙 + 𝟎𝟎𝟎𝟎.𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 = 𝟎𝟎𝟎𝟎.𝟐𝟐𝟐𝟐

Steps Example

- Label the equations as (1) & (2). � 𝑥𝑥𝑥𝑥 − 2𝑦𝑦𝑦𝑦 = 3 (1) 0.4𝑥𝑥𝑥𝑥 + 0. 2𝑦𝑦𝑦𝑦 = 0.2 (2)

- Clear the decimals. 4x + 2y = 2 (3) Multiply equation (2) by 10.

- Add equations (1) and (3) to eliminate y. x – 2y = 3 + 4x + 2y = 2 5x = 5 x = 1

- Substitute x into (1). 1 − 2𝑦𝑦𝑦𝑦 = 3 Replace x by 1.

- Solve for y. y = -1 Solution: (1, -1)

• System involving fractions

Example: Solve the system of equations by elimination. �𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

𝒚𝒚𝒚𝒚 – 𝟏𝟏𝟏𝟏 = 𝟎𝟎𝟎𝟎

- 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

𝒚𝒚𝒚𝒚 − 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏 = 𝟎𝟎𝟎𝟎

Steps Solution

- Write in standard form and label as (1) and (2). � 23

𝑥𝑥𝑥𝑥 + 12

𝑦𝑦𝑦𝑦 = 1 (𝟏𝟏𝟏𝟏)−12

𝑥𝑥𝑥𝑥 − 23

𝑦𝑦𝑦𝑦 = -1 (𝟐𝟐𝟐𝟐)

- Clear the fractions. 6 �23𝑥𝑥𝑥𝑥� + 6 �1

2𝑦𝑦𝑦𝑦� = 1∙ 6 Multiply (1) by the LCD.

4x + 3y = 6 (3) 6 �- 1

2𝑥𝑥𝑥𝑥� – 6 �2

3𝑦𝑦𝑦𝑦� = -1∙ 6 Multiply (2) by the LCD.

-3x – 4y = -6 (4)

- Multiply equation (3) by 3 and (4) by 4, 12x + 9y = 18 Multiply (3) by 3. and add them. + -12x – 16y = -24 Multiply (4) by 4.

-7y = -6 y = 𝟔𝟔𝟔𝟔

𝟕𝟕𝟕𝟕

- Substitute 67 for y in equation (1), and solve for x.

23 x + 1

2�6

7� = 1, 2

3 x + 3

7 = 1 Multiply by the LCD: 21

14 x + 9 = 21 , x = 𝟔𝟔𝟔𝟔𝟕𝟕𝟕𝟕

Solution: �𝟔𝟔𝟔𝟔𝟕𝟕𝟕𝟕

, 𝟔𝟔𝟔𝟔𝟕𝟕𝟕𝟕 �





Applications

Example: Todd bought 5 apples and 4 oranges for $3 . Susan bought 7 apples and 4 oranges for

$4 . How much does one apple and orange cost?

- List the facts .Apple Orange Price

Todd 5 4 $ 3 .00Susan 7 4 $ 4 .00

- Label x and y . Let x = the cost of one apple, y = the cost of one orange .

- Write system of equations . �𝟓𝟓𝒙𝒙 + 𝟒𝟒𝒚𝒚 = 𝟑𝟑 (1)𝟕𝟕𝒙𝒙 + 𝟒𝟒𝒚𝒚 = 𝟒𝟒 (2)

- Solve equations . 5x + 4y = 3 (1) – (2)– 7x + 4y = 4

-2x = -1x = 1

2= $𝟎𝟎.𝟓𝟓𝟎𝟎

- Substitute x = 12

into equation (1) 5 �12� + 4y = 3 Multiply 2 for each term .

and solve for y. 5 + 8y = 6y = 1

8≈ $𝟎𝟎.𝟏𝟏𝟑𝟑

- Answer: One apple costs $0 .50 and one orange costs $0 .13 . (x, y) = ($0.50, $0.13)

Example: The perimeter of a rectangular field is 400m . The length is 40m less than twice the

width . Determine the dimension of the rectangular field .

- Facts: width wlength 2w – 40 = lperimeter P = 400m (P = 2l + 2w)

- Equations: �𝟐𝟐𝒍𝒍 + 𝟐𝟐𝒘𝒘 = 𝟒𝟒𝟎𝟎𝟎𝟎 𝒍𝒍 = 𝟐𝟐𝒘𝒘 – 𝟒𝟒𝟎𝟎

- Standard form: �2𝑙𝑙 + 2𝑤𝑤 = 400 (1) 𝑙𝑙 − 2𝑤𝑤 = -40 (2)

- (1) + (2) : 2l + 2w = 400 + l – 2w = - 40

3l = 360l = 360

3= 120 l = 120m

- Substitute l = 120 into (1) 2 (120) + 2w = 400and solve for w . 2w = 400 – 240

w = 80 m- Answer: Length = 120 m, Width = 80 m

w

l

Page 4-7

Applications

Example: Todd bought 5 apples and 4 oranges for $3 . Susan bought 7 apples and 4 oranges for

$4 . How much does one apple and orange cost?

- List the facts .Apple Orange Price

Todd 5 4 $ 3 .00Susan 7 4 $ 4 .00

- Label x and y . Let x = the cost of one apple, y = the cost of one orange .

- Write system of equations . �𝟓𝟓𝒙𝒙 + 𝟒𝟒𝒚𝒚 = 𝟑𝟑 (1)𝟕𝟕𝒙𝒙 + 𝟒𝟒𝒚𝒚 = 𝟒𝟒 (2)

- Solve equations . 5x + 4y = 3 (1) – (2)– 7x + 4y = 4

-2x = -1x = 1

2= $𝟎𝟎.𝟓𝟓𝟎𝟎

- Substitute x = 12

into equation (1) 5 �12� + 4y = 3 Multiply 2 for each term .

and solve for y. 5 + 8y = 6y = 1

8≈ $𝟎𝟎.𝟏𝟏𝟑𝟑

- Answer: One apple costs $0 .50 and one orange costs $0 .13 . (x, y) = ($0.50, $0.13)

Example: The perimeter of a rectangular field is 400m . The length is 40m less than twice the

width . Determine the dimension of the rectangular field .

- Facts: width wlength 2w – 40 = lperimeter P = 400m (P = 2l + 2w)

- Equations: �𝟐𝟐𝒍𝒍 + 𝟐𝟐𝒘𝒘 = 𝟒𝟒𝟎𝟎𝟎𝟎 𝒍𝒍 = 𝟐𝟐𝒘𝒘 – 𝟒𝟒𝟎𝟎

- Standard form: �2𝑙𝑙 + 2𝑤𝑤 = 400 (1) 𝑙𝑙 − 2𝑤𝑤 = -40 (2)

- (1) + (2) : 2l + 2w = 400 + l – 2w = - 40

3l = 360l = 360

3= 120 l = 120m

- Substitute l = 120 into (1) 2 (120) + 2w = 400and solve for w . 2w = 400 – 240

w = 80 m- Answer: Length = 120 m, Width = 80 m

w

l

Page 4-7





4-3 SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES

Linear Inequalities in Two Variables

• A linear inequality: a mathematical statement with an inequality symbol in which the

highest power of the variable is one .

• Inequality symbols review

Symbol Indication> greater than< less than≥ greater than or equal to≤ less than or equal to

• A linear inequality in two variables: a linear inequality contains two variables .

• Standard linear inequality in two variables

Standard Inequality ExampleAx + By > C 2x + 3y > 6Ax + By < C 3x – 5y < 17Ax + By ≥ C 4x + 5y ≥ 10Ax + By ≤ C 6x – 11y ≤ 21

• Solutions of linear inequalities in two variables: an ordered pair (x, y) that makes the

inequality true .

Example: Determine if (1, 2) satisfies the inequality 3x + 5y < 17 .

?

3(1) + 5(2) < 17 Replace (x, y) by (1, 2) .

?

3 + 10 < 17 √

13 < 17 Correct!

Yes, (1, 2) is a solution A true statement .

Page 4-8

4-3 SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES

Linear Inequalities in Two Variables



• Inequality symbols review

Symbol Indication> greater than< less than≥ greater than or equal to≤ less than or equal to

• A linear inequality in two variables: a linear inequality contains two variables .

• Standard linear inequality in two variables

Standard Inequality ExampleAx + By > C 2x + 3y > 6Ax + By < C 3x – 5y < 17Ax + By ≥ C 4x + 5y ≥ 10Ax + By ≤ C 6x – 11y ≤ 21


inequality true .

Example: Determine if (1, 2) satisfies the inequality 3x + 5y < 17 .

?

3(1) + 5(2) < 17 Replace (x, y) by (1, 2) .

?

3 + 10 < 17 √

13 < 17 Correct!

Yes, (1, 2) is a solution A true statement .

Page 4-8





Graphing Linear Inequality in Two Variables

Procedure for graphing a linear inequality in two variables

Steps Example: Graph x – y ≥ 3.

- Change the inequality symbol to an equal sign . x – y = 3

- Graph the boundary line of the corresponding equation .(Using x and y intercepts)

x y = x – 3 (x, y)0 -3 (0, -3)3 0 (3, 0)

- Draw a solid line if the inequality symbol is ≤ or ≥ .

- Draw a dashed line if the inequality symbol is < or > .

- Choose a test point such as (0, 0) . Choose (0, 0)

- If the test point satisfies the inequality, shade x – y ≥ 3 ?

the side of the line that contains the test point . 0 – 0 ≥ 3 False

- If the test point does not satisfy the inequality, shade the

side of the line that does not contain the test point .

Shade the region that does not contain (0, 0) .

Example: Graph the inequality 2x – y < 8 .

- “ < ” changes to “=” 2x – y = 8

- Draw a dashed boundary line (∵ the inequality symbol is < )

x y = 2x – 8 (x, y)0 -8 (0, -8)4 0 (4, 0)

- Test point: choose (0, 0) . 2x – y < 8?

2(0) – 0 < 8 √

0 < 8 True . Shade the region that contains (0, 0) .

∙(3, 0)

y

x

∙ (0, -8)

y

∙(0, 0)

∙ (0, -3)

≥ : Solid line

∙(4, 0)

∙(0, 0)

x

x

y

Page 4-9

Graphing Linear Inequality in Two Variables

Procedure for graphing a linear inequality in two variables

Steps Example: Graph x – y ≥ 3.

- Change the inequality symbol to an equal sign . x – y = 3

- Graph the boundary line of the corresponding equation .(Using x and y intercepts)

x y = x – 3 (x, y)0 -3 (0, -3)3 0 (3, 0)

- Draw a solid line if the inequality symbol is ≤ or ≥ .

- Draw a dashed line if the inequality symbol is < or > .

- Choose a test point such as (0, 0) . Choose (0, 0)

- If the test point satisfies the inequality, shade x – y ≥ 3 ?

the side of the line that contains the test point . 0 – 0 ≥ 3 False

- If the test point does not satisfy the inequality, shade the

side of the line that does not contain the test point .

Shade the region that does not contain (0, 0) .

Example: Graph the inequality 2x – y < 8 .

- “ < ” changes to “=” 2x – y = 8

- Draw a dashed boundary line (∵ the inequality symbol is < )

x y = 2x – 8 (x, y)0 -8 (0, -8)4 0 (4, 0)

- Test point: choose (0, 0) . 2x – y < 8?

2(0) – 0 < 8 √

0 < 8 True . Shade the region that contains (0, 0) .

∙(3, 0)

y

x

∙ (0, -8)

y

∙(0, 0)

∙ (0, -3)

≥ : Solid line

∙(4, 0)

∙(0, 0)

x

x

y

Page 4-9





Page 4-10

Do not choose the point to test if it is on the solution line.

Example: Graph the inequality y ≥ 5x.

- “ ≥ ” changes to “=”

- Draw a solid boundary line. (∵ the inequality symbol is ≥ )

x y = 5x (x, y) 0 0 (0, 0) 1 5 (1, 5)

- Test point: choose (1, 1) y ≥ 5x

(0, 0) is on the solution line. ?

1 ≥ 5 ∙ 1 ?

1 > 5 False Shade the region that not contains (1, 1).

Example: Write the linear inequalities whose graph is the shaded region.

1.

y < -2 A dashed line: <

2.

x ≥ 2 A solid line: ≥.

∙ -2

x

y

∙ 0

∙ 0

x

y

∙ 2

∙ (1, 5)

y

It is on the solution line.

y = 5x

x ∙ (0, 0)

∙ (1, 1)





Page 4-11

Systems of Linear Inequalities

• A system of linear inequalities in two variables: a group of two or more inequalities with

the same two variables.

Standard Form Example

�𝐴𝐴𝐴𝐴1𝑥𝑥𝑥𝑥 + 𝐵𝐵𝐵𝐵1𝑦𝑦𝑦𝑦 > 𝐶𝐶𝐶𝐶1 𝐴𝐴𝐴𝐴2𝑥𝑥𝑥𝑥 + 𝐵𝐵𝐵𝐵2𝑦𝑦𝑦𝑦 ≥ 𝐶𝐶𝐶𝐶2

2x + 3y < 4 3x – 5y ≥ 8

< , > , ≤ , ≥

• Solutions of a system of linear inequalities: an ordered pair that satisfies both inequalities.

• Graphing a system of linear inequalities in two variables

Steps Example: Graph �𝒙𝒙𝒙𝒙 –𝒚𝒚𝒚𝒚 ≥ 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 + 𝒚𝒚𝒚𝒚 < 𝟒𝟒𝟒𝟒

- Change the inequality symbols to equal signs � 𝑥𝑥𝑥𝑥 – 𝑦𝑦𝑦𝑦 = 2 (1) 2𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 = 4 (2)

and label with (1) and (2).

- Graph the boundary lines of the corresponding equations.

Draw a dashed line if the symbol is < or > . Draw a solid line if the symbol is ≤ or ≥ .

- Choose a test point (0, 0). Test point (0, 0) x – y ≥ 2 2x + y < 4

? ?

0 − 0 ≥ 2 2∙0 + 0 < 4 × √

0 ≥ 2 False 0 < 4 True Shade the region that does not contain Shade the region that contains (0, 0), i.e. below x – y = 2. (0, 0), i.e below 2x + y = 4.

- The solution set is the region where the shading overlaps. The vertex is (2, 0). The vertex is formed by an intersection of two boundary lines.

Example: Graph the system of inequalities �𝒙𝒙𝒙𝒙 ≤ 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐 ≤ 𝐲𝐲𝐲𝐲 ≤ 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 > 𝟐𝟐𝟐𝟐

. Find the coordinates of any

vertices formed. 𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 = 𝟐𝟐𝟐𝟐

(1) (2) x y = x – 2 x y = 4 –2x

0 -2 0 4 2 0 2 0

x y = 1 – 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙

0 1 2 0

x

y

x – y = 2 ( ≥ )

∙ (0, 0)

2x + y = 4 (< )

∙ (0, -2)

∙ (0, 4)

y

x

x = 3

y = 4

y = 2

0

x + 2y = 2

∙ (-2, 2)

∙ (-5, 4) ∙ (3, 4)

∙ (3, 2)

∙

∙

Vertices (-5, 4) (-2, 2) (3, 4) (3, 2)

∙ (2, 0)

Page 4-11

Systems of Linear Inequalities

• A system of linear inequalities in two variables: a group of two or more inequalities with

the same two variables.


�𝐴𝐴𝐴𝐴1𝑥𝑥𝑥𝑥 + 𝐵𝐵𝐵𝐵1𝑦𝑦𝑦𝑦 > 𝐶𝐶𝐶𝐶1 𝐴𝐴𝐴𝐴2𝑥𝑥𝑥𝑥 + 𝐵𝐵𝐵𝐵2𝑦𝑦𝑦𝑦 ≥ 𝐶𝐶𝐶𝐶2

2x + 3y < 4 3x – 5y ≥ 8

< , > , ≤ , ≥

• Solutions of a system of linear inequalities: an ordered pair that satisfies both inequalities.


Steps Example: Graph �𝒙𝒙𝒙𝒙 –𝒚𝒚𝒚𝒚 ≥ 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 + 𝒚𝒚𝒚𝒚 < 𝟒𝟒𝟒𝟒

- Change the inequality symbols to equal signs � 𝑥𝑥𝑥𝑥 – 𝑦𝑦𝑦𝑦 = 2 (1) 2𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 = 4 (2)

and label with (1) and (2).

- Graph the boundary lines of the corresponding equations.

Draw a dashed line if the symbol is < or > . Draw a solid line if the symbol is ≤ or ≥ .

- Choose a test point (0, 0). Test point (0, 0) x – y ≥ 2 2x + y < 4

? ?

0 − 0 ≥ 2 2∙0 + 0 < 4 × √

0 ≥ 2 False 0 < 4 True Shade the region that does not contain Shade the region that contains (0, 0), i.e. below x – y = 2. (0, 0), i.e below 2x + y = 4.

- The solution set is the region where the shading overlaps. The vertex is (2, 0). The vertex is formed by an intersection of two boundary lines.

Example: Graph the system of inequalities �𝒙𝒙𝒙𝒙 ≤ 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐 ≤ 𝐲𝐲𝐲𝐲 ≤ 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 > 𝟐𝟐𝟐𝟐

. Find the coordinates of any

vertices formed. 𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 = 𝟐𝟐𝟐𝟐

(1) (2) x y = x – 2 x y = 4 –2x

0 -2 0 4 2 0 2 0

x y = 1 – 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙

0 1 2 0

x

y

x – y = 2 ( ≥ )

∙ (0, 0)

2x + y = 4 (< )

∙ (0, -2)

∙ (0, 4)

y

x

x = 3

y = 4

y = 2

0

x + 2y = 2

∙ (-2, 2)

∙ (-5, 4) ∙ (3, 4)

∙ (3, 2)

∙

∙

Vertices (-5, 4) (-2, 2) (3, 4) (3, 2)

∙ (2, 0)





Unit 4 Summary

• A system of linear equationsProperty Numbers of Solution Graph


�𝐴𝐴1𝑥𝑥 + 𝐵𝐵1𝑦𝑦 = 𝐶𝐶1𝐴𝐴2𝑥𝑥 + 𝐵𝐵2𝑦𝑦 = 𝐶𝐶2

�3𝑥𝑥 − 4𝑦𝑦 = 5 5𝑥𝑥 + 7𝑦𝑦 = -2



�2𝑥𝑥 − 3𝑦𝑦 + 4𝑧𝑧 = 7

4𝑥𝑥 – 2𝑦𝑦 − 𝑧𝑧 = 3 5𝑥𝑥 − 6𝑦𝑦 + 2𝑧𝑧 = 2


the system true .



system .

• Properties of linear equationsProperty Numbers of Solution Lines Graph

consistent& independent one solution lines intersect

consistent& dependent infinite number of solutions lines coincide

(the same line)


• Substitution method: Solve for one variable in one equation and substitute the result into

the other equation to solve another variable . (Objective is to eliminate one of the unknown

variables) .

• Elimination method: Add or subtract the equations to eliminate one of the variables

(unknowns), then solve the resulting equation in one variable .



• Standard linear inequalities in two variables

Standard Inequality ExampleAx + By > C 2x + 3y > 6Ax + By < C 3x − 5y < 17Ax + By ≥ C 4x + 5y ≥ 10Ax + By ≤ C 6x − 11y ≤ 21

Page 4-12

Unit 4 Summary

• A system of linear equationsProperty Numbers of Solution Graph


�𝐴𝐴1𝑥𝑥 + 𝐵𝐵1𝑦𝑦 = 𝐶𝐶1𝐴𝐴2𝑥𝑥 + 𝐵𝐵2𝑦𝑦 = 𝐶𝐶2

�3𝑥𝑥 − 4𝑦𝑦 = 5 5𝑥𝑥 + 7𝑦𝑦 = -2



�2𝑥𝑥 − 3𝑦𝑦 + 4𝑧𝑧 = 7

4𝑥𝑥 – 2𝑦𝑦 − 𝑧𝑧 = 3 5𝑥𝑥 − 6𝑦𝑦 + 2𝑧𝑧 = 2


the system true .



system .

• Properties of linear equationsProperty Numbers of Solution Lines Graph

consistent& independent one solution lines intersect

consistent& dependent infinite number of solutions lines coincide

(the same line)


• Substitution method: Solve for one variable in one equation and substitute the result into

the other equation to solve another variable . (Objective is to eliminate one of the unknown

variables) .

• Elimination method: Add or subtract the equations to eliminate one of the variables

(unknowns), then solve the resulting equation in one variable .



• Standard linear inequalities in two variables

Standard Inequality ExampleAx + By > C 2x + 3y > 6Ax + By < C 3x − 5y < 17Ax + By ≥ C 4x + 5y ≥ 10Ax + By ≤ C 6x − 11y ≤ 21

Page 4-12





Page 4-13


inequality true.

• Procedure for graphing a linear inequality in two variables

Change the inequality symbol to an equal sign.

Graph the boundary line of the corresponding equation.

o Draw a solid line if the inequality symbol is ≤ or ≥ . o Draw a dashed line if the inequality symbol is < or >.

Choose a test point such as (0, 0). o If the test point satisfies the inequality, shade the side of the line that contains the test point. o If the test point does not satisfy the inequality, shade the side of the line that does not contain

the test point.

• Do not choose the point to test if it is on the solution line.

• A system of linear inequalities in two variables


� 𝐴𝐴𝐴𝐴1𝑥𝑥𝑥𝑥 + 𝐵𝐵𝐵𝐵1𝑦𝑦𝑦𝑦 > 𝐶𝐶𝐶𝐶1 𝐴𝐴𝐴𝐴2𝑥𝑥𝑥𝑥 + 𝐵𝐵𝐵𝐵2𝑦𝑦𝑦𝑦 ≥ 𝐶𝐶𝐶𝐶2

2x + 3y < 4 3x − 5y ≥ 8

< , > , ≤ , ≥


Change the inequality symbols to equal signs and label with (1) and (2).

Graph the boundary lines of the corresponding equations.

o Draw a solid line if the inequality symbol is ≤ or ≥ . o Draw a dashed line if the inequality symbol is < or >.

Choose a test point (0, 0).

o If the test point satisfies the inequality, shade the side of the line that contains the test point. o If the test point does not satisfy the inequality, shade the side of the line that does not contain

the test point.

The solution set is the region where the shading overlaps.

A vertex is formed by an intersection of two boundary lines.





PRACTICE QUIZ

Unit 4 Systems of Equations & Inequalities

1. Solve the following system graphically .

�2𝑥𝑥 − 𝑦𝑦 = 1 3𝑥𝑥 + 𝑦𝑦 = 4

2. a. Solve the following system by substitution .

�2𝑥𝑥 + 𝑦𝑦 = 3 3𝑥𝑥 − 2𝑦𝑦 = 1

b. Solve the following system by elimination .

� 1

3𝑥𝑥 + 1

2𝑦𝑦 − 3 = 0

3

4𝑥𝑥 + 1

3𝑦𝑦 − 2 = 0

3. The perimeter of a rectangle field is 140m . The length is 10m more than

4 times the width . Determine the dimension of the rectangle field .

4 . Graph the inequality .

a . 3x + y > 4

b . y ≤ 3x

5. Graph the system of inequalities � 2𝑥𝑥 – 𝑦𝑦 ≤ -2 4𝑥𝑥 + 𝑦𝑦 > 2

.

Page 7

PRACTICE QUIZ

Unit 4 Systems of Equations & Inequalities

1. Solve the following system graphically .

�2𝑥𝑥 − 𝑦𝑦 = 1 3𝑥𝑥 + 𝑦𝑦 = 4

2. a. Solve the following system by substitution .

�2𝑥𝑥 + 𝑦𝑦 = 3 3𝑥𝑥 − 2𝑦𝑦 = 1

b. Solve the following system by elimination .

� 1

3𝑥𝑥 + 1

2𝑦𝑦 − 3 = 0

3

4𝑥𝑥 + 1

3𝑦𝑦 − 2 = 0

3. The perimeter of a rectangle field is 140m . The length is 10m more than

4 times the width . Determine the dimension of the rectangle field .

4 . Graph the inequality .

a . 3x + y > 4

b . y ≤ 3x

5. Graph the system of inequalities � 2𝑥𝑥 – 𝑦𝑦 ≤ -2 4𝑥𝑥 + 𝑦𝑦 > 2

.

Page 7




Algebra I & II Key Concepts, Practice, and Quizzes Unit 5 – Polynomial Functions

UNIT 5 POLYNOMIAL FUNCTIONS

5-1 ADDITION & SUBTRACTION OF POLYNOMIALS

Polynomials

• Review basic algebraic terms

Algebraic Term Description Examplealgebraic expression A mathematical phrase that contains numbers,

variables, and arithmetic operations . 9x2 – x + 3

coefficient The number in front of a variable . 9, -1

termA term can be a constant, variable, or the product of a number and variable .(Terms are separated by a plus or minus sign .)

9x2 , - x, 3

• Monomial: an algebraic expression consisting of just one term . (The prefix “mono” means one .)

Example: 3x, 7y2

• Binomial: an algebraic expression consisting of two terms . (The prefix “bi-” means two .)

Example: ax + b , 9t2 – 2t

• Trinomial: an algebraic expression consisting of three terms . (The prefix “tri-” means three .)

Example: ax2 + bx + c , - 4qp2 + 3q + 5

• Polynomial: an algebraic expression consisting of two or more terms . (The prefix “poly-”

means many .)

Example: 5x2 – 2x + 6y + 1 , -2a2 – 2b + 6ab + a – 5

• Summary

Name Example Coefficientmonomial (one term) 7a 7binomial (two terms) 3x – 5 3trinomial (three terms) -4x2 + xy + 7 -4, 1polynomial (two or more terms) 2pq + 4p3 + 11 + p 2, 4, 1

Note: A polynomial uses only the operations of addition, subtraction, multiplication (no division), and non-negative integer exponents .

Example: 2𝑥𝑥+7𝑦𝑦−2

, 3𝑎𝑎𝑎𝑎4𝑎𝑎2𝑎𝑎+5+𝑎𝑎3

, and 3𝑥𝑥𝑥𝑥2 − 1𝑥𝑥𝑦𝑦-2 are algebraic expressions but not polynomials .

division negative exponent: 1𝑥𝑥

= 𝑥𝑥𝑥𝑥−1

-x = (-1)(x)

Page 5-1



Polynomials






9x2 , - x, 3


Example: 3x, 7y2






means many .)


• Summary








-x = (-1)(x)

Page 5-1



Polynomials






9x2 , - x, 3


Example: 3x, 7y2






means many .)


• Summary








-x = (-1)(x)

Page 5-1





Degree of Polynomial

• The degree of a term with one variable: the exponent (power) of its variable .

Example: 5x2 degree: 2-3t7 degree: 7

• The degree of a term with more variables: the sum of the exponents of its variables .

Example: -3x3 y5 z2 degree: 3 + 5 + 2 = 10

• The degree of a polynomial with more variables: the highest degree of any individual term .Example: 4ab3 + 3a2b2c3 – 5a + 1 degree: 7

• The leading term of a polynomial: the term with the highest degree in the polynomial .

Example: 4ab3 + 3ab2c3 – 5a + 1 leading term: 3ab2c3

• The leading coefficient: the coefficient of the leading term .

Example: 4ab3 + 3ab2c3 – 5a + 1 leading coefficient: 3

• Examples of polynomial

• Descending order: the power of a variable decreases for each succeeding term.

Example: 2𝒙𝒙𝟑𝟑 + 5𝒙𝒙𝟐𝟐 − 𝒙𝒙 + 2

-13𝑎𝑎𝒃𝒃𝟒𝟒 + 21𝒃𝒃𝟑𝟑 − 𝑎𝑎𝒃𝒃𝟐𝟐 + 𝒃𝒃 − 34 The descending order of power b.

• Ascending order: the power of a variable increases for each succeeding term .

Example: -9 + 7𝒚𝒚 + 4𝒚𝒚𝟐𝟐 − 3𝒚𝒚𝟑𝟑

1 + 23𝒕𝒕𝑢𝑢 + 2𝒕𝒕𝟐𝟐𝑢𝑢3 − 5𝒕𝒕𝟑𝟑 + 𝒕𝒕𝟒𝟒 The ascending order of power t.

Polynomial 3t2 + t3 – 5 2p2q3 + 5r – 7p3q2rterm 3t2 , t3, - 5 2p2q3 , 5r , - 7p3q2r

degree of the term 2 , 3, 0 5 , 1 , 6

degree of the polynomial 3 6

leading term t3 - 7p3q2r

leading coefficient 1 -7

2 + 2 + 3 = 7

1 + 2 + 3 = 6

a = a1

Page 5-2

Degree of Polynomial










• Examples of polynomial


Example: 2𝒙𝒙𝟑𝟑 + 5𝒙𝒙𝟐𝟐 − 𝒙𝒙 + 2

-13𝑎𝑎𝒃𝒃𝟒𝟒 + 21𝒃𝒃𝟑𝟑 − 𝑎𝑎𝒃𝒃𝟐𝟐 + 𝒃𝒃 − 34 The descending order of power b.


Example: -9 + 7𝒚𝒚 + 4𝒚𝒚𝟐𝟐 − 3𝒚𝒚𝟑𝟑

1 + 23𝒕𝒕𝑢𝑢 + 2𝒕𝒕𝟐𝟐𝑢𝑢3 − 5𝒕𝒕𝟑𝟑 + 𝒕𝒕𝟒𝟒 The ascending order of power t.

Polynomial 3t2 + t3 – 5 2p2q3 + 5r – 7p3q2rterm 3t2 , t3, - 5 2p2q3 , 5r , - 7p3q2r

degree of the term 2 , 3, 0 5 , 1 , 6

degree of the polynomial 3 6

leading term t3 - 7p3q2r

leading coefficient 1 -7

2 + 2 + 3 = 7

1 + 2 + 3 = 6

a = a1

Page 5-2





Evaluating Polynomial Functions

• Polynomial function: The expression used to describe the function is a polynomial .

Example: f (x) = 2x3 –3x 2 + 7x +8

g(x) = -3x4+ 5x2 – 2

• Evaluating polynomial functions

Example: 1. If f (x) = 2x3 + 1, find f (2) and f (-1) .

f (2) = 2(2)3 + 1 = 16 + 1 = 17 Replace x with 2 .

f (-1) = 2(-1)3 + 1 = -2 + 1 = -1 Replace x with -1 .

2. If R (x) = -8x3 + x2 + 2, find R (0) and R �12� .

R (0) = - 8(0)3 + (0)2 + 2 = 2 Replace x with 0 .

R �𝟏𝟏𝟐𝟐� = - 8�1

2�3+ �1

2�2

+ 2 Replace x with 12

.

= - 1 + 14

+ 2 = 𝟓𝟓𝟒𝟒

Example: The polynomial function C (x) = 3,000 + 0 .5x2 can be used to determine the

total cost (in dollars) of producing x laptops in an electronics firm .

1. What is the total cost of producing 10 laptops?

2. Use the following graph to estimate C (40) .

Solution: 1. C (10) = 3,000 + 0 .5(10)2 C (x) = 3,000 + 0 .5x2 , replace x with 10 .

= $3,050

2. C (40): locate x = 40 on the x axis, move vertically to the graph, and then

move horizontally to the C(x) axis . Thus C(40) ≈ $3,800 .

x

∙ (40, 3,800)

1,000 ∙

4,000 ∙

∙10

∙20

∙30

∙40

2,000 ∙

3,000 ∙

C (x)

Polynomials 𝑓𝑓𝑓𝑓(𝑥𝑥𝑥𝑥) & g(x) are functions .

x = number of laptops

Page 5-3

Evaluating Polynomial Functions

• Polynomial function: The expression used to describe the function is a polynomial .

Example: f (x) = 2x3 –3x 2 + 7x +8

g(x) = -3x4+ 5x2 – 2

• Evaluating polynomial functions

Example: 1. If f (x) = 2x3 + 1, find f (2) and f (-1) .

f (2) = 2(2)3 + 1 = 16 + 1 = 17 Replace x with 2 .

f (-1) = 2(-1)3 + 1 = -2 + 1 = -1 Replace x with -1 .

2. If R (x) = -8x3 + x2 + 2, find R (0) and R �12� .

R (0) = - 8(0)3 + (0)2 + 2 = 2 Replace x with 0 .

R �𝟏𝟏𝟐𝟐� = - 8�1

2�3+ �1

2�2

+ 2 Replace x with 12

.

= - 1 + 14

+ 2 = 𝟓𝟓𝟒𝟒

Example: The polynomial function C (x) = 3,000 + 0 .5x2 can be used to determine the

total cost (in dollars) of producing x laptops in an electronics firm .

1. What is the total cost of producing 10 laptops?

2. Use the following graph to estimate C (40) .

Solution: 1. C (10) = 3,000 + 0 .5(10)2 C (x) = 3,000 + 0 .5x2 , replace x with 10 .

= $3,050

2. C (40): locate x = 40 on the x axis, move vertically to the graph, and then

move horizontally to the C(x) axis . Thus C(40) ≈ $3,800 .

x

∙ (40, 3,800)

1,000 ∙

4,000 ∙

∙10

∙20

∙30

∙40

2,000 ∙

3,000 ∙

C (x)

Polynomials 𝑓𝑓𝑓𝑓(𝑥𝑥𝑥𝑥) & g(x) are functions .

x = number of laptops

Page 5-3





Adding and Subtracting Polynomials

• Adding or subtracting polynomials

Example: Find the sum of 2x3 – 3x2 + x – 4 and x3 + 4x2 + 2x + 1 .

Steps Solution

(2x3 – 3x2 + x – 4) + (x3 + 4x2 + 2x + 1)

- Regroup like terms . = (2x3 + x3) + (-3x2 + 4x2) + (x + 2x) + (-4 + 1)

- Combine like terms . = 3x3 + x2 + 3x – 3

Example: Find the difference of 5x2 + 4x – 2 and 2x2 – 3x + 13 .

Steps Solution

(5x2 + 4x – 2) – (2x2 – 3x + 13)

- Remove parentheses . = 5x2 + 4x – 2 – 2x2 + 3x – 13(Reverse each sign in second parentheses .)

- Regroup like terms . = (5x2 – 2x2) + (4x + 3x) + (-2 – 13)

- Combine like terms . = 3x2 + 7x – 15

• Column method

Example: Find the sum of 3x3 – 5x2 + 7x – 3 and 2x3 + 3x + 5 .

Steps Solution

- Line up like terms in columns . 3x3 – 5x2 + 7x – 3- Add . + 2x3 + 3x + 5 Leave space for the missing term .

5x3 – 5x2 +10x + 2

Example: Find the difference of (5x2 – 2x + 3) – (2x2 – 5) .

Steps Solution

- Line up like terms in columns: 5x2 – 2x + 3 Subtrahend

- Change signs in minuend and add: + - 2x2 + 5 Minuend

(Leave space for the missing term .) 3x2 – 2x + 8 Difference

• The opposite of the polynomial: - p: the opposite of the polynomialp: polynomial p + (-p) = 0

Example: Write two expressions for the opposite of the polynomial .7a4b2 – 3a3b – 4a2

Solution: opposite expression: - (7a4b2 – 3a3b – 4a2)or -7a4b2 + 3a3b + 4a2

Replace each term with its opposite .

Page 5-4

Adding and Subtracting Polynomials

• Adding or subtracting polynomials

Example: Find the sum of 2x3 – 3x2 + x – 4 and x3 + 4x2 + 2x + 1 .

Steps Solution

(2x3 – 3x2 + x – 4) + (x3 + 4x2 + 2x + 1)

- Regroup like terms . = (2x3 + x3) + (-3x2 + 4x2) + (x + 2x) + (-4 + 1)

- Combine like terms . = 3x3 + x2 + 3x – 3

Example: Find the difference of 5x2 + 4x – 2 and 2x2 – 3x + 13 .

Steps Solution

(5x2 + 4x – 2) – (2x2 – 3x + 13)

- Remove parentheses . = 5x2 + 4x – 2 – 2x2 + 3x – 13(Reverse each sign in second parentheses .)

- Regroup like terms . = (5x2 – 2x2) + (4x + 3x) + (-2 – 13)

- Combine like terms . = 3x2 + 7x – 15

• Column method

Example: Find the sum of 3x3 – 5x2 + 7x – 3 and 2x3 + 3x + 5 .

Steps Solution

- Line up like terms in columns . 3x3 – 5x2 + 7x – 3- Add . + 2x3 + 3x + 5 Leave space for the missing term .

5x3 – 5x2 +10x + 2

Example: Find the difference of (5x2 – 2x + 3) – (2x2 – 5) .

Steps Solution

- Line up like terms in columns: 5x2 – 2x + 3 Subtrahend

- Change signs in minuend and add: + - 2x2 + 5 Minuend

(Leave space for the missing term .) 3x2 – 2x + 8 Difference

• The opposite of the polynomial: - p: the opposite of the polynomialp: polynomial p + (-p) = 0

Example: Write two expressions for the opposite of the polynomial .7a4b2 – 3a3b – 4a2

Solution: opposite expression: - (7a4b2 – 3a3b – 4a2)or -7a4b2 + 3a3b + 4a2

Replace each term with its opposite .

Page 5-4





Page 5-5

5-2 MULTIPLYING POLYNOMIALS

Multiplication of Polynomials

• Multiplying monomials

Example: (-2x3 y2)(3x2

y4) = (-2 ∙ 3)(x3∙ x2)(y2 ∙ y4) Multiply the coefficients and add exponents.

= -6x5 y6 am ∙ an = am+n

• Multiplying monomial and binomial

Example: 3x3 (5x2 – 2x) = (3x3)(5x2) – (3x3)(2x) Distributive property: a (b + c) = ab + ac

= (3∙5)(x3+2) – (3∙2)(x3+1) Multiply the coefficients and add exponents.

= 15x5 – 6x4 am ∙ an = am+n

Note: The distributive property can be used to multiply a polynomial by a monomial.

Example: 5ab2 (2a2b + ab2 – a) Distribute

= (5ab2)(2a2b) + (5ab2)(ab2) + (5ab2)(-a) Multiply the coefficients and add exponents.

= 10a3b3 + 5a2b4 – 5a2b2 am ∙ an = am+n

• Multiplying binomial and polynomial

Example: (3a2 + 5) (2a2 + a – 3)

= (3a2) (2a2) + (3a2) a + (3a2)(-3) + 5(2a2) + 5a + 5(-3) Distribute

= 6a4 + 3a3 – 9a2 + 10a2 + 5a – 15 Combine like terms.

= 6a4 + 3a3 + a2 + 5a – 15 Tip: The distributive property is handy to get rid of parentheses.

• Column method Example: Find the product: 2a2 + a – 3 and 3a2 + 5.

Steps Solution

- Line up like terms in columns. 2a2 + a – 3 (Leave space for the missing term.) × 3a2 + 5

10a2 + 5a –15 5 times (2a2 + a – 3) - Multiply. + 6a4 + 3a3 – 9a2

3a2 times (2a2 + a – 3)

6a4 + 3a3 + a2 + 5a – 15

Tip: the same as 213 × 102

426 + 213 21726

Page 5-5






= -6x5 y6 am ∙ an = am+n




= 15x5 – 6x4 am ∙ an = am+n




= 10a3b3 + 5a2b4 – 5a2b2 am ∙ an = am+n


Example: (3a2 + 5) (2a2 + a – 3)





Steps Solution



3a2 times (2a2 + a – 3)

6a4 + 3a3 + a2 + 5a – 15


426 + 213 21726

Page 5-5






= -6x5 y6 am ∙ an = am+n




= 15x5 – 6x4 am ∙ an = am+n




= 10a3b3 + 5a2b4 – 5a2b2 am ∙ an = am+n


Example: (3a2 + 5) (2a2 + a – 3)





Steps Solution



3a2 times (2a2 + a – 3)

6a4 + 3a3 + a2 + 5a – 15


426 + 213 21726





FOIL Method to Multiply Binomials

• The FOIL method: an easy way to find the product of two binomials .

(a + b) (c + d) = ac + ad + bc + bd F O I L Example

F - First terms first term × first term (a + b) (c + d) (x + 2) (x + 3)O - Outer terms outside term × outside term (a + b) (c + d) (x + 2) (x + 3)I - Inner terms inside term × inside term (a + b) (c + d) (x + 2) (x + 3)L - Last terms last term × last term (a + b) (c + d) (x + 2) (x + 3)

FOIL Method Example(a + b) (c + d) = ac + ad + bc + bd F O I L

(x + 2) (x + 3) = x·x + x·3 + 2·x + 2∙3 = x2 + 5x + 6 F O I L

Tip: - Multiplication of binomials also uses distributive property .

- Each term of one binomial multiplied by each term of the other by repeatedly using the distributive property . (x + 2)(x + 3) = x(x + 3) + 2(x + 3) = x·x + x∙3 + 2∙x + 2∙3

• Multiplying binomials (2 terms × 2 terms)

Example: Find the following products .

1. (2x – 3) (3x – 4) = 2x ∙ 3x + 2x (- 4) – 3 ∙ 3x – 3 (-4) FOIL

= 6x2 – 8x – 9x + 12 anam = a n+ m

= 6x2 – 17x + 12 Combine like terms .

2. (3r – t) (5r + t2) = 3r ∙ 5r + 3r ∙ t2 – t ∙ 5r – t ∙ t2 FOIL

= 15r2 + 3r t2 – 5r t – t3anam = a n+ m

3. (xy2 + y) (2x2y + x) = xy2 ∙ 2x2 y + xy2 ∙ x + y ∙ 2x2 y + y ∙ x FOIL

= 2x3 y3 + x2y2 + 2x2 y2 + x y anam = a n+ m

= 2x3 y3 + 3x2y2 + x y Combine like terms .

4. (a + 2) (a + 1) (a – 1) = (a2 + 3a + 2) (a – 1) FOIL: (a + 2) (a + 1)

= a3– a2 + 3a2 – 3a + 2a – 2 Distribute

= a3 + 2a2 – a – 2 Combine like terms .

F O I L

Page 5-6

FOIL Method to Multiply Binomials


(a + b) (c + d) = ac + ad + bc + bd F O I L Example

F - First terms first term × first term (a + b) (c + d) (x + 2) (x + 3)O - Outer terms outside term × outside term (a + b) (c + d) (x + 2) (x + 3)I - Inner terms inside term × inside term (a + b) (c + d) (x + 2) (x + 3)L - Last terms last term × last term (a + b) (c + d) (x + 2) (x + 3)

FOIL Method Example(a + b) (c + d) = ac + ad + bc + bd F O I L

(x + 2) (x + 3) = x·x + x·3 + 2·x + 2∙3 = x2 + 5x + 6 F O I L

Tip: - Multiplication of binomials also uses distributive property .

- Each term of one binomial multiplied by each term of the other by repeatedly using the distributive property . (x + 2)(x + 3) = x(x + 3) + 2(x + 3) = x·x + x∙3 + 2∙x + 2∙3

• Multiplying binomials (2 terms × 2 terms)


1. (2x – 3) (3x – 4) = 2x ∙ 3x + 2x (- 4) – 3 ∙ 3x – 3 (-4) FOIL

= 6x2 – 8x – 9x + 12 anam = a n+ m

= 6x2 – 17x + 12 Combine like terms .

2. (3r – t) (5r + t2) = 3r ∙ 5r + 3r ∙ t2 – t ∙ 5r – t ∙ t2 FOIL

= 15r2 + 3r t2 – 5r t – t3anam = a n+ m

3. (xy2 + y) (2x2y + x) = xy2 ∙ 2x2 y + xy2 ∙ x + y ∙ 2x2 y + y ∙ x FOIL

= 2x3 y3 + x2y2 + 2x2 y2 + x y anam = a n+ m

= 2x3 y3 + 3x2y2 + x y Combine like terms .

4. (a + 2) (a + 1) (a – 1) = (a2 + 3a + 2) (a – 1) FOIL: (a + 2) (a + 1)

= a3– a2 + 3a2 – 3a + 2a – 2 Distribute

= a3 + 2a2 – a – 2 Combine like terms .

F O I L

Page 5-6





Page 5-7

Special Binomial Products

• Special binomial products – squaring binominals Special Products Formula Initial Expansion Example

difference of squares

(a + b)(a – b) = a2 – b2

It does not matter if (a – b) comes first (a + b)(a – b) = a2 – ab + ba – b2

= a2 – b2 (x + 2)(x – 2) = x2 –22 = x2 – 4 or (x – 2)(x + 2) = x2 –22 = x2 – 4

square of sum (a + b)2 = a2 + 2ab + b2

A perfect square trinomial (a + b)2 = (a + b)(a + b)

= a2 + ab + ba + b2 = a2 + 2ab + b2

(x + 3)2 = x2 + 2∙ x∙ 3 + 32

= x2 + 6x + 9

square of difference

(a – b)2 = a2 – 2ab + b2

A perfect square trinomial (a – b)2 = (a – b)(a – b)

= a2 –ab – ba + b2 = a2 – 2ab + b2

(x – 4)2 = x2 – 2∙ x∙ 4 + 42

= x2 – 8x + 16

• Special binomial products: special forms of binomial products that are worth memorizing.

• Memory aid: (a ± b)2 = (a2 ± 2ab + b2)

Example: Find the following products.

1. (3y + 4)(3y – 4) = (3y)2 – 42 (a + b) (a – b) = a2 – b2 = 9y2 – 16 a = 3y , b = 4

2. �𝟓𝟓𝟓𝟓𝒕𝒕𝒕𝒕 + 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐�2

= (5t)2 + 2(5t)�12� + �1

2�2

(a + b)2 = a2 + 2ab + b2 = 25t2 + 5t + 𝟏𝟏𝟏𝟏

𝟒𝟒𝟒𝟒 a = 5t , b = 1

2

3. (3q – 𝟏𝟏𝟏𝟏𝟔𝟔𝟔𝟔𝒑𝒑𝒑𝒑) 2 = (3q)2 – 2(3q)� 1

6𝑝𝑝𝑝𝑝� + � 1

6𝑝𝑝𝑝𝑝�

2

(a – b)2 = a2 – 2ab + b2

= 9q2 – q p + 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑𝟔𝟔𝟔𝟔𝒑𝒑𝒑𝒑2 a = 3q , b = 1

6𝑝𝑝𝑝𝑝

4. (t + 𝟏𝟏𝟏𝟏)3 = (t + 1)2 (t + 1) anam = a n+ m

= (t2 + 2t + 1) (t + 1) (a + b)2 = a2 + 2ab + b2

= t3 + t2 + 2t2 + 2t + t + 1 Distribute

= t3 + 3t2 + 3t + 1 Combine like terms.

5. (2A – 3 + 4B)(2A – 3 – 4B) = (2A – 3)2 – (4B)2 (a + b) (a – b) = a2 - b2 : a = 2A – 3, b = 4B

= (2A)2 – 2(2A)∙3 + 32 – 16B2 (a – b)2 = a2 – 2ab + b2 : a = 2A, b = 3

= 4A2 – 12A + 9 – 16B2 Simplify

• Using function notation:

Example: Given f (x) = -3x + x2 , find and simplify 1. f (u – 1) , and 2. f (a + h) – f (a) .

1. f (u – 1) = -3(u – 1) + (u – 1)2 Replace x with (u – 1)

= -3u + 3 + u2 – 2u + 1 (a – b)2 = a2 –2ab + b2 = u2- 5u + 4 Combine like terms.

2. f (a + h) – f (a) = [-3(a + h) + (a + h)2] – (-3a + a2) Replace x with (a + h) and a.

= -3a – 3h + a2 + 2ah + h2 + 3a – a2 Remove parentheses.

= h2 + 2ah – 3h Combine like terms.

a

a b a b

b

( a = x , b = 2 )

Page 5-7

Special Binomial Products

• Special binomial products – squaring binominals Special Products Formula Initial Expansion Example

difference of squares

(a + b)(a – b) = a2 – b2

It does not matter if (a – b) comes first (a + b)(a – b) = a2 – ab + ba – b2

= a2 – b2 (x + 2)(x – 2) = x2 –22 = x2 – 4 or (x – 2)(x + 2) = x2 –22 = x2 – 4

square of sum (a + b)2 = a2 + 2ab + b2

A perfect square trinomial (a + b)2 = (a + b)(a + b)

= a2 + ab + ba + b2 = a2 + 2ab + b2

(x + 3)2 = x2 + 2∙ x∙ 3 + 32

= x2 + 6x + 9

square of difference

(a – b)2 = a2 – 2ab + b2

A perfect square trinomial (a – b)2 = (a – b)(a – b)

= a2 –ab – ba + b2 = a2 – 2ab + b2

(x – 4)2 = x2 – 2∙ x∙ 4 + 42

= x2 – 8x + 16

• Special binomial products: special forms of binomial products that are worth memorizing.

• Memory aid: (a ± b)2 = (a2 ± 2ab + b2)

Example: Find the following products.

1. (3y + 4)(3y – 4) = (3y)2 – 42 (a + b) (a – b) = a2 – b2 = 9y2 – 16 a = 3y , b = 4

2. �𝟓𝟓𝟓𝟓𝒕𝒕𝒕𝒕 + 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐�2

= (5t)2 + 2(5t)�12� + �1

2�2

(a + b)2 = a2 + 2ab + b2 = 25t2 + 5t + 𝟏𝟏𝟏𝟏

𝟒𝟒𝟒𝟒 a = 5t , b = 1

2

3. (3q – 𝟏𝟏𝟏𝟏𝟔𝟔𝟔𝟔𝒑𝒑𝒑𝒑) 2 = (3q)2 – 2(3q)� 1

6𝑝𝑝𝑝𝑝� + � 1

6𝑝𝑝𝑝𝑝�

2

(a – b)2 = a2 – 2ab + b2

= 9q2 – q p + 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑𝟔𝟔𝟔𝟔𝒑𝒑𝒑𝒑2 a = 3q , b = 1

6𝑝𝑝𝑝𝑝

4. (t + 𝟏𝟏𝟏𝟏)3 = (t + 1)2 (t + 1) anam = a n+ m

= (t2 + 2t + 1) (t + 1) (a + b)2 = a2 + 2ab + b2

= t3 + t2 + 2t2 + 2t + t + 1 Distribute

= t3 + 3t2 + 3t + 1 Combine like terms.

5. (2A – 3 + 4B)(2A – 3 – 4B) = (2A – 3)2 – (4B)2 (a + b) (a – b) = a2 - b2 : a = 2A – 3, b = 4B

= (2A)2 – 2(2A)∙3 + 32 – 16B2 (a – b)2 = a2 – 2ab + b2 : a = 2A, b = 3

= 4A2 – 12A + 9 – 16B2 Simplify

• Using function notation:

Example: Given f (x) = -3x + x2 , find and simplify 1. f (u – 1) , and 2. f (a + h) – f (a) .

1. f (u – 1) = -3(u – 1) + (u – 1)2 Replace x with (u – 1)

= -3u + 3 + u2 – 2u + 1 (a – b)2 = a2 –2ab + b2 = u2- 5u + 4 Combine like terms.

2. f (a + h) – f (a) = [-3(a + h) + (a + h)2] – (-3a + a2) Replace x with (a + h) and a.

= -3a – 3h + a2 + 2ah + h2 + 3a – a2 Remove parentheses.

= h2 + 2ah – 3h Combine like terms.

a

a b a b

b

( a = x , b = 2 )





5-3 FACTORING

Highest / Greatest Common Factor

• Greatest / highest common factor (GCF or HCF): the largest factor of two or more terms .

Example Factors GCF or HCF369

3 3 ∙ 23 ∙ 3

3

2x2y8x2y3

2 ∙ x2 ∙ y 2 ∙ 4 ∙ x2 ∙ y ∙ y2 2 x2 y

• Factoring a polynomial: express a polynomial as a product of other polynomials . It is the

reverse of multiplying . Multiplying (or expanding)

a (b + c + d) = ab + ac + ad Factoring

ExampleMultiplying Factoring GCF

2ab (4a –3ab + 1) 8a2b – 6a2b2 + 2ab

= 8a2b – 6a2b2 + 2ab = 2ab (4a) – 2ab (3ab) + 2ab ∙ 1 2ab

= 2ab (4a – 3ab + 1)

Examples

Expression Factoring GCF-8x2 – 4x -4x · 2x – 4x · 1 = -4x (2x + 1) -4x3y5 – 9y3 + 6y 3y· y 4 – 3· 3y· y 2 + 3y · 2 = 3y ( y 4 – 3y 2 + 2) 3y5a3b + 10a2b – 15ab 5ab · a 2 + 5ab·2a – 5ab · 3 = 5ab(a 2 + 2a – 3) 5ab7x2 ( x + 5) – (3x + 15) 7x2 ( x + 5) – 3 ( x + 5) = (x + 5)(7x2 – 3) ( x + 5)2x2 + 3y + 4 Not factorable No

Tip: Factor each term and pull out the GCF .

• Negative of the greatest common factor

Factoring GCF Factor Out a Negative GCF Negative GCF2x – 4x2 = 2x (1 – 2x) 2x 2x – 4x2 = -2x (-1 + 2x) - 2x

3ab – 9ab2 + 6a2b = 3ab (1– 3b + 2a) 3ab 3ab – 9ab2 + 6a2b = -3ab (-1+ 3b – 2a) -3ab

Page 5-8

5-3 FACTORING




3 3 ∙ 23 ∙ 3

3

2x2y8x2y3

2 ∙ x2 ∙ y 2 ∙ 4 ∙ x2 ∙ y ∙ y2 2 x2 y





2ab (4a –3ab + 1) 8a2b – 6a2b2 + 2ab


= 2ab (4a – 3ab + 1)

Examples






Page 5-8

5-3 FACTORING




3 3 ∙ 23 ∙ 3

3

2x2y8x2y3

2 ∙ x2 ∙ y 2 ∙ 4 ∙ x2 ∙ y ∙ y2 2 x2 y





2ab (4a –3ab + 1) 8a2b – 6a2b2 + 2ab


= 2ab (4a – 3ab + 1)

Examples






Page 5-8





Factoring Polynomials by Grouping

Steps for factoring by grouping:

Steps Example: 8y2 – 2 y + 12y – 3

- Group terms with the GCF . 8y2 – 2 y + 12y – 3 = (8y2 – 2 y) + (12y – 3)

- Factor out the GCF from each group . = 2y (4y –1) + 3(4y –1)

- Factor out the GCF again from the last step . = (4y – 1)(2y + 3)

Factoring completely: Continue factoring until no further factors can be found .

Example: Factor the following completely .

1. 6ab2 – 3a2b + 2b – a = (6ab2 – 3a2b ) + (2b – a) Group terms with the GCF .

= 3ab (2b – a) + (2b – a) ∙ 1 Factor out the GCF ; (2b – a) = (2b – a) ∙ 1

= (2b – a) (3ab + 1) Factor out the GCF again .

2. 2ab + bc – 2bc + 4ab = (2ab + 4ab) + (bc – 2bc)

= 6ab – bc Combine like terms .

= b(6a – c) Factor out the GCF .

3. x3 – xy2 – x2y + y3 = (x3 – x2y) – (xy2 – y3) Group

= x2 (x – y) – y2 (x – y) Factor out the GCF .

= (x – y) (x2 – y2) a2 – b2 = (a + b)(a – b)

= (x – y) (x + y)(x – y) Keep factoring until cannot factor any further .

= (x – y)2 (x + y)

Tip: Recognize factoring patterns, such as 2b – a, x – y, …

4. 32x3y – 2xy3 = 2xy (16x2 – y2) Factor out the GCF .

= 2xy [(4x)2 – y2)] a2 – b2 = (a + b)(a – b)

= 2xy (4x + y) (4x – y)

Rearrange and group terms with the same pattern .

(4x + y) and (4x – y) cannot be factored further .

Page 5-9

Factoring Polynomials by Grouping

Steps for factoring by grouping:

Steps Example: 8y2 – 2 y + 12y – 3

- Group terms with the GCF . 8y2 – 2 y + 12y – 3 = (8y2 – 2 y) + (12y – 3)

- Factor out the GCF from each group . = 2y (4y –1) + 3(4y –1)

- Factor out the GCF again from the last step . = (4y – 1)(2y + 3)

Factoring completely: Continue factoring until no further factors can be found .


1. 6ab2 – 3a2b + 2b – a = (6ab2 – 3a2b ) + (2b – a) Group terms with the GCF .

= 3ab (2b – a) + (2b – a) ∙ 1 Factor out the GCF ; (2b – a) = (2b – a) ∙ 1

= (2b – a) (3ab + 1) Factor out the GCF again .

2. 2ab + bc – 2bc + 4ab = (2ab + 4ab) + (bc – 2bc)

= 6ab – bc Combine like terms .

= b(6a – c) Factor out the GCF .

3. x3 – xy2 – x2y + y3 = (x3 – x2y) – (xy2 – y3) Group

= x2 (x – y) – y2 (x – y) Factor out the GCF .

= (x – y) (x2 – y2) a2 – b2 = (a + b)(a – b)

= (x – y) (x + y)(x – y) Keep factoring until cannot factor any further .

= (x – y)2 (x + y)

Tip: Recognize factoring patterns, such as 2b – a, x – y, …

4. 32x3y – 2xy3 = 2xy (16x2 – y2) Factor out the GCF .

= 2xy [(4x)2 – y2)] a2 – b2 = (a + b)(a – b)

= 2xy (4x + y) (4x – y)

Rearrange and group terms with the same pattern .

(4x + y) and (4x – y) cannot be factored further .

Page 5-9





Factoring x2 + bx + c

Factoring x2 + bx + c : Cross-multiplication methodSteps Standard form Example

x2 + bx + c x2 + 3x + 2- Setting up two sets of parentheses . = ( ) ( ) = ( ) ( )

x2 + bx + c x2 + 3x + 2- Factor the first term x2: x2 = x · x x c1 x 1- Factor the last term c (by trial and error): c = c1· c2 x c2 x 2

x · x = x2 c1 · c2 = c x · x = x2 1∙ 2 = 2

- Cross multiply and then add up to the middle term . (c1)(x) + (c2)(x) = bx 1∙ x + 2 ∙ x = 3x - Complete the parenthesis with x + c1 and x + c2. x2 + bx + c x2 + 3x + 2

= (x + c1)(x + c2) = (x + 1)(x + 2)F O I L

- Check using FOIL . (x + 1)(x + 2) = x2 + 2x + x + 2(x + 1)(x + 2) = x2 + 3x + 2 √

Factoring x2 + bx + c Using the Cross-Multiplication MethodIn general

x2 + bx + c = ( ) ( )x c1 x c2

x · x = x2 c1 · c2 = c

(c1)(x) + (c2)(x) = bx

x2 + bx + c= (x + c1 )(x + c2)

Examplex2 – 5 x + 6 = ( ) ( )x -2x -3

x · x = x2 (-2)(-3) = 6

-2∙x + (-3)x = -5x yes!x2 – 5x + 6 = (x – 2) (x – 3)

Summary: Factoring x2 + bx + c Example: x2 – 5x + 6x2 + (c1 + c2) x + c1c2 = (x + c1)(x + c2)x c1x c2Check: c1x+ c2 x = bx

x2 + [-2 + (-3)] x + 6 = (x – 2)(x – 3) x -2 x -3

Check: -2x + (-3x) = -5x yes!

Tip: Cross multiply and then add up to the middle term .

Example: Factor the following: Trial and Error Process1. x2 – 6x + 8 = ( ) ( ) x2 – 6x + 8 x2 – 6x + 8

x -2 x 2 x 1 x -4 x 4 x 8

x · x = x2 (-2)(-4) = 8

-2∙x + (-4)x = -6x yes! 2x + 4x = -6x no 1∙ x + 8x = -6x no

Answer: x2 – 6x + 8 = (x – 2)( x – 4) Check: -2 + (-4) = -6 √

2. a2 + 5a – 6 = ( ) ( ) a2 + 5a – 6 a2 + 5a – 6 a - 1 a 2 a 1

a 6 a -3 a - 6 a · a = a2 (-1)(6) = -6

(-1) a + 6a = 5a yes! 2a + (-3)a = 5a no 1∙ a + (-6 )a = 5a noAnswer: a2 + 5a – 6 = (a – 1)(a + 6) Check: -1 + 6 = 5 √

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Page 5-10

Factoring x2 + bx + c

Factoring x2 + bx + c : Cross-multiplication methodSteps Standard form Example

x2 + bx + c x2 + 3x + 2- Setting up two sets of parentheses . = ( ) ( ) = ( ) ( )

x2 + bx + c x2 + 3x + 2- Factor the first term x2: x2 = x · x x c1 x 1- Factor the last term c (by trial and error): c = c1· c2 x c2 x 2

x · x = x2 c1 · c2 = c x · x = x2 1∙ 2 = 2

- Cross multiply and then add up to the middle term . (c1)(x) + (c2)(x) = bx 1∙ x + 2 ∙ x = 3x - Complete the parenthesis with x + c1 and x + c2. x2 + bx + c x2 + 3x + 2

= (x + c1)(x + c2) = (x + 1)(x + 2)F O I L

- Check using FOIL . (x + 1)(x + 2) = x2 + 2x + x + 2(x + 1)(x + 2) = x2 + 3x + 2 √

Factoring x2 + bx + c Using the Cross-Multiplication MethodIn general

x2 + bx + c = ( ) ( )x c1 x c2

x · x = x2 c1 · c2 = c

(c1)(x) + (c2)(x) = bx

x2 + bx + c= (x + c1 )(x + c2)

Examplex2 – 5 x + 6 = ( ) ( )x -2x -3

x · x = x2 (-2)(-3) = 6

-2∙x + (-3)x = -5x yes!x2 – 5x + 6 = (x – 2) (x – 3)

Summary: Factoring x2 + bx + c Example: x2 – 5x + 6x2 + (c1 + c2) x + c1c2 = (x + c1)(x + c2)x c1x c2Check: c1x+ c2 x = bx

x2 + [-2 + (-3)] x + 6 = (x – 2)(x – 3) x -2 x -3

Check: -2x + (-3x) = -5x yes!

Tip: Cross multiply and then add up to the middle term .

Example: Factor the following: Trial and Error Process1. x2 – 6x + 8 = ( ) ( ) x2 – 6x + 8 x2 – 6x + 8

x -2 x 2 x 1 x -4 x 4 x 8

x · x = x2 (-2)(-4) = 8

-2∙x + (-4)x = -6x yes! 2x + 4x = -6x no 1∙ x + 8x = -6x no

Answer: x2 – 6x + 8 = (x – 2)( x – 4) Check: -2 + (-4) = -6 √

2. a2 + 5a – 6 = ( ) ( ) a2 + 5a – 6 a2 + 5a – 6 a - 1 a 2 a 1

a 6 a -3 a - 6 a · a = a2 (-1)(6) = -6

(-1) a + 6a = 5a yes! 2a + (-3)a = 5a no 1∙ a + (-6 )a = 5a noAnswer: a2 + 5a – 6 = (a – 1)(a + 6) Check: -1 + 6 = 5 √

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Page 5-10





Page 5-11

5-4 FACTORING ax 2 + bx + c

Factoring Trinomials: ax2 + bx + c

Procedure for factoring ax2 + bx + c using the cross-multiplication method

Steps In general Example

ax2 + bx + c 2x2 + x – 3 - Setting up two sets of parenthesis. = ( ) ( ) = ( )( )

ax2 + bx + c 2x2 + x – 3 - Factor the first term ax2: ax2 = a1x ∙ a2x a1x c1 x -1

- Factor the last term c (by trial and error): a2x c2 2x 3 c = c1 ∙ c2 a1x ∙ a2x = ax2 c1 ∙ c2 = c x ∙ 2x = 2x2 -1∙ 3 = -3

- Cross-multiply, then add up to the middle term. (a1x) (c2) + (a2x)(c1) = bx 3 ∙ x + (-1)(2x) = x - Complete the parenthesis with (a1x + c1) and (a2x + c2). ax2 + bx + c 2x2 + x – 3

= (a1x + c1)(a2x + c2) = (x – 1)(2x + 3) F O I L

- Check using FOIL. (x – 1)(2x + 3) = 2x2 + 3x – 2x – 3 √

(x – 1)(2x + 3) = 2x2 + x – 3

Factoring ax2 + bx + c Using the Cross-Multiplication Method In general

ax2 + bx + c = ( ) ( ) a1x c1 a2x c2

a1x ∙ a2x = ax2 c = c1∙ c2

(a1x) (c2) + (a2x)(c1) = bx ax2 + bx + c = (a1x + c1)(a2x + c2)

Example 3x2 + 10 x + 8 = ( ) ( ) 3x 4 x 2

3x ∙ x = 3x2 4 ∙ 2 = 8

3x ∙ 2 + 4 ∙ x = 10x yes! 3x2 + 10x + 8 = (3x + 4) (x + 2)

Summary: Factoring ax2 + bx + c a1 a2x2 + (a1c2 + c1a2)x + c1 c2 = (a1x + c1)(a2x + c2) a1x c1

a2x c2 Tip: Cross-multiply and then add up to the middle term.

(Original expression)

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Page 5-11

5-4 FACTORING ax 2 + bx + c

Factoring Trinomials: ax2 + bx + c

Procedure for factoring ax2 + bx + c using the cross-multiplication method

Steps In general Example

ax2 + bx + c 2x2 + x – 3 - Setting up two sets of parenthesis. = ( ) ( ) = ( )( )

ax2 + bx + c 2x2 + x – 3 - Factor the first term ax2: ax2 = a1x ∙ a2x a1x c1 x -1

- Factor the last term c (by trial and error): a2x c2 2x 3 c = c1 ∙ c2 a1x ∙ a2x = ax2 c1 ∙ c2 = c x ∙ 2x = 2x2 -1∙ 3 = -3

- Cross-multiply, then add up to the middle term. (a1x) (c2) + (a2x)(c1) = bx 3 ∙ x + (-1)(2x) = x - Complete the parenthesis with (a1x + c1) and (a2x + c2). ax2 + bx + c 2x2 + x – 3

= (a1x + c1)(a2x + c2) = (x – 1)(2x + 3) F O I L

- Check using FOIL. (x – 1)(2x + 3) = 2x2 + 3x – 2x – 3 √

(x – 1)(2x + 3) = 2x2 + x – 3

Factoring ax2 + bx + c Using the Cross-Multiplication Method In general

ax2 + bx + c = ( ) ( ) a1x c1 a2x c2

a1x ∙ a2x = ax2 c = c1∙ c2

(a1x) (c2) + (a2x)(c1) = bx ax2 + bx + c = (a1x + c1)(a2x + c2)

Example 3x2 + 10 x + 8 = ( ) ( ) 3x 4 x 2

3x ∙ x = 3x2 4 ∙ 2 = 8

3x ∙ 2 + 4 ∙ x = 10x yes! 3x2 + 10x + 8 = (3x + 4) (x + 2)

Summary: Factoring ax2 + bx + c a1 a2x2 + (a1c2 + c1a2)x + c1 c2 = (a1x + c1)(a2x + c2) a1x c1

a2x c2 Tip: Cross-multiply and then add up to the middle term.

(Original expression)

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5-4 FACTORING ax2 + bx + c





More Examples for Factoring ax2 + bx + c

Example: Factor the following . Trial and Error Process

8x2 + 10x – 25 = ( ) ( ) 1 . 8x2 + 10x – 252x 5 x -54x -5 8x 5

2x ∙ 4x = 8x 2 5(-5) = -25 x ∙ 5 + (-5)(8x) = 10x no

? 2 . 8x2 + 10x – 25 (2x)(-5) + 5(4x) = 10x yes! 4x 5

2x -5

8x2 + 10x – 25 = (2x + 5) (4x – 5) (4x)(-5) + 5(2x) = 10x no

3 . 8x2 + 10x – 258x -25

x 1

8x ∙1 + (-25)x = 10x no

Tip: Write the factors with their appropriate signs (+ or –) to get the right middle term .

Check: (2x + 5)(4x – 5) = 8x2 – 10x + 20x – 25F O I L

(2x + 5)(4x – 5) = 8x2 + 10x – 25 Correct!


1. 3y2 (y + 4) + (y + 4) (y – 2) = (y + 4)[3y2 + (y – 2)] Factor out the GCF (y + 4) .

= (y + 4)(3y2 + y – 2) y 1

3y -2 -2y + 3y = y √ = (y + 4)(y + 1)(3y – 2)

2. m2 – m = - 𝟏𝟏𝟒𝟒

m2 – m + 14

= ( ) ( ) Write in standard form

m - 12

m - 12 - 1

2𝑚𝑚 − 1

2𝑚𝑚 = − 𝑚𝑚

m2 – m + 14

= �𝑚𝑚 – 12� �𝑚𝑚 – 1

2�

= �𝒎𝒎 – 𝟏𝟏𝟐𝟐�𝟐𝟐

3. 3s2 – 2st – 5t2 = ( ) ( )s t -5st + 3st = -2st √

3s -5t 3s2 – 2st – 5t2 = (s + t) (3s – 5t)

4. 2q4 + 14q2 + 20 = 2(q4 + 7q2 + 10) = 2 ( ) ( ) Factor out the GCF (2) .q2 2 q2 5

2q4 + 14q2 + 20 = 2(q2 + 2) (q2 + 5) 5q2 + 2q2 = 7q2 √

?

?

?

√

(ax2 + bx + c) .

Page 5-12

More Examples for Factoring ax2 + bx + c

Example: Factor the following . Trial and Error Process

8x2 + 10x – 25 = ( ) ( ) 1 . 8x2 + 10x – 252x 5 x -54x -5 8x 5

2x ∙ 4x = 8x 2 5(-5) = -25 x ∙ 5 + (-5)(8x) = 10x no

? 2 . 8x2 + 10x – 25 (2x)(-5) + 5(4x) = 10x yes! 4x 5

2x -5

8x2 + 10x – 25 = (2x + 5) (4x – 5) (4x)(-5) + 5(2x) = 10x no

3 . 8x2 + 10x – 258x -25

x 1

8x ∙1 + (-25)x = 10x no

Tip: Write the factors with their appropriate signs (+ or –) to get the right middle term .

Check: (2x + 5)(4x – 5) = 8x2 – 10x + 20x – 25F O I L

(2x + 5)(4x – 5) = 8x2 + 10x – 25 Correct!


1. 3y2 (y + 4) + (y + 4) (y – 2) = (y + 4)[3y2 + (y – 2)] Factor out the GCF (y + 4) .

= (y + 4)(3y2 + y – 2) y 1

3y -2 -2y + 3y = y √ = (y + 4)(y + 1)(3y – 2)

2. m2 – m = - 𝟏𝟏𝟒𝟒

m2 – m + 14

= ( ) ( ) Write in standard form

m - 12

m - 12 - 1

2𝑚𝑚 − 1

2𝑚𝑚 = − 𝑚𝑚

m2 – m + 14

= �𝑚𝑚 – 12� �𝑚𝑚 – 1

2�

= �𝒎𝒎 – 𝟏𝟏𝟐𝟐�𝟐𝟐

3. 3s2 – 2st – 5t2 = ( ) ( )s t -5st + 3st = -2st √

3s -5t 3s2 – 2st – 5t2 = (s + t) (3s – 5t)

4. 2q4 + 14q2 + 20 = 2(q4 + 7q2 + 10) = 2 ( ) ( ) Factor out the GCF (2) .q2 2 q2 5

2q4 + 14q2 + 20 = 2(q2 + 2) (q2 + 5) 5q2 + 2q2 = 7q2 √

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√

(ax2 + bx + c) .

Page 5-12





Factoring Trinomials: AC Method

AC method for factoring trinomials: ax2 + bx + c

Factoring ax2 + bx + c = 0 by Grouping Example

Steps Solve 14x – 6 = -12x2

• Convert to standard form if necessary . 12x2 + 14x – 6 = 0• Factor out the greatest common factor (GCF) . 2(6x2 + 7x – 3) = 0• Multiply a and c in ax2 + bx + c . ac = 6 (-3) = -18• Factor the product ac that sum to the middle coefficient b . 9 (-2) = -18, 9 + (-2) = 7• Rewrite the middle term as the sum using the 2(6x2 + 7x – 3) = 0

factors found in last step . 2 (6x2 + 9x – 2x – 3) = 0• Factor by grouping . 2[3x(2x + 3) – (2x + 3)] = 0

2 (2x + 3)(3x – 1) = 0 Factor out (2x + 3) .

Example: Factor 6x2 – 8 = 2x using ac method .

Steps Solution

6x2 – 8 = 2x

- Write in standard form: 6x2 – 2x – 8 = 0

- Factor out the greatest common factor: 2(3x2 – 1x – 4) = 0

- Multiply a and c in ax2 + bx + c : ac = 3 ∙ (-4) = -12- Factor the product ac that sum to the middle coefficient b .

(There are different pairs to get the product of ac of -12 . Try to find twonumbers that multiply to ac and added to obtain b = -1 .)

Some Factors of ac (-12) Sum of Factors ( b = -1)-2 & 6 -2 + 6 = 4

-1 & 12 -1 + 12 = 11-3 & 4 -3 + 4 = 13 & -4 3 + (- 4) = -1 Correct!

The right choices are 3 and -4, since they both add up to b = -1 . 3 (-4) = -12, 3 + (-4) = -1

- Rewrite the middle term as 3x – 4x . 2(3x2 – 1x – 4) = 0

2(3x2 + 3x – 4x – 4) = 0

- Factor by grouping . Factor out (x + 1) 2 [3x (x + 1) – 4(x + 1) = 0

2(x + 1)(3x – 4) = 0

Page 5-13

Factoring Trinomials: AC Method

AC method for factoring trinomials: ax2 + bx + c





2 (2x + 3)(3x – 1) = 0 Factor out (2x + 3) .

Example: Factor 6x2 – 8 = 2x using ac method .

Steps Solution

6x2 – 8 = 2x

- Write in standard form: 6x2 – 2x – 8 = 0

- Factor out the greatest common factor: 2(3x2 – 1x – 4) = 0

- Multiply a and c in ax2 + bx + c : ac = 3 ∙ (-4) = -12- Factor the product ac that sum to the middle coefficient b .

(There are different pairs to get the product of ac of -12 . Try to find twonumbers that multiply to ac and added to obtain b = -1 .)

Some Factors of ac (-12) Sum of Factors ( b = -1)-2 & 6 -2 + 6 = 4

-1 & 12 -1 + 12 = 11-3 & 4 -3 + 4 = 13 & -4 3 + (- 4) = -1 Correct!

The right choices are 3 and -4, since they both add up to b = -1 . 3 (-4) = -12, 3 + (-4) = -1

- Rewrite the middle term as 3x – 4x . 2(3x2 – 1x – 4) = 0

2(3x2 + 3x – 4x – 4) = 0

- Factor by grouping . Factor out (x + 1) 2 [3x (x + 1) – 4(x + 1) = 0

2(x + 1)(3x – 4) = 0

Page 5-13





5-5 FACTORING SPECIAL PRODUCTS

Special Factoring

• Special factoring formulas

Name Formula Exampledifference of squares a2 – b2 = (a + b)(a – b) n2 –25 = n2 – 52 = (n + 5)(n – 5)

square of sum(perfect square trinomial)

a2 + 2ab + b2 = (a + b)2

x2 + 6x + 9 = (x + 3)2 a = x, b = 3x 3x 3Check: (x + 3)2 = x2 + 2 ∙ x ∙ 3 + 32 = x2 + 6x + 9 √

square of difference(perfect square trinomial) a2 – 2ab + b2 = (a – b)2

25t2 – 20t + 4 = (5t – 2)2 a = 5t, b = 25t -25t -2

Check: (5t – 2)2 = (5t)2 – 2(5t)(2) + 22 = 25t2 – 20t + 4 √

Note: The quickest way to factor an expression is to recognize it as a special product .

• Memory aid: (a2 ± ab + b2) = (a ± b)2 Notice the plus or minus sign in the second term .

• To use perfect square trinomial formulas: Use cross-multiplication method to factor a

perfect square . Then use the square formula to check .


1. 9x2 – 16y2 = 32x2 – 42y2 = (3x)2 – (4y)2 anbn = (a b)n

= (3x + 4y)(3x – 4y) a2 – b2 = (a + b)(a – b) : a = 3x , b = 4y

2. 12x + 9 + 4x2 = 4x2 + 12x + 9 Rewrite in standard form: ax2 + bx + c2x 3 2x 3

= (2x + 3)23(2x) + 3(2x) = 12x

Check: (2x + 3)2 = (2x)2 + 2∙2x ∙ 3 + 32 = 4x2 + 12x + 9 √ a2 + 2ab + b2 = (a + b)2

3. 25A2 – 20AB + 4B2 = (5A – 2B)2

5A -2B (5A) (-2B) + (5A) (-2B) = -20AB

5A -2B

Check: (5A – 2B)2 = (5A)2 – 2(5A)(2B) + (2B)2 = 25A2 – 20AB + 4B2 √ a2 – 2ab + b2 = (a – b)2 : a = 5A , b = 2B

4. 𝟐𝟐𝟏𝟏𝟏𝟏

x2 – 𝟐𝟐𝟗𝟗

y2 = 2 � 142𝑥𝑥𝑥𝑥2 – 1

32 𝑦𝑦2� Factor out the GCF (2) . 𝑎𝑎

𝑛𝑛

𝑏𝑏𝑛𝑛= �𝑎𝑎

𝑏𝑏�𝑛𝑛

= 2 ��𝒙𝒙𝟒𝟒�2

– �𝒚𝒚𝟑𝟑�2� a2 – b2 = (a + b)(a – b) : a = 𝑥𝑥

4, b = 𝑦𝑦

3

= 2 ��𝒙𝒙𝟒𝟒

+ 𝒚𝒚𝟑𝟑� �𝒙𝒙

𝟒𝟒− 𝒚𝒚

𝟑𝟑��

5. -12x2 + 36xy – 27y2 = -3(4x2 – 12xy + 9y2) Factor out the GCF (-3) . 2x - 3y

2x - 3y (2x) (-3y) + (2x) (-3y) = -12xy = -3(2x – 3y)2

Check: (2x – 3y)2 = (2x)2 – 2(2x)(3y) + (3y)2 = 4x2 – 12xy + 9y2 √ a2 – 2ab + b2 = (a – b)2 : a = 2x , b = 3y

a b

Factoring (L R)

a2 – b2 = (a + b)(a – b)

Multiplying (L R)

Page 5-14

5-5 FACTORING SPECIAL PRODUCTS

Special Factoring




a2 + 2ab + b2 = (a + b)2


square of difference(perfect square trinomial) a2 – 2ab + b2 = (a – b)2

25t2 – 20t + 4 = (5t – 2)2 a = 5t, b = 25t -25t -2

Check: (5t – 2)2 = (5t)2 – 2(5t)(2) + 22 = 25t2 – 20t + 4 √

Note: The quickest way to factor an expression is to recognize it as a special product .

• Memory aid: (a2 ± ab + b2) = (a ± b)2 Notice the plus or minus sign in the second term .

• To use perfect square trinomial formulas: Use cross-multiplication method to factor a

perfect square . Then use the square formula to check .


1. 9x2 – 16y2 = 32x2 – 42y2 = (3x)2 – (4y)2 anbn = (a b)n

= (3x + 4y)(3x – 4y) a2 – b2 = (a + b)(a – b) : a = 3x , b = 4y

2. 12x + 9 + 4x2 = 4x2 + 12x + 9 Rewrite in standard form: ax2 + bx + c2x 3 2x 3

= (2x + 3)23(2x) + 3(2x) = 12x

Check: (2x + 3)2 = (2x)2 + 2∙2x ∙ 3 + 32 = 4x2 + 12x + 9 √ a2 + 2ab + b2 = (a + b)2

3. 25A2 – 20AB + 4B2 = (5A – 2B)2

5A -2B (5A) (-2B) + (5A) (-2B) = -20AB

5A -2B

Check: (5A – 2B)2 = (5A)2 – 2(5A)(2B) + (2B)2 = 25A2 – 20AB + 4B2 √ a2 – 2ab + b2 = (a – b)2 : a = 5A , b = 2B

4. 𝟐𝟐𝟏𝟏𝟏𝟏

x2 – 𝟐𝟐𝟗𝟗

y2 = 2 � 142𝑥𝑥𝑥𝑥2 – 1

32 𝑦𝑦2� Factor out the GCF (2) . 𝑎𝑎

𝑛𝑛


𝑏𝑏�𝑛𝑛

= 2 ��𝒙𝒙𝟒𝟒�2

– �𝒚𝒚𝟑𝟑�2� a2 – b2 = (a + b)(a – b) : a = 𝑥𝑥

4, b = 𝑦𝑦

3

= 2 ��𝒙𝒙𝟒𝟒

+ 𝒚𝒚𝟑𝟑� �𝒙𝒙

𝟒𝟒− 𝒚𝒚

𝟑𝟑��

5. -12x2 + 36xy – 27y2 = -3(4x2 – 12xy + 9y2) Factor out the GCF (-3) . 2x - 3y

2x - 3y (2x) (-3y) + (2x) (-3y) = -12xy = -3(2x – 3y)2

Check: (2x – 3y)2 = (2x)2 – 2(2x)(3y) + (3y)2 = 4x2 – 12xy + 9y2 √ a2 – 2ab + b2 = (a – b)2 : a = 2x , b = 3y

a b

Factoring (L R)

a2 – b2 = (a + b)(a – b)

Multiplying (L R)

Page 5-14





More Examples for Special Factoring

• Factoring by grouping

Example:

1. p3 – p2q – pq2 + q3 = (p3 – p2q) – (pq2 – q3) Group .

= p2 (p – q) – q2 (p – q) Factor out the GCF (p2 and q2) .

= (p – q) (p2 – q2 ) a2 – b2 = (a + b)(a – b): a = p , b = q

= (p – q) (p + q)(p – q)

= (p – q)2 (p + q)

2. 3s2 – 18st + 27t2 – 75u2 = 3(s2 – 6st + 9t2 – 25u2) Factor out the GCF (3) .

= 3[(s2 – 6st + 9t2) – 52u2] s - 3t -3st + (-3st) = -6st s - 3t amn = (am)n

= 3[(s – 3t)2 – (5u)2] a2 – b2 = (a + b)(a – b)

= 3[(s – 3t) + 5u] [(s – 3t) – 5u] a = s – 3t , b = 5u

• Special factoring of higher degree


1. p10 + 22p5+ 121 = p10 + 22p5 + 121p5 11 p5 11 11p5 + 11p5 = 22 p5

= (p5 + 11)2

Check: (p5 + 11)2 = (p5)2 + 2(p5)(11) + 112 = p10 + 22p5 + 121 √ a2 + 2ab + b2 = (a + b)2 : a = p5 , b = 11

2. u4 – 81 = u2∙ 2 – 92 = (u2)2 – 92amn = (am)n

= (u2 + 9)(u2 – 9) a2 – b2 = (a + b)( a – b): a = u2 , b = 9

= (u2 + 9)(u2 – 32) a2– b2 = (a + b)( a – b) : a = u , b = 3

= (u2 + 9)(u + 3) )(u – 3)

3. 16y8 – x8 = 42y4 ∙ 2 – x4 ∙ 2 = (4y4)2 – (x4)2anbn = (ab)n , amn = (am)n

= (4y4 + x4)(4y4 – x4) a2– b2 = (a + b)( a – b) : a = 4y4 , b = x4

= (4y4 + x4)[(2y2)2 – (x2)2] 4y4 = 22y2 ∙ 2 = (2y2)2

= (4y4 + x4)(2y2 + x2)(2y2 – x2) a2 – b2 = (a + b)( a – b): a = 2y2 , b = x2

a b

a b

a b

a b

Keep factoring until no further factors can be found .

Page 5-15

More Examples for Special Factoring

• Factoring by grouping

Example:

1. p3 – p2q – pq2 + q3 = (p3 – p2q) – (pq2 – q3) Group .

= p2 (p – q) – q2 (p – q) Factor out the GCF (p2 and q2) .

= (p – q) (p2 – q2 ) a2 – b2 = (a + b)(a – b): a = p , b = q

= (p – q) (p + q)(p – q)

= (p – q)2 (p + q)

2. 3s2 – 18st + 27t2 – 75u2 = 3(s2 – 6st + 9t2 – 25u2) Factor out the GCF (3) .

= 3[(s2 – 6st + 9t2) – 52u2] s - 3t -3st + (-3st) = -6st s - 3t amn = (am)n

= 3[(s – 3t)2 – (5u)2] a2 – b2 = (a + b)(a – b)

= 3[(s – 3t) + 5u] [(s – 3t) – 5u] a = s – 3t , b = 5u

• Special factoring of higher degree


1. p10 + 22p5+ 121 = p10 + 22p5 + 121p5 11 p5 11 11p5 + 11p5 = 22 p5

= (p5 + 11)2

Check: (p5 + 11)2 = (p5)2 + 2(p5)(11) + 112 = p10 + 22p5 + 121 √ a2 + 2ab + b2 = (a + b)2 : a = p5 , b = 11

2. u4 – 81 = u2∙ 2 – 92 = (u2)2 – 92amn = (am)n

= (u2 + 9)(u2 – 9) a2 – b2 = (a + b)( a – b): a = u2 , b = 9

= (u2 + 9)(u2 – 32) a2– b2 = (a + b)( a – b) : a = u , b = 3

= (u2 + 9)(u + 3) )(u – 3)

3. 16y8 – x8 = 42y4 ∙ 2 – x4 ∙ 2 = (4y4)2 – (x4)2anbn = (ab)n , amn = (am)n

= (4y4 + x4)(4y4 – x4) a2– b2 = (a + b)( a – b) : a = 4y4 , b = x4

= (4y4 + x4)[(2y2)2 – (x2)2] 4y4 = 22y2 ∙ 2 = (2y2)2

= (4y4 + x4)(2y2 + x2)(2y2 – x2) a2 – b2 = (a + b)( a – b): a = 2y2 , b = x2

a b

a b

a b

a b

Keep factoring until no further factors can be found .

Page 5-15





Factoring the Sum & Difference of Cubes

• Factoring a sum or difference of two cubes

Name Formula Example

sum of cubes a3 + b3 = (a + b)(a2 – ab + b2)27 + x3 = 33 + x3 = (3 + x) (32 – 3 x + x2)

difference of cubes a3 – b3 = (a – b)(a2 + ab + b2) 27 – y3 = 33 – y3 = (3 – y)(32 + 3y + y2)

Note: a3 – b3 ≠ (a – b)3

• Memory aid: a3 ± b3 = (a ± b)(a2 ∓ ab + b2 )


1. x3 + 125 = x3 + 53 = (x + 5) (x2 – 5x + 52) a3 + b3 = (a + b)(a2 – ab + b2): a = x, b = 5

= (x + 5) (x2 – 5x + 25)

2. -2t3 + 54 = -2(t3 – 27) = -2(t3 – 33) Factor out -2 . (-2)(-27) = 54

= -2(t – 3) (t2 + 3t + 9) a3 – b3 = (a – b)(a2 + ab + b2): a = t, b = 3

3. -wu4 – 0.001wu = -wu (u3 + 0 .001) Factor out -wu.

= -wu (u3 + 0.13) 0 .13 = 0 .001

= - wu (u + 0.1) (u2 – 0.1u + 0.01) a3 + b3 = (a + b)(a2 – ab + b2) : a = u, b = 0 .1

4. p6 – 27q6 = p2 ∙ 3 – 33(q2)3 = (p2)3 – (3q2)3 amn = (am)n

= (p2 – 3q2) [(p2)2 + 3p2q2 + (3q2)2] a3 – b3 = (a – b)(a2 + ab + b2): a = p2, b = 3q2

= (p2 – 3q2) (p4 + 3p2q2 + 9q4) (am)n = amn

5. 7y5 – 7y2 = 7y2 (y3 – 1) = 7y2 (y3 – 13) Factor out 7y2 .

= 7y2 (y – 1)(y2 + y + 1) a3 – b3 = (a – b)(a2 + ab + b2 ): a = y , b = 1

6. 𝟏𝟏𝟔𝟔𝟔𝟔

+ z3 = 13

𝟔𝟔𝟑𝟑+ z3

= �𝟏𝟏𝟔𝟔�3

+ z3 𝑎𝑎𝑛𝑛


𝑏𝑏�𝑛𝑛

= �𝟏𝟏𝟔𝟔

+ 𝒛𝒛� � 𝟏𝟏𝟔𝟔𝟐𝟐− 𝟏𝟏

𝟔𝟔𝒛𝒛 + 𝒛𝒛𝟐𝟐� a3 + b3 = (a + b)(a2 - ab + b2 ): a = 1

4 , b = z

Not factorable

a b 0 .1 2

= 0 .01

a b

a b

a b a b a2 a b b2

Notice the reversed plus or minus sign in thesecond term .

Page 5-16

Factoring the Sum & Difference of Cubes





Note: a3 – b3 ≠ (a – b)3

• Memory aid: a3 ± b3 = (a ± b)(a2 ∓ ab + b2 )


1. x3 + 125 = x3 + 53 = (x + 5) (x2 – 5x + 52) a3 + b3 = (a + b)(a2 – ab + b2): a = x, b = 5

= (x + 5) (x2 – 5x + 25)

2. -2t3 + 54 = -2(t3 – 27) = -2(t3 – 33) Factor out -2 . (-2)(-27) = 54

= -2(t – 3) (t2 + 3t + 9) a3 – b3 = (a – b)(a2 + ab + b2): a = t, b = 3

3. -wu4 – 0.001wu = -wu (u3 + 0 .001) Factor out -wu.

= -wu (u3 + 0.13) 0 .13 = 0 .001

= - wu (u + 0.1) (u2 – 0.1u + 0.01) a3 + b3 = (a + b)(a2 – ab + b2) : a = u, b = 0 .1

4. p6 – 27q6 = p2 ∙ 3 – 33(q2)3 = (p2)3 – (3q2)3 amn = (am)n

= (p2 – 3q2) [(p2)2 + 3p2q2 + (3q2)2] a3 – b3 = (a – b)(a2 + ab + b2): a = p2, b = 3q2

= (p2 – 3q2) (p4 + 3p2q2 + 9q4) (am)n = amn

5. 7y5 – 7y2 = 7y2 (y3 – 1) = 7y2 (y3 – 13) Factor out 7y2 .

= 7y2 (y – 1)(y2 + y + 1) a3 – b3 = (a – b)(a2 + ab + b2 ): a = y , b = 1

6. 𝟏𝟏𝟔𝟔𝟔𝟔

+ z3 = 13

𝟔𝟔𝟑𝟑+ z3

= �𝟏𝟏𝟔𝟔�3

+ z3 𝑎𝑎𝑛𝑛


𝑏𝑏�𝑛𝑛

= �𝟏𝟏𝟔𝟔

+ 𝒛𝒛� � 𝟏𝟏𝟔𝟔𝟐𝟐− 𝟏𝟏

𝟔𝟔𝒛𝒛 + 𝒛𝒛𝟐𝟐� a3 + b3 = (a + b)(a2 - ab + b2 ): a = 1

4 , b = z

Not factorable

a b 0 .1 2

= 0 .01

a b

a b

a b a b a2 a b b2

Notice the reversed plus or minus sign in thesecond term .

Page 5-16





Unit 5 Summary






9x2 , - x, 3

• Polynomial













• The opposite of the polynomial: - p: the opposite of the polynomial

p: polynomial p + (-p) = 0

-x = (-1)(x)

Page 5-17

Unit 5 Summary






9x2 , - x, 3

• Polynomial













• The opposite of the polynomial: - p: the opposite of the polynomial

p: polynomial p + (-p) = 0

-x = (-1)(x)

Page 5-17






Example

(a + b) (c + d) = ac + ad + bc + bd (x + 2) (x + 3) = x·x + x·3 + 2x + 2∙3F O I L F O I L

= x2 + 5x + 6





• Factoring x2 + bx + c

Factoring x2 + bx + c Example: x2 – 5x + 6x2 + (c1 + c2) x + c1c2 = (x + c1)(x + c2)x c1x c2Check: c1x+ c2 x = bx

x2 + [-2 + (-3)] x + 6 = (x – 2)(x – 3) x -2 x -3

Check: -2x + (-3x) = -5x yes! .

• Factoring ax2 + bx + c

Summary: Factoring ax2 + bx + ca1a2x2 + (a1c2 + c1a2)x + c1c2 = (a1x + c1)(a2x + c2)

a1x c1 a2x c2

• AC method for factoring trinomials: ax2 + bx + c





2 (2x + 3)(3x – 1) = 0 Factor out (2x + 3) .

? ?

Page 5-18








a2 + 2ab + b2 = (a + b)2


square of difference(perfect square trinomial)

a2 – 2ab + b2 = (a – b)2

25t2 – 20t + 4 = (5t – 2)2 a = 5t, b = 25t -25t -2

Check: (5t – 2)2 = (5t)2 – 2(5t)(2) + 22 = 25t2 – 20t + 4 √





Factoring (L R)

a2 – b2 = (a + b)(a – b)

Multiplying (L R)

a b a b a2 a b b2

Page 5-19





PRACTICE QUIZ

Unit 5 Polynomial Functions

1. The function f (x) = 2,000 + 0 .6x2 can be used to determine

the cost of producing x machines in a factory . a. What is the total cost of producing 40 machines?b. Use the following graph to estimate f (30) .

2 . a . Find the sum of 5𝑥𝑥3 + 2𝑥𝑥2 − 4𝑥𝑥 + 1 = 0and 2𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 5 = 0

b. Find the difference of 7𝑥𝑥3 + 5𝑥𝑥2 + 𝑥𝑥 − 5 = 0and 3𝑥𝑥3 − 2𝑥𝑥2 + 2𝑥𝑥 − 3 = 0

3. Find the following products .

a . �2𝑡𝑡 + 13�(6t − 9)

b. (u + 2)3

4 . Given f (x) = 2x – x2 , find f (b + 2) .

5. Factor the following completely .

a. 4c2d – 2cd 2 + 2c – d

b . 27x3y – 3xy3

c . x2 – 2x – 3

d . 3x2 –17x + 24 = 0e . t2 + 2

3𝑡𝑡 = - 1

9

6. Factor the following completely .a . 4x2 – 9y2

b . 29 u2 – 2

25 𝑣𝑣2

c . t4 – 16d . x6 – 8y6

f (x)

x

Page 8

PRACTICE QUIZ

Unit 5 Polynomial Functions

1. The function f (x) = 2,000 + 0 .6x2 can be used to determine

the cost of producing x machines in a factory . a. What is the total cost of producing 40 machines?b. Use the following graph to estimate f (30) .

2 . a . Find the sum of 5𝑥𝑥3 + 2𝑥𝑥2 − 4𝑥𝑥 + 1 = 0and 2𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 5 = 0

b. Find the difference of 7𝑥𝑥3 + 5𝑥𝑥2 + 𝑥𝑥 − 5 = 0and 3𝑥𝑥3 − 2𝑥𝑥2 + 2𝑥𝑥 − 3 = 0

3. Find the following products .

a . �2𝑡𝑡 + 13�(6t − 9)

b. (u + 2)3

4 . Given f (x) = 2x – x2 , find f (b + 2) .

5. Factor the following completely .

a. 4c2d – 2cd 2 + 2c – d

b . 27x3y – 3xy3

c . x2 – 2x – 3

d . 3x2 –17x + 24 = 0e . t2 + 2

3𝑡𝑡 = - 1

9

6. Factor the following completely .a . 4x2 – 9y2

b . 29 u2 – 2

25 𝑣𝑣2

c . t4 – 16d . x6 – 8y6

f (x)

x

Page 8

10 20 30 40

3,000

2,000

1,000




Algebra I & II Key Concepts, Practice, and Quizzes Unit 6 – Rational Expressions

UNIT 6 RATIONAL EXPRESSIONS6-1 RATIONAL EXPRESSIONS & MULTIPLICATION

Rational Functions

Example

• Rational number �𝒂𝒂𝒃𝒃�: the ratio or quotient of two numbers (a fraction) . 2

3 , -1

5

• Rational expression: an expression that is a ratio or quotient of two polynomials . 2𝑥𝑥−37𝑥𝑥+5

• Rational function: a function that is a ratio or quotient of two polynomials .

Rational Function Example

𝑓𝑓(𝑥𝑥) = 𝑎𝑎(𝑥𝑥)𝑏𝑏(𝑥𝑥)

b(x) ≠ 0 𝑓𝑓(𝑥𝑥) = 4𝑥𝑥+35𝑥𝑥−7

, 𝑓𝑓(𝑥𝑥) = 3𝑦𝑦2−2𝑦𝑦+45𝑦𝑦−6

Example: For the following functions, identify their domains .

Rational Function Restriction DomanSet-Builder Notation Interval Notation

𝒇𝒇(𝒙𝒙) = 𝟑𝟑𝒙𝒙 + 𝟏𝟏𝒙𝒙 − 𝟓𝟓

x ≠ 5 �3𝑥𝑥+15−5

= 3𝑥𝑥+10

is undefined� { x | x ≠ 5 } (-∞, 5) ∪ (5, ∞)

𝒇𝒇(𝒙𝒙) =𝟓𝟓𝒚𝒚𝒚𝒚𝟐𝟐 + 𝟐𝟐𝒚𝒚 − 𝟏𝟏

𝟏𝟏𝟏𝟏𝒚𝒚t ≠ 0 �5𝑦𝑦𝑦𝑦2+2𝑦𝑦−1

0 is undefined� { t | t ≠ 0 } (-∞, 0) ∪ (0, ∞)

𝒇𝒇(𝒙𝒙) = 𝟕𝟕𝒚𝒚+𝟒𝟒𝟓𝟓𝒚𝒚−𝟑𝟑 y ≠ 3

5 (5 ∙

3

5− 3 = 0) � 𝑦𝑦 | 𝑦𝑦 ≠

3 5

� �-∞, 3 5� ∪ � 3

5,∞�

• Reducing rational expressions to lowest forms Example

A rational expression reduced to lowest terms means that no common

factors other than 1 occur in its top and bottom polynomials . 𝟐𝟐 𝒒𝒒𝟐𝟐

𝟔𝟔 𝒒𝒒𝟒𝟒= 2 𝑞𝑞2

6 𝑞𝑞2𝑞𝑞2= 1

3 𝑞𝑞2

Example: Reduce to lowest terms (simplify) . Lowest terms

1. 𝟑𝟑𝒃𝒃+𝟗𝟗𝟑𝟑

= 3𝑏𝑏3

+ 93 = b + 3

2. 𝟑𝟑𝒙𝒙𝟐𝟐+ 𝟐𝟐𝒙𝒙𝟓𝟓𝒙𝒙𝟑𝟑−𝟕𝟕𝒙𝒙𝟐𝟐

= 𝑥𝑥(3𝑥𝑥+2)𝑥𝑥2(5𝑥𝑥−7)

= 𝟑𝟑𝒙𝒙+𝟐𝟐𝒙𝒙(𝟓𝟓𝒙𝒙−𝟕𝟕)

Factor, and then reduce .

3. 𝟗𝟗𝒚𝒚𝟐𝟐−𝟑𝟑𝒚𝒚𝟏𝟏𝟏𝟏𝒚𝒚

= 3𝑦𝑦(3𝑦𝑦−1)18𝑦𝑦

= 3𝑦𝑦−16

= 3𝑦𝑦6− 1

6= 𝒚𝒚

𝟐𝟐− 𝟏𝟏

𝟔𝟔

4. 𝟑𝟑𝒚𝒚𝟐𝟐−𝟏𝟏𝟐𝟐(𝒚𝒚−𝟐𝟐)𝟐𝟐

= 3(𝑦𝑦2−4)(𝑦𝑦−2)2

= 3(𝑦𝑦2−22)(𝑦𝑦−2)2

= 3(𝑦𝑦+2)(𝑦𝑦−2)(𝑦𝑦−2)2

= 𝟑𝟑(𝒚𝒚+𝟐𝟐)𝒚𝒚−𝟐𝟐

a2 – b2 = (a + b) (a – b)

5. 𝒂𝒂(𝟐𝟐𝒂𝒂−𝟏𝟏)(𝟐𝟐𝒂𝒂+𝟓𝟓)𝟒𝟒𝒂𝒂𝟐𝟐+ 𝟏𝟏𝒂𝒂−𝟓𝟓

= 𝑎𝑎(2𝑎𝑎−1)(2𝑎𝑎+5)(2𝑎𝑎−1)(2𝑎𝑎+5)

= a 4a2 + 8a –5 = (2a – 1) (2a + 5)

6. 𝟒𝟒(𝒙𝒙𝟑𝟑−𝒚𝒚𝟑𝟑)𝟏𝟏𝟐𝟐(𝒙𝒙−𝒚𝒚)

= 4(𝑥𝑥−𝑦𝑦)(𝑥𝑥2+𝑥𝑥𝑦𝑦+𝑦𝑦2)12(𝑥𝑥−𝑦𝑦)

= 𝟏𝟏𝟑𝟑

(𝒙𝒙𝟐𝟐 + 𝒙𝒙𝒚𝒚 + 𝒚𝒚𝟐𝟐) a3 – b3 = (a – b)(a2 + ab + b2 )

6

3

2

12a -12a 5

3

1

1 1

polynomials

35

Page 6-1

UNIT 6 RATIONAL EXPRESSIONS6-1 RATIONAL EXPRESSIONS & MULTIPLICATION

Rational Functions

Example

• Rational number �𝒂𝒂𝒃𝒃�: the ratio or quotient of two numbers (a fraction) . 2

3 , -1

5

• Rational expression: an expression that is a ratio or quotient of two polynomials . 2𝑥𝑥−37𝑥𝑥+5




b(x) ≠ 0 𝑓𝑓(𝑥𝑥) = 4𝑥𝑥+35𝑥𝑥−7

, 𝑓𝑓(𝑥𝑥) = 3𝑦𝑦2−2𝑦𝑦+45𝑦𝑦−6

Example: For the following functions, identify their domains .

Rational Function Restriction DomanSet-Builder Notation Interval Notation

𝒇𝒇(𝒙𝒙) = 𝟑𝟑𝒙𝒙 + 𝟏𝟏𝒙𝒙 − 𝟓𝟓

x ≠ 5 �3𝑥𝑥+15−5

= 3𝑥𝑥+10

is undefined� { x | x ≠ 5 } (-∞, 5) ∪ (5, ∞)

𝒇𝒇(𝒙𝒙) =𝟓𝟓𝒚𝒚𝒚𝒚𝟐𝟐 + 𝟐𝟐𝒚𝒚 − 𝟏𝟏

𝟏𝟏𝟏𝟏𝒚𝒚t ≠ 0 �5𝑦𝑦𝑦𝑦2+2𝑦𝑦−1

0 is undefined� { t | t ≠ 0 } (-∞, 0) ∪ (0, ∞)

𝒇𝒇(𝒙𝒙) = 𝟕𝟕𝒚𝒚+𝟒𝟒𝟓𝟓𝒚𝒚−𝟑𝟑 y ≠ 3

5 (5 ∙

3

5− 3 = 0) � 𝑦𝑦 | 𝑦𝑦 ≠

3 5

� �-∞, 3 5� ∪ � 3

5,∞�

• Reducing rational expressions to lowest forms Example

A rational expression reduced to lowest terms means that no common

factors other than 1 occur in its top and bottom polynomials . 𝟐𝟐 𝒒𝒒𝟐𝟐

𝟔𝟔 𝒒𝒒𝟒𝟒= 2 𝑞𝑞2

6 𝑞𝑞2𝑞𝑞2= 1

3 𝑞𝑞2

Example: Reduce to lowest terms (simplify) . Lowest terms

1. 𝟑𝟑𝒃𝒃+𝟗𝟗𝟑𝟑

= 3𝑏𝑏3

+ 93 = b + 3

2. 𝟑𝟑𝒙𝒙𝟐𝟐+ 𝟐𝟐𝒙𝒙𝟓𝟓𝒙𝒙𝟑𝟑−𝟕𝟕𝒙𝒙𝟐𝟐

= 𝑥𝑥(3𝑥𝑥+2)𝑥𝑥2(5𝑥𝑥−7)

= 𝟑𝟑𝒙𝒙+𝟐𝟐𝒙𝒙(𝟓𝟓𝒙𝒙−𝟕𝟕)

Factor, and then reduce .

3. 𝟗𝟗𝒚𝒚𝟐𝟐−𝟑𝟑𝒚𝒚𝟏𝟏𝟏𝟏𝒚𝒚

= 3𝑦𝑦(3𝑦𝑦−1)18𝑦𝑦

= 3𝑦𝑦−16

= 3𝑦𝑦6− 1

6= 𝒚𝒚

𝟐𝟐− 𝟏𝟏

𝟔𝟔

4. 𝟑𝟑𝒚𝒚𝟐𝟐−𝟏𝟏𝟐𝟐(𝒚𝒚−𝟐𝟐)𝟐𝟐

= 3(𝑦𝑦2−4)(𝑦𝑦−2)2

= 3(𝑦𝑦2−22)(𝑦𝑦−2)2

= 3(𝑦𝑦+2)(𝑦𝑦−2)(𝑦𝑦−2)2

= 𝟑𝟑(𝒚𝒚+𝟐𝟐)𝒚𝒚−𝟐𝟐

a2 – b2 = (a + b) (a – b)

5. 𝒂𝒂(𝟐𝟐𝒂𝒂−𝟏𝟏)(𝟐𝟐𝒂𝒂+𝟓𝟓)𝟒𝟒𝒂𝒂𝟐𝟐+ 𝟏𝟏𝒂𝒂−𝟓𝟓

= 𝑎𝑎(2𝑎𝑎−1)(2𝑎𝑎+5)(2𝑎𝑎−1)(2𝑎𝑎+5)

= a 4a2 + 8a –5 = (2a – 1) (2a + 5)

6. 𝟒𝟒(𝒙𝒙𝟑𝟑−𝒚𝒚𝟑𝟑)𝟏𝟏𝟐𝟐(𝒙𝒙−𝒚𝒚)

= 4(𝑥𝑥−𝑦𝑦)(𝑥𝑥2+𝑥𝑥𝑦𝑦+𝑦𝑦2)12(𝑥𝑥−𝑦𝑦)

= 𝟏𝟏𝟑𝟑

(𝒙𝒙𝟐𝟐 + 𝒙𝒙𝒚𝒚 + 𝒚𝒚𝟐𝟐) a3 – b3 = (a – b)(a2 + ab + b2 )

6

3

2

12a -12a 5

3

1

1 1

polynomials

35

Page 6-1





Multiplying Rational Expressions

• Multiplying fractions: �23� �4

5� = 2 ∙ 4

3 ∙ 5= 8

15

• Multiplying rational expressions: 𝑁𝑁1𝐷𝐷1

∙ 𝑁𝑁2𝐷𝐷2

= 𝑁𝑁1𝑁𝑁2𝐷𝐷1𝐷𝐷2

𝑁𝑁1𝐷𝐷1

𝑎𝑎𝑎𝑎𝑎𝑎 𝑁𝑁2𝐷𝐷2

are rational expressions . (D1D2 ≠ 0)

Steps- Multiply the numerators .

- Multiply the denominators .

- Simplify (cancel or reduce common factors) if possible .

Example: 2𝑥𝑥2

3𝑦𝑦3∙ 3𝑥𝑥4𝑦𝑦

= (2𝑥𝑥2)(3𝑥𝑥)(3𝑦𝑦3)(4𝑦𝑦)

= 6𝑥𝑥3

12𝑦𝑦4= 𝒙𝒙𝟑𝟑

𝟐𝟐𝟐𝟐𝟒𝟒a n a m = a n + m

• Note: It is more efficient to reduce or cancel common factors before multiplying . 1 1

Example:38

∙ 49

= 1 ∙ 12 ∙ 3

= 𝟏𝟏𝟔𝟔

2 3

𝑥𝑥2

𝑦𝑦∙ 𝑦𝑦

4

4𝑥𝑥= 𝒙𝒙𝟐𝟐𝟑𝟑

𝟒𝟒

Example: Perform the indicated operations and simplify .1

1. 𝟐𝟐𝟐𝟐𝟐𝟐𝒃𝒃𝒃𝒃𝟐𝟐

∙ 𝒃𝒃𝟑𝟑

𝟒𝟒𝟐𝟐𝟐𝟐𝒃𝒃= 1 ∙ 𝑏𝑏

2 = 𝒃𝒃𝟐𝟐

2

2. 𝒙𝒙𝟐𝟐−𝒙𝒙𝟐𝟐+𝟐𝟐𝟐𝟐

(𝒙𝒙−𝟐𝟐)𝟐𝟐∙ 𝒙𝒙

𝟐𝟐−𝟐𝟐𝟐𝟐

𝒙𝒙𝟑𝟑+𝟐𝟐𝟑𝟑= (𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)

(𝑥𝑥−𝑦𝑦)2∙ (𝑥𝑥+𝑦𝑦)(𝑥𝑥−𝑦𝑦)

𝑥𝑥3+𝑦𝑦3a2 – b2 = (a + b) (a – b)

= (𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)(𝑥𝑥−𝑦𝑦)

∙ (𝑥𝑥+𝑦𝑦)𝑥𝑥3+𝑦𝑦3

(a + b)(a2 – ab + b2 ) = a3 + b3

= 𝟏𝟏𝒙𝒙−𝟐𝟐

3. 𝒃𝒃𝟐𝟐−𝟐𝟐𝒃𝒃− 𝟖𝟖𝒃𝒃−𝟐𝟐

∙ 𝒃𝒃𝟐𝟐−𝟒𝟒𝒃𝒃+𝟐𝟐

= (𝑏𝑏+2)(𝑏𝑏−4)𝑏𝑏−2

∙ (𝑏𝑏+2)(𝑏𝑏−2)𝑏𝑏+2

a2 – b2 = (a + b) (a – b)

= (b - 4)(b + 2)

4. 𝟑𝟑𝒙𝒙𝟑𝟑+𝟑𝟑𝟐𝟐𝟑𝟑

𝟒𝟒�𝒙𝒙𝟐𝟐−𝟐𝟐𝟐𝟐�∙ (𝒙𝒙+𝟐𝟐)𝟑𝟑�𝒙𝒙𝟐𝟐−𝒙𝒙𝟐𝟐+𝟐𝟐𝟐𝟐�

= 3�𝑥𝑥3+ 𝑦𝑦3�4(𝑥𝑥+𝑦𝑦)(𝑥𝑥−𝑦𝑦)

∙ (𝑥𝑥+𝑦𝑦)3(𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)

a2 – b2 = (a + b) (a – b)

= �𝑥𝑥3+ 𝑦𝑦3�4(𝑥𝑥−𝑦𝑦)

∙ 1(𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)

= (𝑥𝑥+𝑦𝑦)�𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2�4(𝑥𝑥−𝑦𝑦)

∙ 1𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2

a3 + b3 = (a + b)(a2 – ab + b2 )

= 𝒙𝒙+𝟐𝟐𝟒𝟒(𝒙𝒙−𝟐𝟐)

1

2

b 2

b -4

Denominator

Numerator

Page 6-2

Multiplying Rational Expressions

• Multiplying fractions: �23� �4

5� = 2 ∙ 4

3 ∙ 5= 8

15




𝑁𝑁1𝐷𝐷1



Steps- Multiply the numerators .

- Multiply the denominators .


Example: 2𝑥𝑥2

3𝑦𝑦3∙ 3𝑥𝑥4𝑦𝑦

= (2𝑥𝑥2)(3𝑥𝑥)(3𝑦𝑦3)(4𝑦𝑦)

= 6𝑥𝑥3

12𝑦𝑦4= 𝒙𝒙𝟑𝟑

𝟐𝟐𝟐𝟐𝟒𝟒a n a m = a n + m

• Note: It is more efficient to reduce or cancel common factors before multiplying . 1 1

Example:38

∙ 49

= 1 ∙ 12 ∙ 3

= 𝟏𝟏𝟔𝟔

2 3

𝑥𝑥2

𝑦𝑦∙ 𝑦𝑦

4

4𝑥𝑥= 𝒙𝒙𝟐𝟐𝟑𝟑

𝟒𝟒

Example: Perform the indicated operations and simplify .1

1. 𝟐𝟐𝟐𝟐𝟐𝟐𝒃𝒃𝒃𝒃𝟐𝟐

∙ 𝒃𝒃𝟑𝟑

𝟒𝟒𝟐𝟐𝟐𝟐𝒃𝒃= 1 ∙ 𝑏𝑏

2 = 𝒃𝒃𝟐𝟐

2

2. 𝒙𝒙𝟐𝟐−𝒙𝒙𝟐𝟐+𝟐𝟐𝟐𝟐

(𝒙𝒙−𝟐𝟐)𝟐𝟐∙ 𝒙𝒙

𝟐𝟐−𝟐𝟐𝟐𝟐

𝒙𝒙𝟑𝟑+𝟐𝟐𝟑𝟑= (𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)

(𝑥𝑥−𝑦𝑦)2∙ (𝑥𝑥+𝑦𝑦)(𝑥𝑥−𝑦𝑦)

𝑥𝑥3+𝑦𝑦3a2 – b2 = (a + b) (a – b)

= (𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)(𝑥𝑥−𝑦𝑦)

∙ (𝑥𝑥+𝑦𝑦)𝑥𝑥3+𝑦𝑦3

(a + b)(a2 – ab + b2 ) = a3 + b3

= 𝟏𝟏𝒙𝒙−𝟐𝟐

3. 𝒃𝒃𝟐𝟐−𝟐𝟐𝒃𝒃− 𝟖𝟖𝒃𝒃−𝟐𝟐

∙ 𝒃𝒃𝟐𝟐−𝟒𝟒𝒃𝒃+𝟐𝟐

= (𝑏𝑏+2)(𝑏𝑏−4)𝑏𝑏−2

∙ (𝑏𝑏+2)(𝑏𝑏−2)𝑏𝑏+2

a2 – b2 = (a + b) (a – b)

= (b - 4)(b + 2)

4. 𝟑𝟑𝒙𝒙𝟑𝟑+𝟑𝟑𝟐𝟐𝟑𝟑

𝟒𝟒�𝒙𝒙𝟐𝟐−𝟐𝟐𝟐𝟐�∙ (𝒙𝒙+𝟐𝟐)𝟑𝟑�𝒙𝒙𝟐𝟐−𝒙𝒙𝟐𝟐+𝟐𝟐𝟐𝟐�

= 3�𝑥𝑥3+ 𝑦𝑦3�4(𝑥𝑥+𝑦𝑦)(𝑥𝑥−𝑦𝑦)

∙ (𝑥𝑥+𝑦𝑦)3(𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)

a2 – b2 = (a + b) (a – b)

= �𝑥𝑥3+ 𝑦𝑦3�4(𝑥𝑥−𝑦𝑦)

∙ 1(𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2)

= (𝑥𝑥+𝑦𝑦)�𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2�4(𝑥𝑥−𝑦𝑦)

∙ 1𝑥𝑥2−𝑥𝑥𝑦𝑦+𝑦𝑦2

a3 + b3 = (a + b)(a2 – ab + b2 )

= 𝒙𝒙+𝟐𝟐𝟒𝟒(𝒙𝒙−𝟐𝟐)

1

2

b 2

b -4

Denominator

Numerator

Page 6-2





Dividing Rational Expressions

Example

• Dividing fractions: to divide by a fraction, multiply by its reciprocal . 34

÷ 52

= 34∙ 25

= 𝟑𝟑𝟏𝟏𝟏𝟏

• Dividing rational expressions: 𝑁𝑁1𝐷𝐷1

÷ 𝑁𝑁2𝐷𝐷2

= 𝑁𝑁1𝐷𝐷1∙ 𝐷𝐷2𝑁𝑁2

D1, D2 , N2 ≠ 0

Steps

- Write as multiplication of the reciprocal . ÷ × , 𝑁𝑁2𝐷𝐷2

𝐷𝐷2𝑁𝑁2


- Multiply .

Example: Perform the indicated operations and simplify .

1. 𝒚𝒚𝟐𝟐

𝒙𝒙𝟑𝟑÷ 𝟏𝟏𝟏𝟏𝒚𝒚

𝟐𝟐= 𝑦𝑦2

𝑥𝑥3∙ 210𝑦𝑦

=𝑦𝑦 ∙ 1𝑥𝑥3 ∙ 5

= 𝒚𝒚𝟓𝟓𝒙𝒙𝟑𝟑

÷ × , 10𝑦𝑦2

210𝑦𝑦

2. 𝟓𝟓𝒂𝒂𝟐𝟐−𝒂𝒂𝒂𝒂𝒂𝒂𝟐𝟐−𝒂𝒂𝟐𝟐

÷ 𝟓𝟓𝒂𝒂𝟐𝟐−𝒂𝒂𝒂𝒂𝒂𝒂−𝒂𝒂

=5𝑎𝑎2−𝑎𝑎𝑎𝑎𝑎𝑎2−𝑎𝑎2

∙ 𝑎𝑎−𝑎𝑎5𝑎𝑎2−𝑎𝑎𝑎𝑎

÷ × , 5𝑎𝑎2−𝑎𝑎𝑎𝑎𝑎𝑎−𝑎𝑎

𝑎𝑎−𝑎𝑎5𝑎𝑎2−𝑎𝑎𝑎𝑎

= 𝑎𝑎−𝑏𝑏(𝑎𝑎+𝑏𝑏)(𝑎𝑎−𝑏𝑏) 𝑎𝑎2−𝑏𝑏2 = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)

=𝟏𝟏

(𝒂𝒂 + 𝒂𝒂)

3. 𝒙𝒙𝟐𝟐−𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐+𝟒𝟒𝒙𝒙+𝟒𝟒

÷ 𝒙𝒙−𝟒𝟒𝟑𝟑𝒙𝒙+𝟏𝟏

=𝑥𝑥2−42

𝑥𝑥2+4𝑥𝑥+4∙ 3𝑥𝑥+6𝑥𝑥−4

÷ × , x−43x+6

3𝑥𝑥+6𝑥𝑥−4

= (𝑥𝑥+4)(𝑥𝑥−4)(𝑥𝑥+2)2 ∙ 3(𝑥𝑥+2)

𝑥𝑥−4 𝑎𝑎2−𝑏𝑏2 = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)

= 𝟑𝟑(𝒙𝒙+𝟒𝟒)𝒙𝒙+𝟐𝟐

4. 𝒙𝒙𝟐𝟐−𝟏𝟏𝟑𝟑

÷ 𝒙𝒙+𝟏𝟏𝟓𝟓

=𝑥𝑥2−13

∙ 5𝑥𝑥+1

÷ × , 𝒙𝒙+𝟏𝟏𝟓𝟓

5𝑥𝑥+1

=(𝑥𝑥+1)(𝑥𝑥−1)

3∙ 5𝑥𝑥+1

𝑎𝑎2−𝑏𝑏2 = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)

= 𝟓𝟓(𝒙𝒙−𝟏𝟏)𝟑𝟑

1

5

x 2

x 2

2

1

𝑁𝑁1𝐷𝐷1


are rational ,

Page 6-3

Dividing Rational Expressions

Example

• Dividing fractions: to divide by a fraction, multiply by its reciprocal . 34

÷ 52

= 34∙ 25

= 𝟑𝟑𝟏𝟏𝟏𝟏




D1, D2 , N2 ≠ 0

Steps

- Write as multiplication of the reciprocal . ÷ × , 𝑁𝑁2𝐷𝐷2

𝐷𝐷2𝑁𝑁2


- Multiply .


1. 𝒚𝒚𝟐𝟐

𝒙𝒙𝟑𝟑÷ 𝟏𝟏𝟏𝟏𝒚𝒚

𝟐𝟐= 𝑦𝑦2

𝑥𝑥3∙ 210𝑦𝑦

=𝑦𝑦 ∙ 1𝑥𝑥3 ∙ 5

= 𝒚𝒚𝟓𝟓𝒙𝒙𝟑𝟑

÷ × , 10𝑦𝑦2

210𝑦𝑦

2. 𝟓𝟓𝒂𝒂𝟐𝟐−𝒂𝒂𝒂𝒂𝒂𝒂𝟐𝟐−𝒂𝒂𝟐𝟐

÷ 𝟓𝟓𝒂𝒂𝟐𝟐−𝒂𝒂𝒂𝒂𝒂𝒂−𝒂𝒂

=5𝑎𝑎2−𝑎𝑎𝑎𝑎𝑎𝑎2−𝑎𝑎2

∙ 𝑎𝑎−𝑎𝑎5𝑎𝑎2−𝑎𝑎𝑎𝑎

÷ × , 5𝑎𝑎2−𝑎𝑎𝑎𝑎𝑎𝑎−𝑎𝑎

𝑎𝑎−𝑎𝑎5𝑎𝑎2−𝑎𝑎𝑎𝑎

= 𝑎𝑎−𝑏𝑏(𝑎𝑎+𝑏𝑏)(𝑎𝑎−𝑏𝑏) 𝑎𝑎2−𝑏𝑏2 = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)

=𝟏𝟏

(𝒂𝒂 + 𝒂𝒂)

3. 𝒙𝒙𝟐𝟐−𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐+𝟒𝟒𝒙𝒙+𝟒𝟒

÷ 𝒙𝒙−𝟒𝟒𝟑𝟑𝒙𝒙+𝟏𝟏

=𝑥𝑥2−42

𝑥𝑥2+4𝑥𝑥+4∙ 3𝑥𝑥+6𝑥𝑥−4

÷ × , x−43x+6

3𝑥𝑥+6𝑥𝑥−4

= (𝑥𝑥+4)(𝑥𝑥−4)(𝑥𝑥+2)2 ∙ 3(𝑥𝑥+2)

𝑥𝑥−4 𝑎𝑎2−𝑏𝑏2 = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)

= 𝟑𝟑(𝒙𝒙+𝟒𝟒)𝒙𝒙+𝟐𝟐

4. 𝒙𝒙𝟐𝟐−𝟏𝟏𝟑𝟑

÷ 𝒙𝒙+𝟏𝟏𝟓𝟓

=𝑥𝑥2−13

∙ 5𝑥𝑥+1

÷ × , 𝒙𝒙+𝟏𝟏𝟓𝟓

5𝑥𝑥+1

=(𝑥𝑥+1)(𝑥𝑥−1)

3∙ 5𝑥𝑥+1

𝑎𝑎2−𝑏𝑏2 = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)

= 𝟓𝟓(𝒙𝒙−𝟏𝟏)𝟑𝟑

1

5

x 2

x 2

2

1

𝑁𝑁1𝐷𝐷1


are rational ,

Page 6-3





6-2 ADDING & SUBTRACTING RATIONAL EXPRESSIONS

Adding / Subtracting Like Rational Expressions

Example

• Like rational expressions: rational expressions that have the 3𝑥𝑥2

𝑥𝑥+2, 5𝑥𝑥

𝑥𝑥+2same denominator .

• Unlike rational expressions: rational expressions that have 7𝑥𝑥+3𝑥𝑥−2

, 3𝑥𝑥−5𝑥𝑥2+5

different denominators .

• Adding or subtracting like rational expressions𝑁𝑁1𝐷𝐷

+ 𝑁𝑁2𝐷𝐷

= 𝑁𝑁1+ 𝑁𝑁2 𝐷𝐷

, 𝑁𝑁1𝐷𝐷− 𝑁𝑁2

𝐷𝐷= 𝑁𝑁1− 𝑁𝑁2

𝐷𝐷 𝑁𝑁1

𝐷𝐷 𝑎𝑎𝑎𝑎𝑎𝑎

𝑁𝑁2

𝐷𝐷are rational expressions . D ≠ 0

Steps Example- Combine the numerators . 3𝑥𝑥

2(𝑥𝑥+4)+ 5𝑥𝑥

2(𝑥𝑥+4)= 3𝑥𝑥+5𝑥𝑥

2(𝑥𝑥+4)

- Denominators do not change . = 8𝑥𝑥2(𝑥𝑥+4)

- Simplify (cancel or reduce common factors) if possible . = 𝟒𝟒𝟒𝟒𝟒𝟒+𝟒𝟒

Example: Add or subtract as indicated and simplify .

1. 𝟒𝟒𝟐𝟐+𝟐𝟐𝟒𝟒𝟐𝟐+𝟐𝟐𝟐𝟐

𝟒𝟒𝟐𝟐−𝟐𝟐𝟐𝟐− 𝟒𝟒+𝟐𝟐

𝟒𝟒𝟐𝟐−𝟐𝟐𝟐𝟐= �𝑥𝑥2+2𝑥𝑥𝑥𝑥+𝑥𝑥2�−(𝑥𝑥+𝑥𝑥)

𝑥𝑥2−𝑥𝑥2Combine numerators .

= (𝟒𝟒+𝟐𝟐)2−(𝟒𝟒+𝟐𝟐) ∙ 1(𝑥𝑥+𝑥𝑥)(𝑥𝑥−𝑥𝑥)

= (𝟒𝟒+𝟐𝟐)[(𝑥𝑥+𝑥𝑥)−1](𝑥𝑥+𝑥𝑥)(𝑥𝑥−𝑥𝑥)

Factor out (x + y) .

= 𝟒𝟒+𝟐𝟐−𝟏𝟏𝟒𝟒−𝟐𝟐

2. 𝟑𝟑𝟑𝟑+𝟗𝟗𝟗𝟗𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗

+ 𝟒𝟒(𝟑𝟑+𝟑𝟑𝟗𝟗)𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗

− 𝟓𝟓𝟑𝟑𝟑𝟑

𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗 = 3𝑎𝑎+9𝑏𝑏+4(𝑎𝑎+3𝑏𝑏)−5𝑎𝑎3

2𝑎𝑎−5𝑏𝑏Combine numerators .

= 3(𝑎𝑎+3𝑏𝑏)+4(𝑎𝑎+3𝑏𝑏)−5𝑎𝑎3

2𝑎𝑎−5𝑏𝑏 Factor out 3 .

= 𝟕𝟕(𝟑𝟑+𝟑𝟑𝟗𝟗)−𝟓𝟓𝟑𝟑𝟑𝟑

𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗Combine like terms .

3. 𝟑𝟑𝒎𝒎− 𝟐𝟐

-𝒎𝒎= 3

𝑚𝑚 + 2

𝑚𝑚= 3+2

𝑚𝑚= 𝟓𝟓

𝒎𝒎Combine numerators .

4. 𝟐𝟐𝟒𝟒−𝟐𝟐

− 𝟏𝟏𝟐𝟐−𝟒𝟒

= 2𝑥𝑥−𝑥𝑥

− 1-(𝑥𝑥−𝑥𝑥)

Factor out (-1) .


+ 1𝑥𝑥−𝑥𝑥

= 𝟑𝟑𝟒𝟒−𝟐𝟐

Combine numerators .

4

1

𝑎𝑎2 + 2𝑎𝑎𝑎𝑎 + 𝑎𝑎2 = (𝑎𝑎 + 𝑎𝑎)2

𝑎𝑎2 − 𝑎𝑎2 = (𝑎𝑎 + 𝑎𝑎)(𝑎𝑎 − 𝑎𝑎)

Page 6-4

6-2 ADDING & SUBTRACTING RATIONAL EXPRESSIONS

Adding / Subtracting Like Rational Expressions

Example

• Like rational expressions: rational expressions that have the 3𝑥𝑥2

𝑥𝑥+2, 5𝑥𝑥

𝑥𝑥+2same denominator .

• Unlike rational expressions: rational expressions that have 7𝑥𝑥+3𝑥𝑥−2

, 3𝑥𝑥−5𝑥𝑥2+5

different denominators .

• Adding or subtracting like rational expressions𝑁𝑁1𝐷𝐷

+ 𝑁𝑁2𝐷𝐷




𝐷𝐷 𝑁𝑁1


𝑁𝑁2

𝐷𝐷are rational expressions . D ≠ 0

Steps Example- Combine the numerators . 3𝑥𝑥

2(𝑥𝑥+4)+ 5𝑥𝑥

2(𝑥𝑥+4)= 3𝑥𝑥+5𝑥𝑥

2(𝑥𝑥+4)

- Denominators do not change . = 8𝑥𝑥2(𝑥𝑥+4)

- Simplify (cancel or reduce common factors) if possible . = 𝟒𝟒𝟒𝟒𝟒𝟒+𝟒𝟒

Example: Add or subtract as indicated and simplify .

1. 𝟒𝟒𝟐𝟐+𝟐𝟐𝟒𝟒𝟐𝟐+𝟐𝟐𝟐𝟐

𝟒𝟒𝟐𝟐−𝟐𝟐𝟐𝟐− 𝟒𝟒+𝟐𝟐

𝟒𝟒𝟐𝟐−𝟐𝟐𝟐𝟐= �𝑥𝑥2+2𝑥𝑥𝑥𝑥+𝑥𝑥2�−(𝑥𝑥+𝑥𝑥)

𝑥𝑥2−𝑥𝑥2Combine numerators .

= (𝟒𝟒+𝟐𝟐)2−(𝟒𝟒+𝟐𝟐) ∙ 1(𝑥𝑥+𝑥𝑥)(𝑥𝑥−𝑥𝑥)

= (𝟒𝟒+𝟐𝟐)[(𝑥𝑥+𝑥𝑥)−1](𝑥𝑥+𝑥𝑥)(𝑥𝑥−𝑥𝑥)

Factor out (x + y) .

= 𝟒𝟒+𝟐𝟐−𝟏𝟏𝟒𝟒−𝟐𝟐

2. 𝟑𝟑𝟑𝟑+𝟗𝟗𝟗𝟗𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗

+ 𝟒𝟒(𝟑𝟑+𝟑𝟑𝟗𝟗)𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗

− 𝟓𝟓𝟑𝟑𝟑𝟑

𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗 = 3𝑎𝑎+9𝑏𝑏+4(𝑎𝑎+3𝑏𝑏)−5𝑎𝑎3

2𝑎𝑎−5𝑏𝑏Combine numerators .

= 3(𝑎𝑎+3𝑏𝑏)+4(𝑎𝑎+3𝑏𝑏)−5𝑎𝑎3

2𝑎𝑎−5𝑏𝑏 Factor out 3 .

= 𝟕𝟕(𝟑𝟑+𝟑𝟑𝟗𝟗)−𝟓𝟓𝟑𝟑𝟑𝟑

𝟐𝟐𝟑𝟑−𝟓𝟓𝟗𝟗Combine like terms .

3. 𝟑𝟑𝒎𝒎− 𝟐𝟐

-𝒎𝒎= 3

𝑚𝑚 + 2

𝑚𝑚= 3+2

𝑚𝑚= 𝟓𝟓

𝒎𝒎Combine numerators .

4. 𝟐𝟐𝟒𝟒−𝟐𝟐

− 𝟏𝟏𝟐𝟐−𝟒𝟒


− 1-(𝑥𝑥−𝑥𝑥)

Factor out (-1) .


+ 1𝑥𝑥−𝑥𝑥

= 𝟑𝟑𝟒𝟒−𝟐𝟐

Combine numerators .

4

1

𝑎𝑎2 + 2𝑎𝑎𝑎𝑎 + 𝑎𝑎2 = (𝑎𝑎 + 𝑎𝑎)2

𝑎𝑎2 − 𝑎𝑎2 = (𝑎𝑎 + 𝑎𝑎)(𝑎𝑎 − 𝑎𝑎)

Page 6-4





Least Common Denominator (LCD)

• Least common multiple (LCM): the lowest number that is divisible by each given number

without a remainder .

Example: The LCM of 2 and 3 is 6 .

- Multiples of 2: 2, 4, 6, 8, 10, 12, …

- Multiples of 3: 3, 6, 9, 12, 15, …

- Common multiples of 2 and 3 are 6 and 12, …

- The least common multiple (LCM) of 2 and 3 is 6 .The common multiple 12 is not the smallest (least) .

• Find the LCM: Use repeated division (or upside-down division) . The product of all the prime numbers around the outside is the LCM .

Example: Find the LCM of 30 and 45 .

5 30 453 6 9 30 ÷ 5 = 6 45 ÷ 5 = 9

2 3 6 ÷ 3 = 2 9 ÷ 3 = 3 (Stop dividing since 2 and 3 are prime numbers)

LCM = 5 × 32 × 2 = 90

• The least common denominator (LCD): the least common multiple (LCM) of the denominators of two or more given fractions .

• Find the LCD: Use repeated division to find the LCM for all denominators of given fractions .

Example: Find the LCD for 422and

165,

84 .

2 8 16 42 2 4 8 21 8 ÷ 2 = 4, 16 ÷ 2 = 8, 42 ÷ 2 = 21

2 2 4 21 4 ÷ 2 = 2, 8 ÷ 2 = 4, move down 21

1 2 21 2 ÷ 2 = 1, 4 ÷ 2 = 2, move down 21

LCD = 24 × 21 = 336

Page 6-5

Least Common Denominator (LCD)

• Least common multiple (LCM): the lowest number that is divisible by each given number

without a remainder .

Example: The LCM of 2 and 3 is 6 .

- Multiples of 2: 2, 4, 6, 8, 10, 12, …

- Multiples of 3: 3, 6, 9, 12, 15, …

- Common multiples of 2 and 3 are 6 and 12, …

- The least common multiple (LCM) of 2 and 3 is 6 .The common multiple 12 is not the smallest (least) .

• Find the LCM: Use repeated division (or upside-down division) . The product of all the prime numbers around the outside is the LCM .

Example: Find the LCM of 30 and 45 .

5 30 453 6 9 30 ÷ 5 = 6 45 ÷ 5 = 9

2 3 6 ÷ 3 = 2 9 ÷ 3 = 3 (Stop dividing since 2 and 3 are prime numbers)

LCM = 5 × 32 × 2 = 90

• The least common denominator (LCD): the least common multiple (LCM) of the denominators of two or more given fractions .

• Find the LCD: Use repeated division to find the LCM for all denominators of given fractions .

Example: Find the LCD for 422and

165,

84 .

2 8 16 42 2 4 8 21 8 ÷ 2 = 4, 16 ÷ 2 = 8, 42 ÷ 2 = 21

2 2 4 21 4 ÷ 2 = 2, 8 ÷ 2 = 4, move down 21

1 2 21 2 ÷ 2 = 1, 4 ÷ 2 = 2, move down 21

LCD = 24 × 21 = 336

Page 6-5





Finding the LCM & LCD for Expressions

• The LCM for algebraic expressions: the smallest expression that is divisible by each of the given expressions . LCM – the least common multiple

• Finding the LCM for expressions

Factor each term . The LCM is the product of all unique factors with the highest exponent .

Example: Find the LCM for 𝟐𝟐𝟐𝟐𝟒𝟒𝒚𝒚𝟐𝟐, 𝟔𝟔𝟐𝟐𝟐𝟐𝒚𝒚𝟑𝟑 and 𝟖𝟖𝟐𝟐𝒚𝒚.Expression Factor Factor With the Highest Exponent

𝟖𝟖𝟐𝟐𝒚𝒚 23 ∙ x ∙ y 23

𝟔𝟔𝟐𝟐𝟐𝟐𝒚𝒚𝟑𝟑 2 ∙ 3 ∙ x2 ∙ y3 3y3

𝟐𝟐𝟐𝟐𝟒𝟒𝒚𝒚𝟐𝟐 2 ∙ x4 ∙ y2 x4

LCM = 23∙ 3y3 ∙ x4 = 24x4y3

Example: Find the LCM for 𝟑𝟑𝟐𝟐(𝟐𝟐𝟐𝟐 − 𝟒𝟒) and 𝟐𝟐𝟐𝟐(𝟐𝟐 + 𝟐𝟐).

Expression Factor Factor With the Highest Exponent𝟑𝟑𝟐𝟐(𝟐𝟐𝟐𝟐 − 𝟒𝟒) 3 ∙ x ∙ (x + 2) ∙ (x – 2) 3 (x + 2) (x – 2)𝟐𝟐𝟐𝟐(𝟐𝟐 + 𝟐𝟐) x2 ∙ (x + 2) x2

LCM = 3x2 (x + 2)(x – 2)

• Finding the LCD for rational expressions LCD – the least common denominator

Factor each given denominator . Find the product of all unique factors with the highest exponent .

Example: Find the LCD for the following fractions .

1. 𝟕𝟕𝟕𝟕𝟗𝟗𝟗𝟗𝟐𝟐

, 𝟓𝟓𝟐𝟐𝟕𝟕𝟕𝟕𝟐𝟐𝟗𝟗

and 𝟒𝟒𝟕𝟕𝟗𝟗𝟔𝟔𝟕𝟕𝟒𝟒𝟗𝟗𝟐𝟐

Denominator Factor Factor With the Highest Exponent𝟗𝟗𝟗𝟗𝟐𝟐 32 ∙ b2 b2

𝟐𝟐𝟕𝟕𝟕𝟕𝟐𝟐𝟗𝟗 33 ∙ a2 ∙ b 33

𝟔𝟔𝟕𝟕𝟒𝟒𝟗𝟗𝟐𝟐 3 ∙ 2 ∙ a4 ∙ b2 2a4

LCD = 2 ∙ 33 ∙ a4 ∙ b2 = 54a4b2

2. 𝟓𝟓𝟕𝟕𝟗𝟗𝟑𝟑𝟕𝟕𝟑𝟑(𝟕𝟕𝟐𝟐−𝟖𝟖𝟕𝟕+𝟏𝟏𝟔𝟔)

and 𝟕𝟕𝟕𝟕𝟐𝟐+𝟗𝟗𝟕𝟕(𝟕𝟕−𝟒𝟒)𝟒𝟒

Denominator Factor Factor With the Highest Exponent𝟑𝟑𝟕𝟕𝟑𝟑(𝟕𝟕𝟐𝟐 − 𝟖𝟖𝟕𝟕 + 𝟏𝟏𝟔𝟔)

a -4a -4

3 ∙ a3 ∙ (a – 4)2 3 a3

𝟕𝟕(𝟕𝟕 − 𝟒𝟒)𝟒𝟒 a ∙ (a – 4)4 (a – 4)4

LCD = 3a3 (a – 4)4

Page 6-6

Finding the LCM & LCD for Expressions

• The LCM for algebraic expressions: the smallest expression that is divisible by each of the given expressions . LCM – the least common multiple

• Finding the LCM for expressions

Factor each term . The LCM is the product of all unique factors with the highest exponent .

Example: Find the LCM for 𝟐𝟐𝟐𝟐𝟒𝟒𝒚𝒚𝟐𝟐, 𝟔𝟔𝟐𝟐𝟐𝟐𝒚𝒚𝟑𝟑 and 𝟖𝟖𝟐𝟐𝒚𝒚.Expression Factor Factor With the Highest Exponent

𝟖𝟖𝟐𝟐𝒚𝒚 23 ∙ x ∙ y 23

𝟔𝟔𝟐𝟐𝟐𝟐𝒚𝒚𝟑𝟑 2 ∙ 3 ∙ x2 ∙ y3 3y3

𝟐𝟐𝟐𝟐𝟒𝟒𝒚𝒚𝟐𝟐 2 ∙ x4 ∙ y2 x4

LCM = 23∙ 3y3 ∙ x4 = 24x4y3

Example: Find the LCM for 𝟑𝟑𝟐𝟐(𝟐𝟐𝟐𝟐 − 𝟒𝟒) and 𝟐𝟐𝟐𝟐(𝟐𝟐 + 𝟐𝟐).

Expression Factor Factor With the Highest Exponent𝟑𝟑𝟐𝟐(𝟐𝟐𝟐𝟐 − 𝟒𝟒) 3 ∙ x ∙ (x + 2) ∙ (x – 2) 3 (x + 2) (x – 2)𝟐𝟐𝟐𝟐(𝟐𝟐 + 𝟐𝟐) x2 ∙ (x + 2) x2

LCM = 3x2 (x + 2)(x – 2)

• Finding the LCD for rational expressions LCD – the least common denominator

Factor each given denominator . Find the product of all unique factors with the highest exponent .

Example: Find the LCD for the following fractions .

1. 𝟕𝟕𝟕𝟕𝟗𝟗𝟗𝟗𝟐𝟐

, 𝟓𝟓𝟐𝟐𝟕𝟕𝟕𝟕𝟐𝟐𝟗𝟗

and 𝟒𝟒𝟕𝟕𝟗𝟗𝟔𝟔𝟕𝟕𝟒𝟒𝟗𝟗𝟐𝟐

Denominator Factor Factor With the Highest Exponent𝟗𝟗𝟗𝟗𝟐𝟐 32 ∙ b2 b2

𝟐𝟐𝟕𝟕𝟕𝟕𝟐𝟐𝟗𝟗 33 ∙ a2 ∙ b 33

𝟔𝟔𝟕𝟕𝟒𝟒𝟗𝟗𝟐𝟐 3 ∙ 2 ∙ a4 ∙ b2 2a4

LCD = 2 ∙ 33 ∙ a4 ∙ b2 = 54a4b2

2. 𝟓𝟓𝟕𝟕𝟗𝟗𝟑𝟑𝟕𝟕𝟑𝟑(𝟕𝟕𝟐𝟐−𝟖𝟖𝟕𝟕+𝟏𝟏𝟔𝟔)

and 𝟕𝟕𝟕𝟕𝟐𝟐+𝟗𝟗𝟕𝟕(𝟕𝟕−𝟒𝟒)𝟒𝟒

Denominator Factor Factor With the Highest Exponent𝟑𝟑𝟕𝟕𝟑𝟑(𝟕𝟕𝟐𝟐 − 𝟖𝟖𝟕𝟕 + 𝟏𝟏𝟔𝟔)

a -4a -4

3 ∙ a3 ∙ (a – 4)2 3 a3

𝟕𝟕(𝟕𝟕 − 𝟒𝟒)𝟒𝟒 a ∙ (a – 4)4 (a – 4)4

LCD = 3a3 (a – 4)4

Page 6-6





Adding/Subtracting Unlike Rational Expressions

• Adding or subtracting unlike fractions (with different denominators)

𝟐𝟐𝟑𝟑

+ 𝟏𝟏𝟒𝟒

= 2 ∙ 𝟒𝟒3 ∙ 𝟒𝟒

+ 1 ∙ 𝟑𝟑 4 ∙ 𝟑𝟑

= 812

+ 312

= 8+312

= 𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐

LCD = 12

𝟓𝟓𝟏𝟏𝟐𝟐− 𝟑𝟑

𝟖𝟖= 5∙212∙2

− 3∙38∙3

= 1024− 9

24 = 10−9

24= 𝟏𝟏𝟐𝟐𝟒𝟒

2 12 8

2 6 4 3 2 LCD = 23 ∙ 3 = 24

• Adding or subtracting unlike rational expressions

Steps Example: 𝟓𝟓𝟓𝟓𝟐𝟐+𝟏𝟏𝟑𝟑𝟓𝟓

+ 𝟕𝟕𝟓𝟓𝟐𝟐−𝟏𝟏𝟒𝟒𝟓𝟓

- Determine the LCD . LCD = 3 ∙ 4 ∙ x = 12x

- Rewrite expressions with the LCD . 𝟓𝟓𝟓𝟓𝟐𝟐+𝟏𝟏𝟑𝟑𝟓𝟓

+ 𝟕𝟕𝟓𝟓𝟐𝟐−𝟏𝟏𝟒𝟒𝟓𝟓

= 𝟒𝟒�5𝑥𝑥2+1�𝟒𝟒 ∙ 3𝑥𝑥

+ 𝟑𝟑(7𝑥𝑥2−1) 𝟑𝟑 ∙ 4𝑥𝑥

- Combine the numerators . = 4�5𝑥𝑥2+1�+3(7𝑥𝑥2−1)12𝑥𝑥

= 20𝑥𝑥2+ 4+ 21𝑥𝑥2−312𝑥𝑥

- Simplify if possible . = 41𝑥𝑥2+ 112𝑥𝑥

= 41𝑥𝑥2

12 𝑥𝑥+ 1

12𝑥𝑥= 𝟒𝟒𝟏𝟏𝟓𝟓

𝟏𝟏𝟐𝟐+ 𝟏𝟏

𝟏𝟏𝟐𝟐𝟓𝟓

Example: Add or subtract and simplify .

1. 𝟑𝟑𝟑𝟑𝟑𝟑−𝟑𝟑

+ 𝟓𝟓𝟑𝟑+𝟑𝟑

− 𝟐𝟐𝟑𝟑𝟐𝟐−𝟗𝟗

= 3𝑎𝑎(𝟑𝟑+𝟑𝟑)

(𝑎𝑎−3)(𝟑𝟑+𝟑𝟑)+ 5(𝟑𝟑−𝟑𝟑)

(𝑎𝑎+3)(𝟑𝟑−𝟑𝟑)− 2

(𝑎𝑎+3)(𝑎𝑎−3)a2 – 9 = a2 – 32 , LCD = (a + 3)(a – 3)

= 3𝑎𝑎(𝑎𝑎+3)+5(𝑎𝑎−3)−2(𝑎𝑎+3)(𝑎𝑎−3)

Combine the numerators .

= 3𝑎𝑎2+9𝑎𝑎+5𝑎𝑎−15−2(𝑎𝑎+3)(𝑎𝑎−3)

= 𝟑𝟑𝟑𝟑𝟐𝟐+𝟏𝟏𝟒𝟒𝟑𝟑−𝟏𝟏𝟕𝟕

(𝟑𝟑+𝟑𝟑)(𝟑𝟑−𝟑𝟑)Combine like terms .

2. 𝟐𝟐𝟓𝟓𝟑𝟑𝟓𝟓𝟐𝟐+ 𝟐𝟐𝟓𝟓−𝟏𝟏

− 𝟑𝟑𝟓𝟓+𝟏𝟏

= 2𝑥𝑥(𝑥𝑥+1)(3𝑥𝑥−1)

− 3𝑥𝑥+1

LCD = (x + 1) (3x – 1)

x 13x - 1

= 2𝑥𝑥(𝑥𝑥+1)(3𝑥𝑥−1)

− 3(3𝑥𝑥−1)(𝑥𝑥+1)(3𝑥𝑥−1)

Rewrite with the LCD .

= 2𝑥𝑥 − [3(3𝑥𝑥−1)](𝑥𝑥+1)(3𝑥𝑥−1)


= 2𝑥𝑥−9𝑥𝑥+3(𝑥𝑥+1)(3𝑥𝑥−1)

Distribute

= -𝟕𝟕𝟓𝟓+𝟑𝟑(𝟓𝟓+𝟏𝟏)(𝟑𝟑𝟓𝟓−𝟏𝟏)


Page 6-7

Adding/Subtracting Unlike Rational Expressions

• Adding or subtracting unlike fractions (with different denominators)

𝟐𝟐𝟑𝟑

+ 𝟏𝟏𝟒𝟒

= 2 ∙ 𝟒𝟒3 ∙ 𝟒𝟒

+ 1 ∙ 𝟑𝟑 4 ∙ 𝟑𝟑

= 812

+ 312

= 8+312

= 𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐

LCD = 12

𝟓𝟓𝟏𝟏𝟐𝟐− 𝟑𝟑

𝟖𝟖= 5∙212∙2

− 3∙38∙3

= 1024− 9

24 = 10−9

24= 𝟏𝟏𝟐𝟐𝟒𝟒

2 12 8

2 6 4 3 2 LCD = 23 ∙ 3 = 24


Steps Example: 𝟓𝟓𝟓𝟓𝟐𝟐+𝟏𝟏𝟑𝟑𝟓𝟓

+ 𝟕𝟕𝟓𝟓𝟐𝟐−𝟏𝟏𝟒𝟒𝟓𝟓

- Determine the LCD . LCD = 3 ∙ 4 ∙ x = 12x

- Rewrite expressions with the LCD . 𝟓𝟓𝟓𝟓𝟐𝟐+𝟏𝟏𝟑𝟑𝟓𝟓

+ 𝟕𝟕𝟓𝟓𝟐𝟐−𝟏𝟏𝟒𝟒𝟓𝟓

= 𝟒𝟒�5𝑥𝑥2+1�𝟒𝟒 ∙ 3𝑥𝑥

+ 𝟑𝟑(7𝑥𝑥2−1) 𝟑𝟑 ∙ 4𝑥𝑥

- Combine the numerators . = 4�5𝑥𝑥2+1�+3(7𝑥𝑥2−1)12𝑥𝑥

= 20𝑥𝑥2+ 4+ 21𝑥𝑥2−312𝑥𝑥

- Simplify if possible . = 41𝑥𝑥2+ 112𝑥𝑥

= 41𝑥𝑥2

12 𝑥𝑥+ 1

12𝑥𝑥= 𝟒𝟒𝟏𝟏𝟓𝟓

𝟏𝟏𝟐𝟐+ 𝟏𝟏

𝟏𝟏𝟐𝟐𝟓𝟓

Example: Add or subtract and simplify .

1. 𝟑𝟑𝟑𝟑𝟑𝟑−𝟑𝟑

+ 𝟓𝟓𝟑𝟑+𝟑𝟑

− 𝟐𝟐𝟑𝟑𝟐𝟐−𝟗𝟗

= 3𝑎𝑎(𝟑𝟑+𝟑𝟑)

(𝑎𝑎−3)(𝟑𝟑+𝟑𝟑)+ 5(𝟑𝟑−𝟑𝟑)

(𝑎𝑎+3)(𝟑𝟑−𝟑𝟑)− 2

(𝑎𝑎+3)(𝑎𝑎−3)a2 – 9 = a2 – 32 , LCD = (a + 3)(a – 3)

= 3𝑎𝑎(𝑎𝑎+3)+5(𝑎𝑎−3)−2(𝑎𝑎+3)(𝑎𝑎−3)


= 3𝑎𝑎2+9𝑎𝑎+5𝑎𝑎−15−2(𝑎𝑎+3)(𝑎𝑎−3)

= 𝟑𝟑𝟑𝟑𝟐𝟐+𝟏𝟏𝟒𝟒𝟑𝟑−𝟏𝟏𝟕𝟕

(𝟑𝟑+𝟑𝟑)(𝟑𝟑−𝟑𝟑)Combine like terms .

2. 𝟐𝟐𝟓𝟓𝟑𝟑𝟓𝟓𝟐𝟐+ 𝟐𝟐𝟓𝟓−𝟏𝟏

− 𝟑𝟑𝟓𝟓+𝟏𝟏

= 2𝑥𝑥(𝑥𝑥+1)(3𝑥𝑥−1)

− 3𝑥𝑥+1

LCD = (x + 1) (3x – 1)

x 13x - 1

= 2𝑥𝑥(𝑥𝑥+1)(3𝑥𝑥−1)

− 3(3𝑥𝑥−1)(𝑥𝑥+1)(3𝑥𝑥−1)

Rewrite with the LCD .

= 2𝑥𝑥 − [3(3𝑥𝑥−1)](𝑥𝑥+1)(3𝑥𝑥−1)


= 2𝑥𝑥−9𝑥𝑥+3(𝑥𝑥+1)(3𝑥𝑥−1)

Distribute

= -𝟕𝟕𝟓𝟓+𝟑𝟑(𝟓𝟓+𝟏𝟏)(𝟑𝟑𝟓𝟓−𝟏𝟏)


Page 6-7





6-3 POLYNOMIAL DIVISION

Dividing Polynomials

• Dividing a monomial by a monomial Monomial: one term

Example: -𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏𝒚𝒚𝟓𝟓

𝟒𝟒𝒙𝒙𝟑𝟑𝒚𝒚𝟏𝟏

Steps Solution

- Divide coefficients . -𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏𝒚𝒚𝟓𝟓

𝟒𝟒𝒙𝒙𝟑𝟑𝒚𝒚𝟏𝟏= �-12

4� �𝑥𝑥

2𝑦𝑦5

𝑥𝑥3𝑦𝑦2�

- Divide like variables . (apply 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= am-n ) = -3 �𝑥𝑥

2

𝑥𝑥3� �𝑦𝑦

5

𝑦𝑦2�

= -𝟑𝟑 �𝒚𝒚𝟑𝟑

𝒙𝒙�

• Dividing a polynomial by a monomial

Example: 12𝑥𝑥2+4𝑥𝑥−2

4𝑥𝑥

Steps Solution

- Split the polynomial into three parts . 𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏+𝟒𝟒𝒙𝒙−𝟏𝟏𝟒𝟒𝒙𝒙

= 12𝑥𝑥2

4𝑥𝑥+ 4𝑥𝑥

4𝑥𝑥− 2

4 𝑥𝑥

- Divide a monomial by a monomial . = 3x + 1 – 𝟏𝟏𝟏𝟏𝒙𝒙

Cancel or reduce common factors .

Example: 𝟑𝟑𝒚𝒚𝟏𝟏+𝟑𝟑𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏

Steps Solution

- Group . 𝟑𝟑𝒚𝒚𝟏𝟏+𝟑𝟑𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏

= (3𝑦𝑦2+3𝑦𝑦)+(2𝑦𝑦+2)𝑦𝑦+1

- Factor out the GCF . = 3𝑦𝑦(𝑦𝑦+1)+2(𝑦𝑦+1)𝑦𝑦+1

- Split the polynomial into two parts . = 3𝑦𝑦(𝑦𝑦+1)𝑦𝑦+1

+ 2(𝑦𝑦+1)𝑦𝑦+1

- Divide a monomial by a monomial . = 3y + 2

1

21

3

Page 6-8

6-3 POLYNOMIAL DIVISION

Dividing Polynomials

• Dividing a monomial by a monomial Monomial: one term

Example: -𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏𝒚𝒚𝟓𝟓

𝟒𝟒𝒙𝒙𝟑𝟑𝒚𝒚𝟏𝟏

Steps Solution

- Divide coefficients . -𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏𝒚𝒚𝟓𝟓

𝟒𝟒𝒙𝒙𝟑𝟑𝒚𝒚𝟏𝟏= �-12

4� �𝑥𝑥

2𝑦𝑦5

𝑥𝑥3𝑦𝑦2�

- Divide like variables . (apply 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= am-n ) = -3 �𝑥𝑥

2

𝑥𝑥3� �𝑦𝑦

5

𝑦𝑦2�

= -𝟑𝟑 �𝒚𝒚𝟑𝟑

𝒙𝒙�

• Dividing a polynomial by a monomial

Example: 12𝑥𝑥2+4𝑥𝑥−2

4𝑥𝑥

Steps Solution

- Split the polynomial into three parts . 𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏+𝟒𝟒𝒙𝒙−𝟏𝟏𝟒𝟒𝒙𝒙

= 12𝑥𝑥2

4𝑥𝑥+ 4𝑥𝑥

4𝑥𝑥− 2

4 𝑥𝑥

- Divide a monomial by a monomial . = 3x + 1 – 𝟏𝟏𝟏𝟏𝒙𝒙

Cancel or reduce common factors .

Example: 𝟑𝟑𝒚𝒚𝟏𝟏+𝟑𝟑𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏

Steps Solution

- Group . 𝟑𝟑𝒚𝒚𝟏𝟏+𝟑𝟑𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏𝒚𝒚+𝟏𝟏

= (3𝑦𝑦2+3𝑦𝑦)+(2𝑦𝑦+2)𝑦𝑦+1

- Factor out the GCF . = 3𝑦𝑦(𝑦𝑦+1)+2(𝑦𝑦+1)𝑦𝑦+1

- Split the polynomial into two parts . = 3𝑦𝑦(𝑦𝑦+1)𝑦𝑦+1

+ 2(𝑦𝑦+1)𝑦𝑦+1

- Divide a monomial by a monomial . = 3y + 2

1

21

3

Page 6-8





Long Division of Polynomials

• Division of whole numbers (long division): Example:

Quotient 3Divisor Dividend 4 15

– – 12 Remainder 3

• Polynomial long division works more conveniently for more general polynomials .

Example: 6𝑥𝑥2+9𝑥𝑥+2

3𝑥𝑥Steps Solution Dividing whole numbers

- Write in divisor dividend form . 3x 6𝑥𝑥2 + 9𝑥𝑥 + 2 3 692

2x 2- Divide the first term . 3x 6𝑥𝑥2 + 9𝑥𝑥 + 2 3 692

– 6x2 (3x)(2x) = 6x

2 – 6 2∙3 = 6

2x + 3 230- Divide the second term . 3x 6𝑥𝑥2 + 9𝑥𝑥 + 2 3 692

6x2 6 Bring 9x down 9x 9 Bring 9 down

(3x)(3) = 9x – 9x – 9 3∙3 = 9

2 2remainder

quotient remainder

Quotient + remainderdivisor

6𝑥𝑥2+9𝑥𝑥+2

3𝑥𝑥= (2x + 3) + 𝟐𝟐

𝟑𝟑𝟑𝟑 692 ÷ 3 = 230 + 𝟐𝟐

𝟑𝟑

divisor

Tip: Continue until the degree of the remainder is less than the degree of the divisor .

(i .e . 2 = 2 ∙ x0 and 3x = 3x1 )

0 < 1

- Check: Dividend = Quotient ∙ Divisor + Remainder ? ?

6𝑥𝑥2 + 9𝑥𝑥 + 2 = (2x + 3)(3x) + 2 692 = 230 ∙ 3 + 2

√ √ 6𝑥𝑥2 + 9𝑥𝑥 + 2 = 6𝑥𝑥2 + 9x + 2 Correct! 692 = 692

Page 6-9

Long Division of Polynomials

• Division of whole numbers (long division): Example:

Quotient 3Divisor Dividend 4 15

– – 12 Remainder 3

• Polynomial long division works more conveniently for more general polynomials .

Example: 6𝑥𝑥2+9𝑥𝑥+2

3𝑥𝑥Steps Solution Dividing whole numbers



– 6x2 (3x)(2x) = 6x

2 – 6 2∙3 = 6

2x + 3 230- Divide the second term . 3x 6𝑥𝑥2 + 9𝑥𝑥 + 2 3 692

6x2 6 Bring 9x down 9x 9 Bring 9 down

(3x)(3) = 9x – 9x – 9 3∙3 = 9

2 2remainder

quotient remainder



3𝑥𝑥= (2x + 3) + 𝟐𝟐

𝟑𝟑𝟑𝟑 692 ÷ 3 = 230 + 𝟐𝟐

𝟑𝟑

divisor

Tip: Continue until the degree of the remainder is less than the degree of the divisor .

(i .e . 2 = 2 ∙ x0 and 3x = 3x1 )

0 < 1

- Check: Dividend = Quotient ∙ Divisor + Remainder ? ?

6𝑥𝑥2 + 9𝑥𝑥 + 2 = (2x + 3)(3x) + 2 692 = 230 ∙ 3 + 2

√ √ 6𝑥𝑥2 + 9𝑥𝑥 + 2 = 6𝑥𝑥2 + 9x + 2 Correct! 692 = 692

Page 6-9

26926





Missing Terms in Long Division

Missing terms in long division: If there is a missing consecutive power term in a polynomial

(i .e . if there are x3 and x, but not x2), insert the missing power term with a coefficient of 0 .

Example: 𝟓𝟓−𝟑𝟑𝒂𝒂𝟐𝟐+𝒂𝒂𝟑𝟑

𝟏𝟏+𝒂𝒂

Steps Solution

- Rewrite both polynomials in descending order .𝑎𝑎3−3𝑎𝑎2+5

𝑎𝑎+1Descending order: 𝐴𝐴𝐴𝐴3 + B𝐴𝐴2 + 𝐶𝐶𝐴𝐴 + 𝐷𝐷 , 𝐴𝐴𝐴𝐴 + 𝐵𝐵

- Write in divisor Dividend form and insert a + 1 𝑎𝑎3 − 3𝑎𝑎2 + 𝟎𝟎𝒂𝒂 + 5

a 0 coefficient for the missing power term . Missing power

𝑎𝑎2 − 4𝑎𝑎 + 4- Divide as usual . a + 1 𝑎𝑎3 − 3𝑎𝑎2 + 0𝑎𝑎 + 5

− 𝑎𝑎3 + 𝑎𝑎2

-4𝑎𝑎2 + 0𝑎𝑎 − -4𝑎𝑎2 − 4𝑎𝑎 4a + 5

− 4a + 41

- Solution . 5−3𝑎𝑎2+𝑎𝑎3

1+𝑎𝑎= (𝒂𝒂𝟐𝟐 − 𝟒𝟒𝒂𝒂 + 𝟒𝟒) +

𝟏𝟏𝒂𝒂+𝟏𝟏


- Check: Dividend = Quotient ∙ Divisor + Remainder ?

5 − 3𝑎𝑎2 + 𝑎𝑎3 = (𝑎𝑎2 − 4𝑎𝑎 + 4)(𝑎𝑎 + 1) + 1

? 5 − 3𝑎𝑎2 + 𝑎𝑎3 = (𝑎𝑎3 + 𝑎𝑎2 − 4𝑎𝑎2 − 4𝑎𝑎 + 4𝑎𝑎 + 4) + 1

√ 5 − 3𝑎𝑎2 + 𝑎𝑎3 = 𝑎𝑎3 − 3𝑎𝑎2 + 5 Correct!

Page 6-10

Missing Terms in Long Division

Missing terms in long division: If there is a missing consecutive power term in a polynomial

(i .e . if there are x3 and x, but not x2), insert the missing power term with a coefficient of 0 .

Example: 𝟓𝟓−𝟑𝟑𝒂𝒂𝟐𝟐+𝒂𝒂𝟑𝟑

𝟏𝟏+𝒂𝒂

Steps Solution

- Rewrite both polynomials in descending order .𝑎𝑎3−3𝑎𝑎2+5

𝑎𝑎+1Descending order: 𝐴𝐴𝐴𝐴3 + B𝐴𝐴2 + 𝐶𝐶𝐴𝐴 + 𝐷𝐷 , 𝐴𝐴𝐴𝐴 + 𝐵𝐵

- Write in divisor Dividend form and insert a + 1 𝑎𝑎3 − 3𝑎𝑎2 + 𝟎𝟎𝒂𝒂 + 5

a 0 coefficient for the missing power term . Missing power

𝑎𝑎2 − 4𝑎𝑎 + 4- Divide as usual . a + 1 𝑎𝑎3 − 3𝑎𝑎2 + 0𝑎𝑎 + 5

− 𝑎𝑎3 + 𝑎𝑎2

-4𝑎𝑎2 + 0𝑎𝑎 − -4𝑎𝑎2 − 4𝑎𝑎 4a + 5

− 4a + 41

- Solution . 5−3𝑎𝑎2+𝑎𝑎3

1+𝑎𝑎= (𝒂𝒂𝟐𝟐 − 𝟒𝟒𝒂𝒂 + 𝟒𝟒) +

𝟏𝟏𝒂𝒂+𝟏𝟏


- Check: Dividend = Quotient ∙ Divisor + Remainder ?

5 − 3𝑎𝑎2 + 𝑎𝑎3 = (𝑎𝑎2 − 4𝑎𝑎 + 4)(𝑎𝑎 + 1) + 1

? 5 − 3𝑎𝑎2 + 𝑎𝑎3 = (𝑎𝑎3 + 𝑎𝑎2 − 4𝑎𝑎2 − 4𝑎𝑎 + 4𝑎𝑎 + 4) + 1

√ 5 − 3𝑎𝑎2 + 𝑎𝑎3 = 𝑎𝑎3 − 3𝑎𝑎2 + 5 Correct!

Page 6-10





Synthetic Division

Synthetic Division: a shortcut method of dividing a polynomial by a binomial of the form

(x – a), by using only the coefficients of the terms .

Steps Example (x – a)

(𝟑𝟑𝒙𝒙𝟒𝟒 − 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝒙𝒙 − 𝟐𝟐𝒙𝒙𝟐𝟐) ÷ (𝒙𝒙 − 𝟐𝟐)

- Rewrite the polynomial in descending order . 𝑥𝑥 − 2 3𝑥𝑥4 + 0𝑥𝑥3 − 2𝑥𝑥2 + 4𝑥𝑥 − 40Insert a zero coefficient for the missing power . Missing power

a in the divisor The coefficient of the dividend- Set up the synthetic coefficients. 2 3 0 -2 4 -40

- Bring down the leading coefficient and multiply 2 3 0 -2 4 -40it by a in (x – a) . Place the product beneath the 2 × 3

second coefficient. 3 6

- Add the numbers in column 2 . 2 3 0 -2 4 -40

63 6 (0 + 6)

- Repeat until the last column done . 2 3 0 -2 4 -40 2 × 6 = 12 , -2 + 12 = 10 6 12 20 482 × 10 = 20 , 4 + 20 = 24 3 6 10 24 82 × 24 = 48 , -40 + 48 = 8

- Write out the answer . 𝑥𝑥3 𝑥𝑥2 x constant remainder


Answer: (𝟑𝟑𝒙𝒙𝟑𝟑 + 𝟔𝟔𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 + 𝟐𝟐𝟒𝟒) + 𝟖𝟖𝒙𝒙−𝟐𝟐

One less than the degree of the dividend

Example: Divide . (𝟔𝟔 + 𝟒𝟒𝒙𝒙𝟑𝟑 − 𝟐𝟐𝒙𝒙 − 𝟐𝟐𝒙𝒙𝟐𝟐) ÷ (𝒙𝒙 + 𝟏𝟏)𝑥𝑥 + 𝟏𝟏 4𝑥𝑥3−2𝑥𝑥2 − 2𝑥𝑥 + 6 Write in descending order .

-1 4 -2 -2 6

-1 × 4 -4 6 -4

4 -6 4 2 𝑥𝑥2 x constant remainder

Answer: (𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟔𝟔𝒙𝒙 + 𝟒𝟒) + 𝟐𝟐𝒙𝒙+𝟏𝟏


+

+

(a = 2)

+ +

2nd coefficient

(x – a): x + 1 = x – (-1) , ∴ a = -1The synthetic coefficients

-1 × 4 = -4 , -2 + (-4) = -6-1 × (-6) = 6 , -2 + 6 = 4-1 × 4 = -4 , 6 + (-4) = 2

Page 6-11

Synthetic Division

Synthetic Division: a shortcut method of dividing a polynomial by a binomial of the form



(𝟑𝟑𝒙𝒙𝟒𝟒 − 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝒙𝒙 − 𝟐𝟐𝒙𝒙𝟐𝟐) ÷ (𝒙𝒙 − 𝟐𝟐)





- Add the numbers in column 2 . 2 3 0 -2 4 -40

63 6 (0 + 6)




Answer: (𝟑𝟑𝒙𝒙𝟑𝟑 + 𝟔𝟔𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 + 𝟐𝟐𝟒𝟒) + 𝟖𝟖𝒙𝒙−𝟐𝟐

One less than the degree of the dividend

Example: Divide . (𝟔𝟔 + 𝟒𝟒𝒙𝒙𝟑𝟑 − 𝟐𝟐𝒙𝒙 − 𝟐𝟐𝒙𝒙𝟐𝟐) ÷ (𝒙𝒙 + 𝟏𝟏)𝑥𝑥 + 𝟏𝟏 4𝑥𝑥3−2𝑥𝑥2 − 2𝑥𝑥 + 6 Write in descending order .

-1 4 -2 -2 6

-1 × 4 -4 6 -4

4 -6 4 2 𝑥𝑥2 x constant remainder

Answer: (𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟔𝟔𝒙𝒙 + 𝟒𝟒) + 𝟐𝟐𝒙𝒙+𝟏𝟏


+

+

(a = 2)

+ +

2nd coefficient

(x – a): x + 1 = x – (-1) , ∴ a = -1The synthetic coefficients

-1 × 4 = -4 , -2 + (-4) = -6-1 × (-6) = 6 , -2 + 6 = 4-1 × 4 = -4 , 6 + (-4) = 2

Page 6-11





6-4 COMPLEX RATIONAL EXPRESSIONS

Simplify Complex Rational ExpressionsMethod I

• Complex fraction: a fraction that contains another fraction in its numerator or denominator(or both) . fractions

Example: 34 32

= 34

÷ 32 ÷

• Complex rational expression: a rational expression whose numerator or denominator (or both) contains rational expressions .

Example

𝑁𝑁1𝐷𝐷1

𝑁𝑁2𝐷𝐷2

𝑁𝑁1𝐷𝐷1

and 𝑁𝑁2𝐷𝐷2

are rational expressions . D1, D2 & N2 ≠ 0 5𝑥𝑥 − 3𝑥𝑥𝑥𝑥

2

𝑥𝑥 3𝑥𝑥 + 4𝑥𝑥𝑥𝑥

• Simplifying a complex rational expression — method I: multiplying the LCD .

Examples: Simplify the following .

1. 𝟏𝟏𝒘𝒘 − 𝟑𝟑 𝟏𝟏𝒘𝒘 + 𝒘𝒘

= � 1𝑤𝑤 − 3 1𝑤𝑤 + 𝑤𝑤

� ∙ 𝒘𝒘𝒘𝒘

= 𝑤𝑤𝑤𝑤 − 3𝑤𝑤 𝑤𝑤𝑤𝑤 + 𝑤𝑤2 Multiply num . & den . by the LCD . (w)

= 𝟏𝟏−𝟑𝟑𝒘𝒘𝟏𝟏+𝒘𝒘𝟐𝟐 Simplify .

2. 𝟏𝟏𝒂𝒂𝟐𝟐

+ 𝟐𝟐𝒃𝒃

𝟑𝟑𝒂𝒂𝒃𝒃𝟐𝟐

− 𝟏𝟏𝒂𝒂𝟐𝟐

=

� 1 𝑎𝑎2

+ 2𝑏𝑏 �(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐)

�3𝑎𝑎 𝑏𝑏2− 1𝑎𝑎2

�(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐) Multiply num . & den . by the LCD . (a2b2)

= 𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐

𝑎𝑎2+ 2(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐)

𝑏𝑏3𝑎𝑎(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐)

𝑏𝑏2 − (𝒂𝒂

𝟐𝟐𝒃𝒃𝟐𝟐) 𝑎𝑎2

= 𝑏𝑏2+ 2𝑎𝑎2𝑏𝑏3𝑎𝑎3 − 𝑏𝑏2

Simplify .

= 𝒃𝒃(𝒃𝒃+𝟐𝟐𝒂𝒂𝟐𝟐)𝟑𝟑𝒂𝒂𝟑𝟑−𝒃𝒃𝟐𝟐

Factor out b.

Page 6-12

6-4 COMPLEX RATIONAL EXPRESSIONS

Simplify Complex Rational ExpressionsMethod I

• Complex fraction: a fraction that contains another fraction in its numerator or denominator(or both) . fractions

Example: 34 32

= 34

÷ 32 ÷

• Complex rational expression: a rational expression whose numerator or denominator (or both) contains rational expressions .

Example

𝑁𝑁1𝐷𝐷1

𝑁𝑁2𝐷𝐷2

𝑁𝑁1𝐷𝐷1


are rational expressions . D1, D2 & N2 ≠ 0 5𝑥𝑥 − 3𝑥𝑥𝑥𝑥

2

𝑥𝑥 3𝑥𝑥 + 4𝑥𝑥𝑥𝑥

• Simplifying a complex rational expression — method I: multiplying the LCD .


1. 𝟏𝟏𝒘𝒘 − 𝟑𝟑 𝟏𝟏𝒘𝒘 + 𝒘𝒘

= � 1𝑤𝑤 − 3 1𝑤𝑤 + 𝑤𝑤

� ∙ 𝒘𝒘𝒘𝒘

= 𝑤𝑤𝑤𝑤 − 3𝑤𝑤 𝑤𝑤𝑤𝑤 + 𝑤𝑤2 Multiply num . & den . by the LCD . (w)

= 𝟏𝟏−𝟑𝟑𝒘𝒘𝟏𝟏+𝒘𝒘𝟐𝟐 Simplify .

2. 𝟏𝟏𝒂𝒂𝟐𝟐

+ 𝟐𝟐𝒃𝒃

𝟑𝟑𝒂𝒂𝒃𝒃𝟐𝟐

− 𝟏𝟏𝒂𝒂𝟐𝟐

=

� 1 𝑎𝑎2

+ 2𝑏𝑏 �(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐)

�3𝑎𝑎 𝑏𝑏2− 1𝑎𝑎2

�(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐) Multiply num . & den . by the LCD . (a2b2)

= 𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐

𝑎𝑎2+ 2(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐)

𝑏𝑏3𝑎𝑎(𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐)

𝑏𝑏2 − (𝒂𝒂

𝟐𝟐𝒃𝒃𝟐𝟐) 𝑎𝑎2

= 𝑏𝑏2+ 2𝑎𝑎2𝑏𝑏3𝑎𝑎3 − 𝑏𝑏2

Simplify .

= 𝒃𝒃(𝒃𝒃+𝟐𝟐𝒂𝒂𝟐𝟐)𝟑𝟑𝒂𝒂𝟑𝟑−𝒃𝒃𝟐𝟐

Factor out b.

Page 6-12





Simplify Complex Rational ExpressionsMethod II

Simplifying a complex rational expression — method II: multiply the reciprocal of the

denominator (÷ → ×) .

𝑁𝑁1𝐷𝐷1

𝑁𝑁2𝐷𝐷2

= 𝑁𝑁1

𝐷𝐷1÷ 𝑵𝑵𝟐𝟐

𝑫𝑫𝟐𝟐= 𝑁𝑁1

𝐷𝐷1∙ 𝑫𝑫𝟐𝟐 𝑵𝑵𝟐𝟐

𝑁𝑁1𝐷𝐷1


are rational expressions . 34 32

= 34

÷ 32

= 34

∙ 23

= 𝟏𝟏𝟐𝟐


1. 𝟏𝟏𝒘𝒘 − 𝟑𝟑 𝟏𝟏𝒘𝒘 + 𝒘𝒘

= 1𝑤𝑤 − 31 1𝑤𝑤 + 𝑤𝑤1

= 1

𝑤𝑤 − 3𝒘𝒘𝒘𝒘 1𝑤𝑤 + 𝑤𝑤

𝟐𝟐𝒘𝒘

Multiply num . & den . by the LCD . LCD = w

= 1−3𝑤𝑤𝑤𝑤 1+ 𝑤𝑤2𝑤𝑤

= 1−3𝑤𝑤𝑤𝑤

÷ 1+𝑤𝑤2

𝑤𝑤 ÷


∙ 𝒘𝒘𝟏𝟏+𝒘𝒘𝟐𝟐

÷ × , 1+𝑤𝑤2 𝑤𝑤

𝑤𝑤1+𝑤𝑤2

= 𝟏𝟏−𝟑𝟑𝒘𝒘𝟏𝟏+𝒘𝒘𝟐𝟐

2. 𝒚𝒚𝟐𝟐−𝟐𝟐𝒚𝒚−𝟖𝟖𝟐𝟐𝒚𝒚+𝟔𝟔𝒚𝒚𝟐𝟐

𝟑𝟑𝒚𝒚𝟐𝟐+𝟒𝟒𝒚𝒚−𝟒𝟒𝟑𝟑𝒚𝒚+𝟏𝟏

= 𝑦𝑦2−2𝑦𝑦−82𝑦𝑦+6𝑦𝑦2

÷ 3𝑦𝑦2+4𝑦𝑦−43𝑦𝑦+1

= 𝑦𝑦2−2𝑦𝑦−8

2𝑦𝑦+6𝑦𝑦2 ∙ 𝟑𝟑𝒚𝒚+𝟏𝟏𝟑𝟑𝒚𝒚𝟐𝟐+𝟒𝟒𝒚𝒚−𝟒𝟒

Factor: 𝑦𝑦2 − 2𝑦𝑦 − 8 , 3𝑦𝑦2 + 4𝑦𝑦 − 4

= (𝑦𝑦+2)(𝑦𝑦−4)2𝑦𝑦(1+3𝑦𝑦)

∙ 3𝑦𝑦+1(𝑦𝑦+2)(3𝑦𝑦−2)

Simplify .

= 𝒚𝒚−𝟒𝟒𝟐𝟐𝒚𝒚(𝟑𝟑𝒚𝒚−𝟐𝟐)

3. 𝟑𝟑𝒕𝒕+𝟐𝟐 − 𝟏𝟏

𝒕𝒕−𝟏𝟏

𝟒𝟒𝒕𝒕=

3(𝒕𝒕−𝟏𝟏)(𝑡𝑡+2)(𝒕𝒕−𝟏𝟏)− 1(𝒕𝒕+𝟐𝟐)

(𝑡𝑡−1)(𝒕𝒕+𝟐𝟐) 4𝑡𝑡1

The LCD = (t + 2)(t – 1) .

= 3(𝑡𝑡−1)−(𝑡𝑡+2)

(𝑡𝑡+2)(𝑡𝑡−1) 4𝑡𝑡1

= 3(𝑡𝑡−1)−(𝑡𝑡+2)(𝑡𝑡+2)(𝑡𝑡−1)

÷ 4𝑡𝑡 1

÷

= 3(𝑡𝑡−1)−(𝑡𝑡+2)(𝑡𝑡+2)(𝑡𝑡−1)

∙ 𝟏𝟏 𝟒𝟒𝒕𝒕

÷ × , 4𝑡𝑡 1

14𝑡𝑡

= 3𝑡𝑡−3−𝑡𝑡−2(𝑡𝑡+2)(𝑡𝑡−1)

∙ 1 4𝑡𝑡

Distribute

= 𝟐𝟐𝒕𝒕−𝟓𝟓𝟒𝟒𝒕𝒕(𝒕𝒕+𝟐𝟐)(𝒕𝒕−𝟏𝟏)


y 2

y -4

y 2

3y -2

÷

÷ × , 3𝑦𝑦2+4𝑦𝑦−4

3𝑦𝑦+1 3𝑦𝑦+1

3𝑦𝑦2+4𝑦𝑦−4

Rewrite to get a single rational expression in the den . & num .


Page 6-13

Simplify Complex Rational ExpressionsMethod II

Simplifying a complex rational expression — method II: multiply the reciprocal of the


𝑁𝑁1𝐷𝐷1

𝑁𝑁2𝐷𝐷2

= 𝑁𝑁1

𝐷𝐷1÷ 𝑵𝑵𝟐𝟐

𝑫𝑫𝟐𝟐= 𝑁𝑁1

𝐷𝐷1∙ 𝑫𝑫𝟐𝟐 𝑵𝑵𝟐𝟐

𝑁𝑁1𝐷𝐷1


are rational expressions . 34 32

= 34

÷ 32

= 34

∙ 23

= 𝟏𝟏𝟐𝟐


1. 𝟏𝟏𝒘𝒘 − 𝟑𝟑 𝟏𝟏𝒘𝒘 + 𝒘𝒘

= 1𝑤𝑤 − 31 1𝑤𝑤 + 𝑤𝑤1

= 1

𝑤𝑤 − 3𝒘𝒘𝒘𝒘 1𝑤𝑤 + 𝑤𝑤

𝟐𝟐𝒘𝒘

Multiply num . & den . by the LCD . LCD = w

= 1−3𝑤𝑤𝑤𝑤 1+ 𝑤𝑤2𝑤𝑤


÷ 1+𝑤𝑤2

𝑤𝑤 ÷


∙ 𝒘𝒘𝟏𝟏+𝒘𝒘𝟐𝟐

÷ × , 1+𝑤𝑤2 𝑤𝑤

𝑤𝑤1+𝑤𝑤2

= 𝟏𝟏−𝟑𝟑𝒘𝒘𝟏𝟏+𝒘𝒘𝟐𝟐

2. 𝒚𝒚𝟐𝟐−𝟐𝟐𝒚𝒚−𝟖𝟖𝟐𝟐𝒚𝒚+𝟔𝟔𝒚𝒚𝟐𝟐

𝟑𝟑𝒚𝒚𝟐𝟐+𝟒𝟒𝒚𝒚−𝟒𝟒𝟑𝟑𝒚𝒚+𝟏𝟏

= 𝑦𝑦2−2𝑦𝑦−82𝑦𝑦+6𝑦𝑦2

÷ 3𝑦𝑦2+4𝑦𝑦−43𝑦𝑦+1

= 𝑦𝑦2−2𝑦𝑦−8

2𝑦𝑦+6𝑦𝑦2 ∙ 𝟑𝟑𝒚𝒚+𝟏𝟏𝟑𝟑𝒚𝒚𝟐𝟐+𝟒𝟒𝒚𝒚−𝟒𝟒

Factor: 𝑦𝑦2 − 2𝑦𝑦 − 8 , 3𝑦𝑦2 + 4𝑦𝑦 − 4

= (𝑦𝑦+2)(𝑦𝑦−4)2𝑦𝑦(1+3𝑦𝑦)

∙ 3𝑦𝑦+1(𝑦𝑦+2)(3𝑦𝑦−2)

Simplify .

= 𝒚𝒚−𝟒𝟒𝟐𝟐𝒚𝒚(𝟑𝟑𝒚𝒚−𝟐𝟐)

3. 𝟑𝟑𝒕𝒕+𝟐𝟐 − 𝟏𝟏

𝒕𝒕−𝟏𝟏

𝟒𝟒𝒕𝒕=

3(𝒕𝒕−𝟏𝟏)(𝑡𝑡+2)(𝒕𝒕−𝟏𝟏)− 1(𝒕𝒕+𝟐𝟐)

(𝑡𝑡−1)(𝒕𝒕+𝟐𝟐) 4𝑡𝑡1

The LCD = (t + 2)(t – 1) .

= 3(𝑡𝑡−1)−(𝑡𝑡+2)

(𝑡𝑡+2)(𝑡𝑡−1) 4𝑡𝑡1

= 3(𝑡𝑡−1)−(𝑡𝑡+2)(𝑡𝑡+2)(𝑡𝑡−1)

÷ 4𝑡𝑡 1

÷

= 3(𝑡𝑡−1)−(𝑡𝑡+2)(𝑡𝑡+2)(𝑡𝑡−1)

∙ 𝟏𝟏 𝟒𝟒𝒕𝒕

÷ × , 4𝑡𝑡 1

14𝑡𝑡

= 3𝑡𝑡−3−𝑡𝑡−2(𝑡𝑡+2)(𝑡𝑡−1)

∙ 1 4𝑡𝑡

Distribute

= 𝟐𝟐𝒕𝒕−𝟓𝟓𝟒𝟒𝒕𝒕(𝒕𝒕+𝟐𝟐)(𝒕𝒕−𝟏𝟏)


y 2

y -4

y 2

3y -2

÷

÷ × , 3𝑦𝑦2+4𝑦𝑦−4

3𝑦𝑦+1 3𝑦𝑦+1

3𝑦𝑦2+4𝑦𝑦−4



Page 6-13





6-5 RATIONAL EQUATIONS

Rational Equations

• Review Example

Expression: a mathematical statement that contains numbers, 2ab2 + 3a

variables, and arithmetic operations (without an equal sign) .

Equation: a mathematical statement that contains two 3x2 + 4x = 2

expressions and separated by an equal sign .

Rational (fractional) expression: an expression that is a ratio 2𝑥𝑥−37𝑥𝑥+5

or quotient of two polynomials .

• Rational (fractional) equation: an equation that contains rational expressions .

Example

Rational Expression Rational Equation5𝑥𝑥2 + 3𝑥𝑥

4 − 𝑥𝑥3𝑥𝑥

+2

5𝑥𝑥= 7

𝑎𝑎2 + 𝑏𝑏2

2𝑎𝑎𝑏𝑏 − 𝑏𝑏−

3𝑎𝑎2

4𝑎𝑎2𝑎𝑎𝑏𝑏 + 3

4𝑎𝑎−

5𝑎𝑎 + 74

=3𝑎𝑎𝑏𝑏5𝑎𝑎

• Solving a rational equation

Steps Example: Solve 𝟏𝟏𝒙𝒙

+ 𝟑𝟑𝟐𝟐𝒙𝒙

= 𝟓𝟓 .

- Find the least common denominator (LCD) . LCD = 2x

- Multiply each term by the LCD . 1𝑥𝑥∙ 2𝑥𝑥 + 3

2𝑥𝑥∙ 2𝑥𝑥 = 5 ∙ 2𝑥𝑥

2 + 3 = 10x

- Solve the variable . 5 = 10x

x = 𝟏𝟏𝟐𝟐

?

- Check . 112

+ 32 ∙ 1 2

= 5 112

= 1÷ 12

= 1 ∙ 21

= 2

√2 + 3 = 5 Correct!

Note: Checking is necessary, not optional (check for a valid solution rather than errors) .

Page 6-14

6-5 RATIONAL EQUATIONS

Rational Equations

• Review Example

Expression: a mathematical statement that contains numbers, 2ab2 + 3a

variables, and arithmetic operations (without an equal sign) .

Equation: a mathematical statement that contains two 3x2 + 4x = 2

expressions and separated by an equal sign .

Rational (fractional) expression: an expression that is a ratio 2𝑥𝑥−37𝑥𝑥+5

or quotient of two polynomials .


Example

Rational Expression Rational Equation5𝑥𝑥2 + 3𝑥𝑥

4 − 𝑥𝑥3𝑥𝑥

+2

5𝑥𝑥= 7

𝑎𝑎2 + 𝑏𝑏2

2𝑎𝑎𝑏𝑏 − 𝑏𝑏−

3𝑎𝑎2

4𝑎𝑎2𝑎𝑎𝑏𝑏 + 3

4𝑎𝑎−

5𝑎𝑎 + 74

=3𝑎𝑎𝑏𝑏5𝑎𝑎


Steps Example: Solve 𝟏𝟏𝒙𝒙

+ 𝟑𝟑𝟐𝟐𝒙𝒙

= 𝟓𝟓 .

- Find the least common denominator (LCD) . LCD = 2x

- Multiply each term by the LCD . 1𝑥𝑥∙ 2𝑥𝑥 + 3

2𝑥𝑥∙ 2𝑥𝑥 = 5 ∙ 2𝑥𝑥

2 + 3 = 10x

- Solve the variable . 5 = 10x

x = 𝟏𝟏𝟐𝟐

?

- Check . 112

+ 32 ∙ 1 2

= 5 112

= 1÷ 12

= 1 ∙ 21

= 2

√2 + 3 = 5 Correct!

Note: Checking is necessary, not optional (check for a valid solution rather than errors) .

Page 6-14





Solving Rational Equations

Example: Solve the following .

1. 𝟑𝟑𝒂𝒂−𝟐𝟐

= 𝟓𝟓𝒂𝒂+𝟒𝟒

LCD = (a – 2)(a + 4)

3𝑎𝑎−2

(𝒂𝒂 − 𝟐𝟐)(𝒂𝒂 + 𝟒𝟒) = 5𝑎𝑎+4

(𝒂𝒂 − 𝟐𝟐)(𝒂𝒂 + 𝟒𝟒) Multiply each term by the LCD .

3(a + 4) = 5(a – 2) Distribute

3a + 12 = 5a – 10 Solve for a: add 10, subtract 3a.

22 = 2a , a = 11 Divide by 2 .

? ?

Check: 311−2

= 511+4

, 39

= 515

√13

= 13

Correct!

2. 𝟐𝟐−𝒕𝒕𝒕𝒕+𝟓𝟓

= 𝟐𝟐 + 𝟕𝟕𝒕𝒕+𝟓𝟓

LCD = t + 5

2−𝑡𝑡𝑡𝑡+5

∙ (𝒕𝒕 + 𝟓𝟓) = 2(𝒕𝒕 + 𝟓𝟓) + 7𝑡𝑡+5

∙ (𝒕𝒕 + 𝟓𝟓) Multiply each term by the LCD .

2 – t = 2t + 10 + 7 Solve for t : add t, subtract 17 .

-15 = 3t , t = -5 Divide by 3 .

?

Check: 2−(-5)−5 + 5 = 2 + 7

-5 + 5, 7

0= 2 + 7

0 No solution (undefined)

3. 𝟓𝟓𝒙𝒙+𝟐𝟐

+ 𝟐𝟐𝒙𝒙𝒙𝒙𝟐𝟐−𝟒𝟒

= 𝟑𝟑𝒙𝒙−𝟐𝟐

x2 – 4 = x2 – 22 = (x + 2)(x – 2)

LCD = (x + 2)(x -2) 5

𝑥𝑥+2∙ (𝑥𝑥 − 2)(𝑥𝑥 + 2) + 2𝑥𝑥

𝑥𝑥2−4(𝑥𝑥 − 2)(𝑥𝑥 + 2) = 3

𝑥𝑥−2(𝑥𝑥 − 2)(𝑥𝑥 + 2)

Multiply each term by the LCD .

5(x – 2) + 2x = 3(x + 2) Distribute

5x – 10 + 2x = 3x + 6 Solve for x: subtract 3x, add 10 .

4x = 16 , x = 4 Divide by 4 .

? ?

Check: 54+2

+ 2∙ 442−4

= 34−2

, 56

+ 812

= 32

? √

5∙26∙2

+ 812

= 1812

, 1812

= 1812

Correct!

1

3

1

3

Page 6-15

Solving Rational Equations

Example: Solve the following .

1. 𝟑𝟑𝒂𝒂−𝟐𝟐

= 𝟓𝟓𝒂𝒂+𝟒𝟒

LCD = (a – 2)(a + 4)

3𝑎𝑎−2

(𝒂𝒂 − 𝟐𝟐)(𝒂𝒂 + 𝟒𝟒) = 5𝑎𝑎+4

(𝒂𝒂 − 𝟐𝟐)(𝒂𝒂 + 𝟒𝟒) Multiply each term by the LCD .

3(a + 4) = 5(a – 2) Distribute

3a + 12 = 5a – 10 Solve for a: add 10, subtract 3a.

22 = 2a , a = 11 Divide by 2 .

? ?

Check: 311−2

= 511+4

, 39

= 515

√13

= 13

Correct!

2. 𝟐𝟐−𝒕𝒕𝒕𝒕+𝟓𝟓

= 𝟐𝟐 + 𝟕𝟕𝒕𝒕+𝟓𝟓

LCD = t + 5

2−𝑡𝑡𝑡𝑡+5

∙ (𝒕𝒕 + 𝟓𝟓) = 2(𝒕𝒕 + 𝟓𝟓) + 7𝑡𝑡+5

∙ (𝒕𝒕 + 𝟓𝟓) Multiply each term by the LCD .

2 – t = 2t + 10 + 7 Solve for t : add t, subtract 17 .

-15 = 3t , t = -5 Divide by 3 .

?

Check: 2−(-5)−5 + 5 = 2 + 7

-5 + 5, 7

0= 2 + 7

0 No solution (undefined)

3. 𝟓𝟓𝒙𝒙+𝟐𝟐

+ 𝟐𝟐𝒙𝒙𝒙𝒙𝟐𝟐−𝟒𝟒

= 𝟑𝟑𝒙𝒙−𝟐𝟐

x2 – 4 = x2 – 22 = (x + 2)(x – 2)

LCD = (x + 2)(x -2) 5

𝑥𝑥+2∙ (𝑥𝑥 − 2)(𝑥𝑥 + 2) + 2𝑥𝑥

𝑥𝑥2−4(𝑥𝑥 − 2)(𝑥𝑥 + 2) = 3

𝑥𝑥−2(𝑥𝑥 − 2)(𝑥𝑥 + 2)

Multiply each term by the LCD .

5(x – 2) + 2x = 3(x + 2) Distribute

5x – 10 + 2x = 3x + 6 Solve for x: subtract 3x, add 10 .

4x = 16 , x = 4 Divide by 4 .

? ?

Check: 54+2

+ 2∙ 442−4

= 34−2

, 56

+ 812

= 32

? √

5∙26∙2

+ 812

= 1812

, 1812

= 1812

Correct!

1

3

1

3

Page 6-15





6-6 APPLICATIONS OF RATIONAL EQUATIONS

Applications

• Mathematical model: uses mathematical language to describe the behavior of a real-lifephenomenon .

• World-problem solving strategy reviewProcedure for Solving Word Problems

- Organize the facts given from the problem .- Identify and label the unknown quantity (let x = unknown) .- Draw a diagram if it will make the problem clearer .- Convert the wording into a mathematical equation .- Solve the equation and find the solution(s) .- Check and state the answer.

Example: Tom plans to plant a flower garden in his backyard . If the size of the garden is as

indicated in the following figure, what is the total area of the garden?

- Diagram . 2m A1 A2

4m

- Organize the facts .

rectangle width (w) = 2m , length (l) = 4m triangle base (b) = 2m , height (h) = 1mtotal area A = A1 + 2A2

Recall: Area of a rectangle A = wl , area of a triangle A = 12𝑏𝑏𝑏

- Equation: A = wl + 2 �𝟏𝟏𝟐𝟐𝒃𝒃𝒃𝒃�

- Solution: A = (2)(4) + 2 ∙ 12

(2)(1) w = 2m , l = 4m , b = 2m , h = 1m

A = 10 m2

?

- Check . 10 = (2)(4) + 2 ∙ 12

(2)(1)√

10 = 8 + 2 Correct!

- Answer: The total area of the garden is 10 m2 .

1m

1 rectangle 2 triangles

1m

2m

Page 6-16

6-6 APPLICATIONS OF RATIONAL EQUATIONS

Applications

• Mathematical model: uses mathematical language to describe the behavior of a real-lifephenomenon .

• World-problem solving strategy reviewProcedure for Solving Word Problems

- Organize the facts given from the problem .- Identify and label the unknown quantity (let x = unknown) .- Draw a diagram if it will make the problem clearer .- Convert the wording into a mathematical equation .- Solve the equation and find the solution(s) .- Check and state the answer.

Example: Tom plans to plant a flower garden in his backyard . If the size of the garden is as

indicated in the following figure, what is the total area of the garden?

- Diagram . 2m A1 A2

4m

- Organize the facts .

rectangle width (w) = 2m , length (l) = 4m triangle base (b) = 2m , height (h) = 1mtotal area A = A1 + 2A2

Recall: Area of a rectangle A = wl , area of a triangle A = 12𝑏𝑏𝑏

- Equation: A = wl + 2 �𝟏𝟏𝟐𝟐𝒃𝒃𝒃𝒃�

- Solution: A = (2)(4) + 2 ∙ 12

(2)(1) w = 2m , l = 4m , b = 2m , h = 1m

A = 10 m2

?

- Check . 10 = (2)(4) + 2 ∙ 12

(2)(1)√

10 = 8 + 2 Correct!

- Answer: The total area of the garden is 10 m2 .

1m

1 rectangle 2 triangles

1m

2m

Page 6-16





Page 6-17

Number Problems

Example: Three divided by one more than a number is equal to the quotient of two and the

same number less than 3. What is the number?

- Facts: 3𝑥𝑥𝑥𝑥+1

23−𝑥𝑥𝑥𝑥

Let x = number

- Equation: 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙+𝟏𝟏𝟏𝟏

= 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑−𝒙𝒙𝒙𝒙

LCD = (𝑥𝑥𝑥𝑥 + 1)(3− 𝑥𝑥𝑥𝑥)

- Solve for x. 3𝑥𝑥𝑥𝑥+1

(𝑥𝑥𝑥𝑥 + 1)(3 − 𝑥𝑥𝑥𝑥) = 23−𝑥𝑥𝑥𝑥

(𝑥𝑥𝑥𝑥 + 1)(3 − 𝑥𝑥𝑥𝑥) × LCD

3(3 − x) = 2 (x + 1) Solve for x.

9 – 3x = 2x + 2 , x = 𝟕𝟕𝟕𝟕𝟓𝟓𝟓𝟓

Example: The quotient of 5 and the product of a number and 4 is equal to the quotient of 7 and

5 times that number less than 2. What is the number?

- Facts: 54𝑥𝑥𝑥𝑥

72−5𝑥𝑥𝑥𝑥

Let x = number

- Equation: 𝟓𝟓𝟓𝟓𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙

= 𝟕𝟕𝟕𝟕𝟐𝟐𝟐𝟐−𝟓𝟓𝟓𝟓𝒙𝒙𝒙𝒙

LCD = 4𝑥𝑥𝑥𝑥(2− 5𝑥𝑥𝑥𝑥)

- Solve for x. 54𝑥𝑥𝑥𝑥∙ 4𝑥𝑥𝑥𝑥(2 − 5𝑥𝑥𝑥𝑥) = 7

2−5𝑥𝑥𝑥𝑥∙ 4𝑥𝑥𝑥𝑥(2 − 5𝑥𝑥𝑥𝑥) × LCD

5(2 − 5x) = 28x 10 −25x = 28x , 𝒙𝒙𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

𝟓𝟓𝟓𝟓𝟑𝟑𝟑𝟑

Example: The difference between the reciprocals of two consecutive positive odd integers is

four over thirty. What are the two integers? English Phrase Algebraic Expression Example

two consecutive integers x , x + 1 If x = 1, x + 1 = 2 two consecutive odd integers x , x + 2 If x = 1, x + 2 = 3 two consecutive even integers x , x + 2

or 2x , 2x + 2 If x = 2, x + 2 = 4 If x = 2, 2x = 4, 2x + 2 = 6

- Equation: 𝟏𝟏𝟏𝟏𝒙𝒙𝒙𝒙− 𝟏𝟏𝟏𝟏

𝒙𝒙𝒙𝒙+𝟐𝟐𝟐𝟐= 𝟒𝟒𝟒𝟒

𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏 Let x = 1st even integer , x+2 = 2nd even integer

- Solve for x. 1𝑥𝑥𝑥𝑥∙ 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2) − 1

𝑥𝑥𝑥𝑥+2∙ 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2) = 4

30∙ 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2) LCD = 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2)

30(x + 2) − 30x = 4x (x + 2)

30x + 60 – 30x = 4x2 + 8x

4x2 + 8x − 60 = 0 , 4(x2 + 2x – 15) = 0 Factor. 4 4

(x2 + 2x – 15) = 04 Divide by 4.

(x – 3)(x + 5) = 0 Factor. x – 3 = 0 or x + 5 = 0 Zero product property. So x = 3 or x = -5 Ignore the negative answer. The two integers are x = 3 and x + 2 = 5

Page 6-17

Number Problems

Example: Three divided by one more than a number is equal to the quotient of two and the

same number less than 3. What is the number?

- Facts: 3𝑥𝑥𝑥𝑥+1

23−𝑥𝑥𝑥𝑥

Let x = number

- Equation: 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙+𝟏𝟏𝟏𝟏

= 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑−𝒙𝒙𝒙𝒙

LCD = (𝑥𝑥𝑥𝑥 + 1)(3− 𝑥𝑥𝑥𝑥)

- Solve for x. 3𝑥𝑥𝑥𝑥+1

(𝑥𝑥𝑥𝑥 + 1)(3 − 𝑥𝑥𝑥𝑥) = 23−𝑥𝑥𝑥𝑥

(𝑥𝑥𝑥𝑥 + 1)(3 − 𝑥𝑥𝑥𝑥) × LCD

3(3 − x) = 2 (x + 1) Solve for x.

9 – 3x = 2x + 2 , x = 𝟕𝟕𝟕𝟕𝟓𝟓𝟓𝟓

Example: The quotient of 5 and the product of a number and 4 is equal to the quotient of 7 and

5 times that number less than 2. What is the number?

- Facts: 54𝑥𝑥𝑥𝑥

72−5𝑥𝑥𝑥𝑥

Let x = number

- Equation: 𝟓𝟓𝟓𝟓𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙

= 𝟕𝟕𝟕𝟕𝟐𝟐𝟐𝟐−𝟓𝟓𝟓𝟓𝒙𝒙𝒙𝒙

LCD = 4𝑥𝑥𝑥𝑥(2− 5𝑥𝑥𝑥𝑥)

- Solve for x. 54𝑥𝑥𝑥𝑥∙ 4𝑥𝑥𝑥𝑥(2 − 5𝑥𝑥𝑥𝑥) = 7

2−5𝑥𝑥𝑥𝑥∙ 4𝑥𝑥𝑥𝑥(2 − 5𝑥𝑥𝑥𝑥) × LCD

5(2 − 5x) = 28x 10 −25x = 28x , 𝒙𝒙𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

𝟓𝟓𝟓𝟓𝟑𝟑𝟑𝟑

Example: The difference between the reciprocals of two consecutive positive odd integers is

four over thirty. What are the two integers? English Phrase Algebraic Expression Example

two consecutive integers x , x + 1 If x = 1, x + 1 = 2 two consecutive odd integers x , x + 2 If x = 1, x + 2 = 3 two consecutive even integers x , x + 2

or 2x , 2x + 2 If x = 2, x + 2 = 4 If x = 2, 2x = 4, 2x + 2 = 6

- Equation: 𝟏𝟏𝟏𝟏𝒙𝒙𝒙𝒙− 𝟏𝟏𝟏𝟏

𝒙𝒙𝒙𝒙+𝟐𝟐𝟐𝟐= 𝟒𝟒𝟒𝟒

𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏 Let x = 1st even integer , x+2 = 2nd even integer

- Solve for x. 1𝑥𝑥𝑥𝑥∙ 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2) − 1

𝑥𝑥𝑥𝑥+2∙ 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2) = 4

30∙ 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2) LCD = 30𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 2)

30(x + 2) − 30x = 4x (x + 2)

30x + 60 – 30x = 4x2 + 8x

4x2 + 8x − 60 = 0 , 4(x2 + 2x – 15) = 0 Factor. 4 4

(x2 + 2x – 15) = 04 Divide by 4.

(x – 3)(x + 5) = 0 Factor. x – 3 = 0 or x + 5 = 0 Zero product property. So x = 3 or x = -5 Ignore the negative answer. The two integers are x = 3 and x + 2 = 5





Page 6-18

Work Problems

The formula for “work” problems that involve two people is:

1𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴

+ 1𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵

= 1 𝑡𝑡𝑡𝑡

Time � 𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 − time required when person 𝐴𝐴𝐴𝐴 works alone 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 − time required when person 𝐵𝐵𝐵𝐵 works alone 𝑡𝑡𝑡𝑡 − time required when two people work together

Rate � 1/ 𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 − person 𝐴𝐴𝐴𝐴 can finish 1 job every 𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 hours 1/ 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 − person 𝐵𝐵𝐵𝐵 can finish 1 job every 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 hours

Example: Tom can paint a room in 2 hours. Susan can paint a room in 3 hours. How long will it take both of them to paint a room together?

- Organize the facts Time to Paint a Room Part of Job Finished in 1 hour Comments Tom: 2 hours 1

2 job If Tom can finish the job in 2 hours, he can finish ½

of the job in 1 hour.

Susan: 3 hours 13 job If Susan can finish the job in 3 hours, she can finish

1/3 of the job in 1 hour.

Together: t hours 12

+ 13 = 1

𝑡𝑡𝑡𝑡

If Tom and Susan work together, they can finish the job in t hours, and they can finish 1/ t of the job in 1 hour.

Let t = time needed to paint a room together.

- Equation: 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

+ 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑

= 𝟏𝟏𝟏𝟏𝒕𝒕𝒕𝒕 1

𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴+ 1

𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵= 1

𝑡𝑡𝑡𝑡 , tA = 2 , tB = 3

- Solve for t. 12∙ 𝟔𝟔𝟔𝟔𝒕𝒕𝒕𝒕 + 1

3∙ 𝟔𝟔𝟔𝟔𝒕𝒕𝒕𝒕 = 1

𝑡𝑡𝑡𝑡∙ 𝟔𝟔𝟔𝟔𝒕𝒕𝒕𝒕 Multiply each term by the LCD. (6t)

3t + 2t = 6 , t = 1.2 hours - Answer: It will take Tom and Susan 1.2 hours to paint a room together.

Example: Jason can plow the snow from the school’s parking lot 3 fewer hours than Shawn. If they work together, they can finish the job in 2 hours. How long will it take each of them to finish the job alone?

- Organize the facts Time to Finish the Job Part of Job Finished in 1 hour Comments Shawn: t hours 1

𝑡𝑡𝑡𝑡 job If Shawn can finish the job in t hours, he can finish 1/t of the job

in 1 hour.

Jason: t – 3 hours 1𝑡𝑡𝑡𝑡−3

job If Jason can finish the job in (t – 3) hours, he can finish 1/(t – 3) of the job in 1 hour.

Together: 2 hours 1𝑡𝑡𝑡𝑡

+ 1𝑡𝑡𝑡𝑡−3

= 12 If Shawn and Jason work together, they can finish the job in 2

hours, and they can finish 1/2 of the job in 1 hour. Let t = time required for Shawn to finish the job alone.

- Equation: 𝟏𝟏𝟏𝟏𝒕𝒕𝒕𝒕

+ 𝟏𝟏𝟏𝟏𝒕𝒕𝒕𝒕−𝟑𝟑𝟑𝟑

= 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 1



𝑡𝑡𝑡𝑡 , tA = t , tB = t – 3. t = 2

- Solve for t. 1𝑡𝑡𝑡𝑡∙ 𝟐𝟐𝟐𝟐𝒕𝒕𝒕𝒕(𝒕𝒕𝒕𝒕 − 𝟑𝟑𝟑𝟑) + 1

𝑡𝑡𝑡𝑡−3∙ 𝟐𝟐𝟐𝟐𝒕𝒕𝒕𝒕(𝒕𝒕𝒕𝒕 − 𝟑𝟑𝟑𝟑) = 1

2∙ 𝟐𝟐𝟐𝟐𝒕𝒕𝒕𝒕(𝒕𝒕𝒕𝒕 − 𝟑𝟑𝟑𝟑) LCD = 2t (t – 3)

2(t − 3) + 2t = t (t – 3) Distribute.

2t – 6 + 2t = t2 – 3t Subtract 4t, add 6. t2 – 7t + 6 = 0 Factor.

(t − 1)(t – 6) = 0 , t – 1 = 0 or t – 6 = 0 Zero product property - Shawn: t = 1 , t = 6 t = 1 hour is not possible

- Jason: t – 3 = 6 – 3 = 3 (2 people can finish the job in 2 hours.)

- Answer: It will take Shawn 6 hours and Jason 3 hours to clean the school’s parking lot.

Page 6-18

Work Problems

The formula for “work” problems that involve two people is:

1𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴

+ 1𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵

= 1 𝑡𝑡𝑡𝑡

Time � 𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 − time required when person 𝐴𝐴𝐴𝐴 works alone 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 − time required when person 𝐵𝐵𝐵𝐵 works alone 𝑡𝑡𝑡𝑡 − time required when two people work together

Rate � 1/ 𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 − person 𝐴𝐴𝐴𝐴 can finish 1 job every 𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 hours 1/ 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 − person 𝐵𝐵𝐵𝐵 can finish 1 job every 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 hours

Example: Tom can paint a room in 2 hours. Susan can paint a room in 3 hours. How long will it take both of them to paint a room together?

- Organize the facts Time to Paint a Room Part of Job Finished in 1 hour Comments Tom: 2 hours 1

2 job If Tom can finish the job in 2 hours, he can finish ½

of the job in 1 hour.

Susan: 3 hours 13 job If Susan can finish the job in 3 hours, she can finish

1/3 of the job in 1 hour.

Together: t hours 12

+ 13 = 1

𝑡𝑡𝑡𝑡

If Tom and Susan work together, they can finish the job in t hours, and they can finish 1/ t of the job in 1 hour.

Let t = time needed to paint a room together.

- Equation: 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

+ 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑

= 𝟏𝟏𝟏𝟏𝒕𝒕𝒕𝒕 1



𝑡𝑡𝑡𝑡 , tA = 2 , tB = 3

- Solve for t. 12∙ 𝟔𝟔𝟔𝟔𝒕𝒕𝒕𝒕 + 1

3∙ 𝟔𝟔𝟔𝟔𝒕𝒕𝒕𝒕 = 1

𝑡𝑡𝑡𝑡∙ 𝟔𝟔𝟔𝟔𝒕𝒕𝒕𝒕 Multiply each term by the LCD. (6t)

3t + 2t = 6 , t = 1.2 hours - Answer: It will take Tom and Susan 1.2 hours to paint a room together.

Example: Jason can plow the snow from the school’s parking lot 3 fewer hours than Shawn. If they work together, they can finish the job in 2 hours. How long will it take each of them to finish the job alone?

- Organize the facts Time to Finish the Job Part of Job Finished in 1 hour Comments Shawn: t hours 1

𝑡𝑡𝑡𝑡 job If Shawn can finish the job in t hours, he can finish 1/t of the job

in 1 hour.

Jason: t – 3 hours 1𝑡𝑡𝑡𝑡−3

job If Jason can finish the job in (t – 3) hours, he can finish 1/(t – 3) of the job in 1 hour.

Together: 2 hours 1𝑡𝑡𝑡𝑡

+ 1𝑡𝑡𝑡𝑡−3

= 12 If Shawn and Jason work together, they can finish the job in 2

hours, and they can finish 1/2 of the job in 1 hour. Let t = time required for Shawn to finish the job alone.

- Equation: 𝟏𝟏𝟏𝟏𝒕𝒕𝒕𝒕

+ 𝟏𝟏𝟏𝟏𝒕𝒕𝒕𝒕−𝟑𝟑𝟑𝟑

= 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 1



𝑡𝑡𝑡𝑡 , tA = t , tB = t – 3. t = 2

- Solve for t. 1𝑡𝑡𝑡𝑡∙ 𝟐𝟐𝟐𝟐𝒕𝒕𝒕𝒕(𝒕𝒕𝒕𝒕 − 𝟑𝟑𝟑𝟑) + 1

𝑡𝑡𝑡𝑡−3∙ 𝟐𝟐𝟐𝟐𝒕𝒕𝒕𝒕(𝒕𝒕𝒕𝒕 − 𝟑𝟑𝟑𝟑) = 1

2∙ 𝟐𝟐𝟐𝟐𝒕𝒕𝒕𝒕(𝒕𝒕𝒕𝒕 − 𝟑𝟑𝟑𝟑) LCD = 2t (t – 3)

2(t − 3) + 2t = t (t – 3) Distribute.

2t – 6 + 2t = t2 – 3t Subtract 4t, add 6. t2 – 7t + 6 = 0 Factor.

(t − 1)(t – 6) = 0 , t – 1 = 0 or t – 6 = 0 Zero product property - Shawn: t = 1 , t = 6 t = 1 hour is not possible

- Jason: t – 3 = 6 – 3 = 3 (2 people can finish the job in 2 hours.)

- Answer: It will take Shawn 6 hours and Jason 3 hours to clean the school’s parking lot.





Proportions

• Ratio, rate, and proportion

Representation Example

Ratio a to b or a:b or 𝑎𝑎𝑏𝑏 with the same unit . 5 to 9 or 5:9 or

m9m5

Rate a to b or a:b or 𝑎𝑎𝑏𝑏 with different units . 3 to 7 or 3:7 or

m7cm3

Proportion𝑎𝑎𝑏𝑏 =

𝑐𝑐𝑑𝑑

an equation with a ratio on each side . m5

cm17cm3

=m

Note: The units for both numerators must match and the units for both denominators must match .

Example: inft

= inft

, minuteshours

= minuteshours

• Solving a proportion: Example

- Cross multiply: multiply along two diagonals .dc

ba=

62

9=

x

- Solve for the unknown . 6 · x = 2 · 9

36

186

92==

⋅= x

• Application

Example: To determine the number of moose in an area, a conservationist catches 50

moose, tags them, and lets them loose . Later, 20 moose are caught; 5 of them

are tagged. How many moose are in the area?

- Equation: 𝑥𝑥 moose50 tagged moose

= 20 moose5 tagged moose

Let x = the numbers of moose

- Cross multiply . 𝑥𝑥 50

= 20 5

- Solve for x . 5·x = 20·50

𝒙𝒙 = 20 ∙ 50 5

= 200

- Answer: There are 200 moose in the area .

Page 6-19

Proportions

• Ratio, rate, and proportion

Representation Example

Ratio a to b or a:b or 𝑎𝑎𝑏𝑏 with the same unit . 5 to 9 or 5:9 or

m9m5


m7cm3


𝑐𝑐𝑑𝑑


cm17cm3

=m

Note: The units for both numerators must match and the units for both denominators must match .

Example: inft

= inft

, minuteshours

= minuteshours

• Solving a proportion: Example

- Cross multiply: multiply along two diagonals .dc

ba=

62

9=

x

- Solve for the unknown . 6 · x = 2 · 9

36

186

92==

⋅= x

• Application

Example: To determine the number of moose in an area, a conservationist catches 50

moose, tags them, and lets them loose . Later, 20 moose are caught; 5 of them

are tagged. How many moose are in the area?

- Equation: 𝑥𝑥 moose50 tagged moose

= 20 moose5 tagged moose

Let x = the numbers of moose

- Cross multiply . 𝑥𝑥 50

= 20 5

- Solve for x . 5·x = 20·50

𝒙𝒙 = 20 ∙ 50 5

= 200

- Answer: There are 200 moose in the area .

Page 6-19





Motion Problems

• Motion formulas

distance = speed ∙ time d = r t t = 𝑑𝑑𝑟𝑟

r = 𝑑𝑑𝑡𝑡

• Table for motion problem

Condition Distance (d) Speed or Rate (r) Time (t) t = 𝑑𝑑𝑟𝑟

A d r dr

B d r 𝑑𝑑𝑟𝑟

Total

Example: Bob biked twice as fast the 150 km to town B than he did the 60 km to town A .

If the total trip took 4.5 hours, then how fast was he biking to town A?

- Table:Condition Distance (d) km Speed (r) km/h Time (t) h t = 𝑑𝑑

𝑟𝑟

To town A 60 km r 60 km𝑟𝑟

To town B 150 km 2r 150 km2𝑟𝑟

Total 4 .5 h

- Equation: 𝟔𝟔𝟔𝟔𝒓𝒓

+ 𝟏𝟏𝟏𝟏𝟔𝟔𝟐𝟐𝒓𝒓

= 4.5 time to A + time to B = 4 .5 h

- Solve for r: 60𝑟𝑟∙ 2𝑟𝑟 + 150

2𝑟𝑟 ∙ 2𝑟𝑟 = (4 .5)(2r) Multiply by the LCD . (2r)

120 + 150 = 9 r Combine like terms .

270 = 9 r r = 30 km/h Divide by 9 .

- Answer: To town A: r = 30 km/h To town B: 2r = (2)(30) = 60 km/h

Example: John boats at a speed of 30 km per hour in still water . The river flows at a

speed of 10 km per hour . How long will it take John to boat 2 km

downstream? 2 km upstream?

Condition Speed (r) Distance (d) Time �𝒕𝒕 = 𝒅𝒅𝒓𝒓�

Downstream r = 30 + 10 = 40 km/h d = 2 km 2 km

40 km/h

Upstream r = 30 − 10 = 20 km /h d = 2 km 2 km

20 km/h

Downstream (fast): speed of boat + speed of riverUpstream (slower): speed of boat – speed of river

Downstream: t = 𝑑𝑑𝑟𝑟

= 2 km40 km/h

= 0.05 h

Upstream: t = 𝑑𝑑𝑟𝑟

= 2 km20 km/h

= 0.1 h

Page 6-20

Motion Problems

• Motion formulas

distance = speed ∙ time d = r t t = 𝑑𝑑𝑟𝑟

r = 𝑑𝑑𝑡𝑡

• Table for motion problem

Condition Distance (d) Speed or Rate (r) Time (t) t = 𝑑𝑑𝑟𝑟

A d r dr

B d r 𝑑𝑑𝑟𝑟

Total

Example: Bob biked twice as fast the 150 km to town B than he did the 60 km to town A .

If the total trip took 4.5 hours, then how fast was he biking to town A?

- Table:Condition Distance (d) km Speed (r) km/h Time (t) h t = 𝑑𝑑

𝑟𝑟

To town A 60 km r 60 km𝑟𝑟

To town B 150 km 2r 150 km2𝑟𝑟

Total 4 .5 h

- Equation: 𝟔𝟔𝟔𝟔𝒓𝒓

+ 𝟏𝟏𝟏𝟏𝟔𝟔𝟐𝟐𝒓𝒓

= 4.5 time to A + time to B = 4 .5 h

- Solve for r: 60𝑟𝑟∙ 2𝑟𝑟 + 150

2𝑟𝑟 ∙ 2𝑟𝑟 = (4 .5)(2r) Multiply by the LCD . (2r)

120 + 150 = 9 r Combine like terms .

270 = 9 r r = 30 km/h Divide by 9 .

- Answer: To town A: r = 30 km/h To town B: 2r = (2)(30) = 60 km/h

Example: John boats at a speed of 30 km per hour in still water . The river flows at a

speed of 10 km per hour . How long will it take John to boat 2 km

downstream? 2 km upstream?

Condition Speed (r) Distance (d) Time �𝒕𝒕 = 𝒅𝒅𝒓𝒓�

Downstream r = 30 + 10 = 40 km/h d = 2 km 2 km

40 km/h

Upstream r = 30 − 10 = 20 km /h d = 2 km 2 km

20 km/h

Downstream (fast): speed of boat + speed of riverUpstream (slower): speed of boat – speed of river

Downstream: t = 𝑑𝑑𝑟𝑟

= 2 km40 km/h

= 0.05 h


= 2 km20 km/h

= 0.1 h

Page 6-20





Unit 6 Summary

• Rational expression: an expression that is a ratio or quotient of two polynomials .




b(x) ≠ 0 𝑓𝑓(𝑥𝑥) = 4𝑥𝑥+35𝑥𝑥−7

, 𝑓𝑓(𝑥𝑥) = 3𝑦𝑦2−2𝑦𝑦+45𝑦𝑦−6




𝑁𝑁1𝐷𝐷1






• Like rational expressions: rational expressions that have the same denominator .

• Unlike rational expressions: rational expressions that have different denominators .

• Adding or subtracting like rational expressions

𝑁𝑁1𝐷𝐷

+ 𝑁𝑁2𝐷𝐷




𝐷𝐷 𝑁𝑁1


𝑁𝑁2

𝐷𝐷are rational expressions . (D ≠ 0)

• Find the LCD Example: Find LCD for ,422and

165

84

2 8 16 42 2 4 8 21 8 ÷ 2 = 4, 16 ÷ 2 = 8, 42 ÷ 2 = 21

2 2 4 21 4 ÷ 2 = 2, 8 ÷ 2 = 4, move down 21

1 2 21 2 ÷ 2 = 1, 4 ÷ 2 = 2, move down 21

LCD = 24 × 21 = 336


- Determine the LCD .

- Rewrite expressions with the LCD .

- Combine the numerators .

- Simplify if possible .

• Complex rational expression: a rational expression whose numerator or denominator

(or both) contains rational expressions .

𝑁𝑁1𝐷𝐷1

𝑁𝑁2𝐷𝐷2

𝑁𝑁1𝐷𝐷1


are rational expressions, D1, D2 & N2 ≠ 0

𝑁𝑁1𝐷𝐷1


are rational expressions . (D1, D2 , N2 ≠ 0)

polynomials

Page 6-21





• Simplifying a complex rational expression — method I: multiply by the LCD .

• Simplifying a complex rational expression — method II: multiply by the reciprocal of the


𝑁𝑁1𝐷𝐷1 𝑁𝑁2𝐷𝐷2

= 𝑁𝑁1𝐷𝐷1

÷ 𝑵𝑵𝟐𝟐𝑫𝑫𝟐𝟐

= 𝑁𝑁1𝐷𝐷1∙ 𝑫𝑫𝟐𝟐 𝑵𝑵𝟐𝟐

𝑁𝑁1𝐷𝐷1


are rational expressions .


• Polynomial long division Example: 6𝑥𝑥2+9𝑥𝑥+2

3𝑥𝑥

Steps Solution Dividing whole numbers



− 6x2 (3x)(2x) = 6x

2 − 6 2∙3 = 6

2x + 3 230

- Divide the second term . 3x 6𝑥𝑥2 + 9𝑥𝑥 + 2 3 692 6x2 6

Bring 9x down 9x 9 Bring 9 down

(3x)(3) = 9x − 9x − 9 3∙3 = 9

2 2Remainder

quotient remainder



3𝑥𝑥= (2x + 3) + 𝟐𝟐

𝟑𝟑𝟑𝟑 692 ÷ 3 = 230 + 𝟐𝟐

𝟑𝟑

divisor

- Check: Dividend = Quotient ∙ Divisor + Remainder Quotient Divisor Dividend

− Remainder

• Missing terms in long division: If there is a missing consecutive power term in a

polynomial, insert the missing power term with a coefficient of 0 .


- Find the least common denominator (LCD) .

- Multiply each term by the LCD .

- Solve the variable .

- Check .

Page 6-22

3 692

2 3 692 – 6

2303 692 6 9 – 9 2





• Synthetic division: a shortcut method of dividing a polynomial by a binomial of the form



(𝟑𝟑𝒙𝒙𝟒𝟒 − 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝒙𝒙 − 𝟐𝟐𝒙𝒙𝟐𝟐) ÷ (𝒙𝒙 − 𝟐𝟐)





- Add column 2 . 2 3 0 -2 4 -40 6

3 6 (0 + 6)




Answer: (𝟑𝟑𝒙𝒙𝟑𝟑 + 𝟔𝟔𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 + 𝟐𝟐𝟒𝟒) + 𝟖𝟖𝒙𝒙−𝟐𝟐

One less than the degree of the dividend .

• The formula for “work” problems that involve two people

1𝑡𝑡𝐴𝐴

+ 1𝑡𝑡𝐵𝐵

= 1 𝑡𝑡

Time � 𝑡𝑡𝐴𝐴 − time required when person 𝐴𝐴 works alone 𝑡𝑡𝐵𝐵 − time required when person 𝐵𝐵 works alone 𝑡𝑡 − time required when two people work together

Rate � 1/ 𝑡𝑡𝐴𝐴 − person 𝐴𝐴 can finish 1 job every 𝑡𝑡𝐴𝐴 hours 1/ 𝑡𝑡𝐵𝐵 − person 𝐵𝐵 can finish 1 job every 𝑡𝑡𝐵𝐵 hours

• Ratio, rate and proportion

Representation ExampleRatio a to b or a:b or 𝑎𝑎

𝑏𝑏 with the same unit . 5 to 9 or 5:9 or



𝑐𝑐𝑑𝑑


cm17cm3

=m

• Motion formulasdistance = speed ∙ time d = r t t = 𝑑𝑑


𝑡𝑡

+

+

(a = 2)

+ +

m9m5

m7cm3

Page 6-23





PRACTICE QUIZ

Unit 6 Rational Expressions

1. Reduce to lowest terms .

a . 2𝑥𝑥2−8𝑥𝑥24𝑥𝑥

b . 𝑏𝑏(𝑏𝑏+1)(3𝑏𝑏−7)3𝑏𝑏2−4𝑏𝑏−7

2 . Perform the indicated operations and simplify .

a . 𝑥𝑥2−4𝑥𝑥−5𝑥𝑥−3

∙ 𝑥𝑥2−9𝑥𝑥−5

b . 𝑦𝑦2−12

÷ 𝑦𝑦+13

c . 5𝑥𝑥− 3

- 𝑥𝑥

3. Find the LCD for the following .

3𝑥𝑥4𝑥𝑥2

, 712𝑥𝑥2𝑦𝑦

and 3𝑥𝑥𝑦𝑦2𝑥𝑥4𝑦𝑦2

4. Perform the indicated operations and simplify .

2𝑏𝑏𝑏𝑏−2

− 1𝑏𝑏+2

+ 3𝑏𝑏2−4

5. Simplify:27𝑦𝑦2−9𝑦𝑦−36

3𝑦𝑦

6. Divide using long division .

a . 6y2 – 3y + 4 by 3yb . x3 + 2x2 + 3 by x – 1

7. Use synthetic division to divide .

2x3 – 3x2 + 5x – 7 by x – 2

Page 9

PRACTICE QUIZ

Unit 6 Rational Expressions

1. Reduce to lowest terms .

a . 2𝑥𝑥2−8𝑥𝑥24𝑥𝑥

b . 𝑏𝑏(𝑏𝑏+1)(3𝑏𝑏−7)3𝑏𝑏2−4𝑏𝑏−7

2 . Perform the indicated operations and simplify .

a . 𝑥𝑥2−4𝑥𝑥−5𝑥𝑥−3

∙ 𝑥𝑥2−9𝑥𝑥−5

b . 𝑦𝑦2−12

÷ 𝑦𝑦+13

c . 5𝑥𝑥− 3

- 𝑥𝑥

3. Find the LCD for the following .

3𝑥𝑥4𝑥𝑥2

, 712𝑥𝑥2𝑦𝑦

and 3𝑥𝑥𝑦𝑦2𝑥𝑥4𝑦𝑦2


2𝑏𝑏𝑏𝑏−2

− 1𝑏𝑏+2

+ 3𝑏𝑏2−4

5. Simplify:27𝑦𝑦2−9𝑦𝑦−36

3𝑦𝑦

6. Divide using long division .

a . 6y2 – 3y + 4 by 3yb . x3 + 2x2 + 3 by x – 1

7. Use synthetic division to divide .

2x3 – 3x2 + 5x – 7 by x – 2

Page 9





8. Simplify the following .

a . 1𝑥𝑥 + 5 1𝑥𝑥 + 𝑥𝑥

b . 2𝑦𝑦+3 − 1

𝑦𝑦−2

3𝑦𝑦

9. Solve the following .

a . 2𝑦𝑦− 3

4𝑦𝑦 = 7

b . 3−𝑥𝑥𝑥𝑥+4

= 3 − 1𝑥𝑥+4

10. It takes Tim 4 hours to clean a house and it takes 3 hours for Amanda to do the same job . How long will it take for both of them to clean the house together?

11. Watermelon is on sale at 2 for $7; how much will it cost for 12 watermelons?

12. Evan walked the 20km to location B twice as fast as he did the 15 km to

location A . If the total trip took 5 hours, then how fast was he walking to

location A?

Page 10




Algebra I & II Key Concepts, Practice, and Quizzes Unit 7 – Radicals

UNIT 7 RADICALS 7-1 ROOTS AND RADICALS

Square Roots

• Square root ( ): a number with the symbol that is the opposite of the square of a

number, such as 24 = and 22 = 4, respectively .Square (22)

2 4 Square root (√4)

• Perfect square: a number that is the exact square of a whole number .

Examples

Square Root Perfect Square864 = 82 = 64

11-121- = -(11) 2 = -121

32

94

94

==22 = 432 = 9

√0.16 = 0.4 0 .42 = 0 .16

√0 = 0 02 = 0

• Using a calculator: ?25 = 2nd F 25 = (The display reads 5 .)

Or 2nd F 25 = for some calculators .

• Each positive number has two square roots, one positive and one negative .

Example: 24 = and 4 = -2 or √4 = ±2

∵ 22 = 4 and (-2)2= 4, so 2 and -2 are both square roots of 4 .

Square Roots ExampleIf x2 = A,

Then � 𝑥𝑥 = √𝐴𝐴 𝑥𝑥 = -√𝐴𝐴

This can be written as Ax ±= . (A ≥ 0)

If x2 = 9,

Then �𝑥𝑥 = √9 = 3 𝑥𝑥 = -√9 = -3

This can be written as 𝑥𝑥 = ±√9 = ±3 .

Note: All even indexed radicals have 2 possible answers – positive and negative roots .

The principal square root (positive root)

Negative root

Page 7-1

UNIT 7 RADICALS 7-1 ROOTS AND RADICALS

Square Roots

• Square root ( ): a number with the symbol that is the opposite of the square of a

number, such as 24 = and 22 = 4, respectively .Square (22)

2 4 Square root (√4)

• Perfect square: a number that is the exact square of a whole number .

Examples

Square Root Perfect Square864 = 82 = 64

11-121- = -(11) 2 = -121

32

94

94

==22 = 432 = 9

√0.16 = 0.4 0 .42 = 0 .16

√0 = 0 02 = 0

• Using a calculator: ?25 = 2nd F 25 = (The display reads 5 .)

Or 2nd F 25 = for some calculators .

• Each positive number has two square roots, one positive and one negative .

Example: 24 = and 4 = -2 or √4 = ±2

∵ 22 = 4 and (-2)2= 4, so 2 and -2 are both square roots of 4 .




If x2 = 9,

Then �𝑥𝑥 = √9 = 3 𝑥𝑥 = -√9 = -3


Note: All even indexed radicals have 2 possible answers – positive and negative roots .


Negative root

Page 7-1





Square Root Functions

• Square root function: f (x) = √𝑥𝑥 Example: Given the function f (x) = √2𝑥𝑥 + 5 ,

1. Determine the function values f (2), f (0), and f (-3) .

f (2) = √2 ∙ 2 + 5 = √9 = 3 Replace x with 2 .

f (0) = √2 ∙ 0 + 5 = √𝟓𝟓 Replace x with 0 .

f (-3) = �2(-3) + 5 = �-𝟏𝟏 It is not a real number. Replace x with -3 .

2. Identify the domain of the function f (x) = √2𝑥𝑥 + 5 .

2𝑥𝑥 + 5 ≥ 02x ≥ -5 Subtract 5 .

x ≥ - 52

Divide by 2 .

Domain = { x | x ≥ - 𝟓𝟓𝟐𝟐� = �- 𝟓𝟓

𝟐𝟐, ∞)

Review: The domain is the set of x-values for which a function is defined .

• Graphing square root function

Example: Graph f (x) = -√2𝑥𝑥 .

x y = -√𝟐𝟐𝟐𝟐 (x, y)0 -√2 ∙ 0 = 0 (0, 0)2 -√2 ∙ 2 = -2 (2, -2)8 -√2 ∙ 8 = -√16 = -4 (8, -4)

• Finding √𝟐𝟐𝟐𝟐 : √𝟐𝟐𝟐𝟐 = |𝟐𝟐| x is any real number Use the absolute value sign to ensure that the positive root is non-negative (x2 never represents a negative number) .

Example: 1. 𝟐𝟐√𝟗𝟗 = 𝟐𝟐√𝟑𝟑𝟐𝟐 = 2 ∙ 3 = 6

2. 𝟐𝟐 ��-𝟑𝟑�𝟐𝟐

= 2 ∙ 3 = 𝟔𝟔

3. √𝟎𝟎𝟐𝟐 = 𝟎𝟎

4. �(-𝟏𝟏𝟏𝟏𝟐𝟐)𝟐𝟐 = �(-17)2𝑥𝑥2 = �(-17)2 √𝑥𝑥2 = 𝟏𝟏𝟏𝟏|𝟐𝟐| Split -17 and x.

5. √𝒂𝒂𝟐𝟐 − 𝟒𝟒𝒂𝒂 + 𝟒𝟒 = �(𝑎𝑎 − 2)2 = |𝒂𝒂 − 𝟐𝟐| Factor , √𝑥𝑥2 = |𝑥𝑥| a - 2

a - 2

x

y

∙ (0, 0)

∙ (2, -2)

∙ (8, -4)

Page 7-2

Square Root Functions

• Square root function: f (x) = √𝑥𝑥 Example: Given the function f (x) = √2𝑥𝑥 + 5 ,

1. Determine the function values f (2), f (0), and f (-3) .

f (2) = √2 ∙ 2 + 5 = √9 = 3 Replace x with 2 .

f (0) = √2 ∙ 0 + 5 = √𝟓𝟓 Replace x with 0 .

f (-3) = �2(-3) + 5 = �-𝟏𝟏 It is not a real number. Replace x with -3 .

2. Identify the domain of the function f (x) = √2𝑥𝑥 + 5 .

2𝑥𝑥 + 5 ≥ 02x ≥ -5 Subtract 5 .

x ≥ - 52

Divide by 2 .

Domain = { x | x ≥ - 𝟓𝟓𝟐𝟐� = �- 𝟓𝟓

𝟐𝟐, ∞)

Review: The domain is the set of x-values for which a function is defined .

• Graphing square root function

Example: Graph f (x) = -√2𝑥𝑥 .

x y = -√𝟐𝟐𝟐𝟐 (x, y)0 -√2 ∙ 0 = 0 (0, 0)2 -√2 ∙ 2 = -2 (2, -2)8 -√2 ∙ 8 = -√16 = -4 (8, -4)

• Finding √𝟐𝟐𝟐𝟐 : √𝟐𝟐𝟐𝟐 = |𝟐𝟐| x is any real number Use the absolute value sign to ensure that the positive root is non-negative (x2 never represents a negative number) .

Example: 1. 𝟐𝟐√𝟗𝟗 = 𝟐𝟐√𝟑𝟑𝟐𝟐 = 2 ∙ 3 = 6

2. 𝟐𝟐 ��-𝟑𝟑�𝟐𝟐

= 2 ∙ 3 = 𝟔𝟔

3. √𝟎𝟎𝟐𝟐 = 𝟎𝟎

4. �(-𝟏𝟏𝟏𝟏𝟐𝟐)𝟐𝟐 = �(-17)2𝑥𝑥2 = �(-17)2 √𝑥𝑥2 = 𝟏𝟏𝟏𝟏|𝟐𝟐| Split -17 and x.

5. √𝒂𝒂𝟐𝟐 − 𝟒𝟒𝒂𝒂 + 𝟒𝟒 = �(𝑎𝑎 − 2)2 = |𝒂𝒂 − 𝟐𝟐| Factor , √𝑥𝑥2 = |𝑥𝑥| a - 2

a - 2

x

y

∙ (0, 0)

∙ (2, -2)

∙ (8, -4)

Page 7-2





Radical

• A radical (root) is an expression that uses a root, such as Examples

square root, cube root, etc . √2𝑥𝑥 + 3 , √5𝑎𝑎3 , �2𝑤𝑤−34𝑡𝑡

7 The “radical” comes from the Latin word “radic”, meaning “root .”

• Radical notation for the nth root √𝒏𝒏 Example

√𝑎𝑎𝑛𝑛 �√ − the radical sign 𝑎𝑎 − the radicand (a real number)

𝑛𝑛 − the index (a positive integer > 1) 1. √5𝑎𝑎3 �5𝑎𝑎 − radicand

3 − index

√𝑎𝑎𝑛𝑛 − radical or radical expression 2. �2𝑤𝑤−34𝑡𝑡

7 �2𝑤𝑤−34𝑡𝑡

− radicand7 − index

• Rational (fractional) exponent notation a Example

A fractional power or a number is raised to a fraction . 7 13

• Rational exponent and radical are both used to indicate the nth root .

nth Root Example

Radical notation √𝑎𝑎𝑛𝑛 = 𝑎𝑎1𝑛𝑛 Rational exponent notation √73 = 7 1 3

Note: if n = 2 , write √𝑎𝑎 rather than √𝑎𝑎2 Omit 2 in √52 , write √5

• nth root to the nth power Example

√𝒂𝒂𝒏𝒏 𝒏𝒏= 𝒂𝒂 ∵ √𝒂𝒂𝒏𝒏 𝒏𝒏

= (𝒂𝒂𝒏𝒏)𝟏𝟏𝒏𝒏 = 𝒂𝒂

𝒏𝒏𝒏𝒏 = 𝒂𝒂𝟏𝟏 = 𝒂𝒂 Note: √𝒂𝒂𝒏𝒏 𝒏𝒏

= √𝒂𝒂𝒏𝒏𝒏𝒏 √𝟕𝟕𝟐𝟐

= (𝟕𝟕𝟐𝟐)𝟏𝟏𝟐𝟐 = 𝟕𝟕

𝟐𝟐𝟐𝟐 = 𝟕𝟕𝟏𝟏 = 𝟕𝟕

• The cube root Example

√𝑎𝑎3 = 𝑏𝑏 means a = b3 √83 = 2 means 8 = 23

∵ 𝑖𝑖𝑖𝑖 √𝑎𝑎3 3= 𝑏𝑏3 , then a = b3 ∵ if √83 3

= 23, then 8 = 23

Example: Find each root .

1. √64 3 = √433 = 𝟒𝟒 √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎 ; 2nd F √3 64 = 4

2. �0.027(𝑎𝑎 + 3)3 3 = �(0.3)3(𝑎𝑎 + 3)3 30 .33 = 0 .027 ; 2nd F √3 .027 = 0 .3

= �(0.3)3 3 �(𝑎𝑎 + 3)3 3√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

= 0.3 (a + 3)• nth root Example

√𝑎𝑎𝑛𝑛 = 𝑏𝑏 means a = bn √164 = 2 means 16 = 24

∵ 𝑖𝑖𝑖𝑖 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑏𝑏𝑛𝑛 , then a = bn ∵ if √164 4

= 24, then 16 = 24

1

n

– a

Page 7-3

Radical

• A radical (root) is an expression that uses a root, such as Examples

square root, cube root, etc . √2𝑥𝑥 + 3 , √5𝑎𝑎3 , �2𝑤𝑤−34𝑡𝑡

7 The “radical” comes from the Latin word “radic”, meaning “root .”

• Radical notation for the nth root √𝒏𝒏 Example

√𝑎𝑎𝑛𝑛 �√ − the radical sign 𝑎𝑎 − the radicand (a real number)

𝑛𝑛 − the index (a positive integer > 1) 1. √5𝑎𝑎3 �5𝑎𝑎 − radicand

3 − index

√𝑎𝑎𝑛𝑛 − radical or radical expression 2. �2𝑤𝑤−34𝑡𝑡

7 �2𝑤𝑤−34𝑡𝑡

− radicand7 − index

• Rational (fractional) exponent notation a Example

A fractional power or a number is raised to a fraction . 7 13

• Rational exponent and radical are both used to indicate the nth root .

nth Root Example


Note: if n = 2 , write √𝑎𝑎 rather than √𝑎𝑎2 Omit 2 in √52 , write √5

• nth root to the nth power Example

√𝒂𝒂𝒏𝒏 𝒏𝒏= 𝒂𝒂 ∵ √𝒂𝒂𝒏𝒏 𝒏𝒏

= (𝒂𝒂𝒏𝒏)𝟏𝟏𝒏𝒏 = 𝒂𝒂

𝒏𝒏𝒏𝒏 = 𝒂𝒂𝟏𝟏 = 𝒂𝒂 Note: √𝒂𝒂𝒏𝒏 𝒏𝒏


= (𝟕𝟕𝟐𝟐)𝟏𝟏𝟐𝟐 = 𝟕𝟕

𝟐𝟐𝟐𝟐 = 𝟕𝟕𝟏𝟏 = 𝟕𝟕

• The cube root Example

√𝑎𝑎3 = 𝑏𝑏 means a = b3 √83 = 2 means 8 = 23

∵ 𝑖𝑖𝑖𝑖 √𝑎𝑎3 3= 𝑏𝑏3 , then a = b3 ∵ if √83 3

= 23, then 8 = 23

Example: Find each root .

1. √64 3 = √433 = 𝟒𝟒 √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎 ; 2nd F √3 64 = 4

2. �0.027(𝑎𝑎 + 3)3 3 = �(0.3)3(𝑎𝑎 + 3)3 30 .33 = 0 .027 ; 2nd F √3 .027 = 0 .3

= �(0.3)3 3 �(𝑎𝑎 + 3)3 3√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

= 0.3 (a + 3)• nth root Example


∵ 𝑖𝑖𝑖𝑖 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑏𝑏𝑛𝑛 , then a = bn ∵ if √164 4

= 24, then 16 = 24

1

n

– a

Page 7-3





Page 7-4

Odd and Even Roots

Example

• If the index n is an even natural number: √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= |𝑎𝑎𝑎𝑎|

• If the index n is an odd natural number: √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎

Natural numbers: 1, 2, 3, …

Example: Find each root.

1. �(-𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐 = �(-2)22

= �-2� = 2 n = 2 is even ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= |𝑎𝑎𝑎𝑎|

2. �(𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑)𝟖𝟖𝟖𝟖 𝟖𝟖𝟖𝟖 = 𝟑𝟑𝟑𝟑|𝐲𝐲𝐲𝐲| Use an | | when a variable is involved.

3. �-𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓

= �-255

= -𝟐𝟐𝟐𝟐 Rewrite 32 as a perfect 5th power. 25

n = 5 is odd; a = -2 ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎

4. �(𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 − 𝟒𝟒𝟒𝟒)𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 − 𝟒𝟒𝟒𝟒 n = 177 is odd ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎

5. �-𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= �(-𝟏𝟏𝟏𝟏)𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= -𝟏𝟏𝟏𝟏 n = 11 is odd; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎 ; 1n = 1

6. � 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

𝟒𝟒𝟒𝟒 = � 1

244 = √14

√244 = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 24 = 16; �𝑎𝑎𝑎𝑎

𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎

𝑛𝑛𝑛𝑛

√𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛

7. √𝟐𝟐𝟐𝟐𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑𝒕𝒕𝒕𝒕𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 = √35 𝑡𝑡𝑡𝑡55 = √35 5 √𝑡𝑡𝑡𝑡5 5 = 3 t 35 = 243; √𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎

5 2nd F √𝑥𝑥𝑥𝑥 243 = 3

or 5 MATH 5 243 Enter 3 (T1-83 Plus)

• Find the function values

Example: Given the function f (x) = - √4𝑥𝑥𝑥𝑥 + 1 3

, determine the function values f (0) and f (31).

f (0) = - √4 ∙ 𝟎𝟎𝟎𝟎 + 1 3 = - √1 3 = -1 Replace y with 0.

f (31) = - √4 ∙ 𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏 + 1 3 Replace y with 31.

= -√125 3 = - √53 3 = -5 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎; 3 2nd F √𝑥𝑥𝑥𝑥 125 = 5

or MATH 4 125 Enter 5 (T1-83 Plus)

�-3𝟒𝟒𝟒𝟒

𝟒𝟒𝟒𝟒

= �-3� = 3

√6𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓= 6 a = 6

�-6𝟓𝟓𝟓𝟓

𝟓𝟓𝟓𝟓

= -6 a = - 6

√0𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓= 0 a = 0

Page 7-4

Odd and Even Roots

Example

• If the index n is an even natural number: √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= |𝑎𝑎𝑎𝑎|

• If the index n is an odd natural number: √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎


Example: Find each root.

1. �(-𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐 = �(-2)22

= �-2� = 2 n = 2 is even ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= |𝑎𝑎𝑎𝑎|

2. �(𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑)𝟖𝟖𝟖𝟖 𝟖𝟖𝟖𝟖 = 𝟑𝟑𝟑𝟑|𝐲𝐲𝐲𝐲| Use an | | when a variable is involved.

3. �-𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓

= �-255

= -𝟐𝟐𝟐𝟐 Rewrite 32 as a perfect 5th power. 25

n = 5 is odd; a = -2 ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎

4. �(𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 − 𝟒𝟒𝟒𝟒)𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 − 𝟒𝟒𝟒𝟒 n = 177 is odd ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎

5. �-𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= �(-𝟏𝟏𝟏𝟏)𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= -𝟏𝟏𝟏𝟏 n = 11 is odd; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎 ; 1n = 1

6. � 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

𝟒𝟒𝟒𝟒 = � 1

244 = √14

√244 = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 24 = 16; �𝑎𝑎𝑎𝑎

𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎

𝑛𝑛𝑛𝑛

√𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛

7. √𝟐𝟐𝟐𝟐𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑𝒕𝒕𝒕𝒕𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 = √35 𝑡𝑡𝑡𝑡55 = √35 5 √𝑡𝑡𝑡𝑡5 5 = 3 t 35 = 243; √𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎

5 2nd F √𝑥𝑥𝑥𝑥 243 = 3

or 5 MATH 5 243 Enter 3 (T1-83 Plus)

• Find the function values

Example: Given the function f (x) = - √4𝑥𝑥𝑥𝑥 + 1 3

, determine the function values f (0) and f (31).

f (0) = - √4 ∙ 𝟎𝟎𝟎𝟎 + 1 3 = - √1 3 = -1 Replace y with 0.

f (31) = - √4 ∙ 𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏 + 1 3 Replace y with 31.

= -√125 3 = - √53 3 = -5 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎; 3 2nd F √𝑥𝑥𝑥𝑥 125 = 5

or MATH 4 125 Enter 5 (T1-83 Plus)

�-3𝟒𝟒𝟒𝟒

𝟒𝟒𝟒𝟒

= �-3� = 3

√6𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓= 6 a = 6

�-6𝟓𝟓𝟓𝟓

𝟓𝟓𝟓𝟓

= -6 a = - 6

√0𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓= 0 a = 0





7-2 RATIONAL EXPONENTS

Powers of Roots

• Review: √𝑎𝑎𝑛𝑛 �√ − the radical sign𝑎𝑎 − the radicand 𝑛𝑛 − the index

• The index of a radical

Index n Read Example Read

𝑎𝑎12 = √𝑎𝑎 the square root of a 3

12 = √3 the square root of 3

𝑎𝑎13 = √𝑎𝑎3 the cube root of a 5

13 = √53 the cube root of 5

𝑎𝑎14 = √𝑎𝑎4 the fourth root of a 7

14 = √74 the fourth root of 7

𝑎𝑎1𝑛𝑛 = √𝑎𝑎𝑛𝑛 the nth root of a 2

111 = √211 the 11th root of 2

𝑎𝑎𝑚𝑚𝑛𝑛 = √𝑎𝑎𝑛𝑛

𝑚𝑚the nth root of a to the mth power 7

56 = √76 5 the 6th root of 7 to the 5th power

• Powers of roots

The nth Root to the mth Power Example𝑎𝑎𝑚𝑚𝑛𝑛 = (√𝑎𝑎𝑛𝑛 )𝑚𝑚 = √𝑎𝑎𝑛𝑛 𝑚𝑚

= √𝑎𝑎𝑚𝑚𝑛𝑛m – power 7

23 = (√73 )2 = √73 2

= √723

Example: Rewrite without rational exponents .

1. (81)14 = √814 = √344 = 3 Rewrite 81 as 34 ; √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

2. (𝑥𝑥2 𝑦𝑦3)17 = �𝒙𝒙𝟐𝟐𝒚𝒚𝟑𝟑𝟕𝟕 𝑎𝑎

1𝑛𝑛 = √𝑎𝑎𝑛𝑛

3. (8)23 = √823 = �√83 �

2 = �√233 �

2= 22 = 4 𝑎𝑎

𝑚𝑚𝑛𝑛 = √𝑎𝑎𝑛𝑛 𝑚𝑚

= (√𝑎𝑎𝑛𝑛 )𝑚𝑚 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

Rewrite 8 as 23

Example: Rewrite the radicals using rational exponents .

1. (√3𝑎𝑎4𝑏𝑏59 )7 = (𝟑𝟑𝟑𝟑𝟒𝟒𝒃𝒃𝟓𝟓)𝟕𝟕𝟗𝟗 � √𝑎𝑎𝑛𝑛 �

𝑚𝑚= 𝑎𝑎

𝑚𝑚𝑛𝑛

2. (�5𝑝𝑝5𝑞𝑞)3 = (𝟓𝟓𝒑𝒑𝟓𝟓𝒒𝒒)𝟑𝟑𝟐𝟐 (�5𝑝𝑝5𝑞𝑞)3 ; � √𝑎𝑎𝑛𝑛 �

𝑚𝑚= 𝑎𝑎

𝑚𝑚𝑛𝑛

3. √2𝑡𝑡3 = (𝟐𝟐𝟐𝟐)𝟏𝟏𝟑𝟑 √𝑎𝑎𝑛𝑛 = 𝑎𝑎

1𝑛𝑛

4. 2w

√7𝑡𝑡5 = 𝟐𝟐𝟐𝟐(𝟕𝟕𝟐𝟐)𝟏𝟏/𝟓𝟓 √𝑎𝑎𝑛𝑛 = 𝑎𝑎

1𝑛𝑛

n = 2

rational exponent notation

radical notation

- Rational exponent notation: 𝑎𝑎1 𝑛𝑛

- Radical notation: √𝑛𝑛

Page 7-5

7-2 RATIONAL EXPONENTS

Powers of Roots

• Review: √𝑎𝑎𝑛𝑛 �√ − the radical sign𝑎𝑎 − the radicand 𝑛𝑛 − the index

• The index of a radical

Index n Read Example Read








111 = √211 the 11th root of 2




• Powers of roots

The nth Root to the mth Power Example𝑎𝑎𝑚𝑚𝑛𝑛 = (√𝑎𝑎𝑛𝑛 )𝑚𝑚 = √𝑎𝑎𝑛𝑛 𝑚𝑚

= √𝑎𝑎𝑚𝑚𝑛𝑛m – power 7

23 = (√73 )2 = √73 2

= √723

Example: Rewrite without rational exponents .

1. (81)14 = √814 = √344 = 3 Rewrite 81 as 34 ; √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

2. (𝑥𝑥2 𝑦𝑦3)17 = �𝒙𝒙𝟐𝟐𝒚𝒚𝟑𝟑𝟕𝟕 𝑎𝑎

1𝑛𝑛 = √𝑎𝑎𝑛𝑛

3. (8)23 = √823 = �√83 �

2 = �√233 �

2= 22 = 4 𝑎𝑎

𝑚𝑚𝑛𝑛 = √𝑎𝑎𝑛𝑛 𝑚𝑚

= (√𝑎𝑎𝑛𝑛 )𝑚𝑚 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

Rewrite 8 as 23

Example: Rewrite the radicals using rational exponents .

1. (√3𝑎𝑎4𝑏𝑏59 )7 = (𝟑𝟑𝟑𝟑𝟒𝟒𝒃𝒃𝟓𝟓)𝟕𝟕𝟗𝟗 � √𝑎𝑎𝑛𝑛 �

𝑚𝑚= 𝑎𝑎

𝑚𝑚𝑛𝑛

2. (�5𝑝𝑝5𝑞𝑞)3 = (𝟓𝟓𝒑𝒑𝟓𝟓𝒒𝒒)𝟑𝟑𝟐𝟐 (�5𝑝𝑝5𝑞𝑞)3 ; � √𝑎𝑎𝑛𝑛 �

𝑚𝑚= 𝑎𝑎

𝑚𝑚𝑛𝑛

3. √2𝑡𝑡3 = (𝟐𝟐𝟐𝟐)𝟏𝟏𝟑𝟑 √𝑎𝑎𝑛𝑛 = 𝑎𝑎

1𝑛𝑛

4. 2w

√7𝑡𝑡5 = 𝟐𝟐𝟐𝟐(𝟕𝟕𝟐𝟐)𝟏𝟏/𝟓𝟓 √𝑎𝑎𝑛𝑛 = 𝑎𝑎

1𝑛𝑛

n = 2


radical notation

- Rational exponent notation: 𝑎𝑎1 𝑛𝑛

- Radical notation: √𝑛𝑛

Page 7-5





Exponents & Rules

• Exponents: review of basic rules

Name Rule Examplezero exponent a0 a0 = 1 (a ≠ 0, 00 is undefined) (15)0 = 1 one exponent a1 a1 = a (But 1n = 1) 71 = 7 , 113 = 1

negative exponent 𝒂𝒂-𝒏𝒏

𝑎𝑎-𝑛𝑛 = 1𝑎𝑎𝑛𝑛

5-2 = 152

= 125

1𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛 1

6-2 = 62 = 36

�𝑎𝑎𝑏𝑏�

-𝑛𝑛= �

𝑏𝑏𝑎𝑎�𝑛𝑛

�45�

-3= �

54�

3

Example: Express each of the following in positive exponential form .

1. (- 0 .1356)0 = 1 a0 = 1

2. 64-1/3 = 1641/3 = 1

√643 = 1

√433 = 𝟏𝟏𝟒𝟒


, 𝑎𝑎1𝑛𝑛 = √𝑎𝑎𝑛𝑛 , √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

3. –5𝑝𝑝–3/4 𝑞𝑞2/5 𝑟𝑟–1/2 = −𝟓𝟓 𝒒𝒒𝟐𝟐/𝟓𝟓

𝒑𝒑𝟑𝟑/𝟒𝟒 𝒓𝒓𝟏𝟏/𝟐𝟐 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

4. �4𝑐𝑐𝑐𝑐3𝑎𝑎𝑎𝑎

�-4/7

= �𝟑𝟑𝒂𝒂𝟑𝟑𝟒𝟒𝟒𝟒𝟒𝟒

�𝟒𝟒/𝟕𝟕

�𝑎𝑎𝑎𝑎�

-𝑛𝑛= �𝑎𝑎

𝑎𝑎�𝑛𝑛

• Exponent rules review

Name Ruleproduct of like bases aman = am + n

quotient of like bases 𝑎𝑎𝑚𝑚


power of a power (am)n = am n power of a product (a ∙ b )n = an bn (am ∙ b n)p = amp bnp

power of a quotient �𝑎𝑎𝑎𝑎�𝑛𝑛

= 𝑎𝑎𝑛𝑛

𝑎𝑎𝑛𝑛 �𝑎𝑎

𝑚𝑚

𝑎𝑎𝑛𝑛�𝑝𝑝


𝑎𝑎𝑛𝑛𝑚𝑚


1. 101/2 10 -2/3 = 101/2 – 2/3 = 103/6 – 4/6 =10-1/6 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏/𝟔𝟔 am an = am + n , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

2. 𝒘𝒘𝟓𝟓/𝟕𝟕

𝒘𝒘𝟐𝟐/𝟕𝟕 = 𝑤𝑤5/7−2/7 = 𝒘𝒘𝟑𝟑/𝟕𝟕 𝑎𝑎𝑚𝑚


3. (y-2/3) 4/5 = y(-2/3) (4/5) = y-8/15 = 𝟏𝟏𝒚𝒚𝟖𝟖/𝟏𝟏𝟓𝟓 (an)m = an m , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

4. (u ∙ v )3/5 = u 3/5 v3/5(a ∙ b)n = an bn

5. (a 2 ∙ 𝟑𝟑 -3)2/5 = a 2 ∙ (2/5) ∙ 𝑏𝑏 -3 ∙ (2/5) = a 4/5 ∙ 𝑏𝑏 -6/5 = 𝒂𝒂𝟒𝟒/𝟓𝟓

𝟑𝟑𝟔𝟔/𝟓𝟓 (am ∙bn)p = a m p b n p , 𝑎𝑎-𝑛𝑛 = 1𝑎𝑎𝑛𝑛

6. �𝒙𝒙𝒚𝒚�−𝟑𝟑

= �𝑦𝑦𝑥𝑥�3

= 𝒚𝒚𝟑𝟑

𝒙𝒙𝟑𝟑 �𝑎𝑎𝑎𝑎�


𝑎𝑎�𝑛𝑛

, �𝑎𝑎𝑎𝑎�𝑛𝑛

= 𝑎𝑎𝑛𝑛

𝑎𝑎𝑛𝑛

7. � 𝒕𝒕𝟓𝟓

𝒖𝒖-𝟐𝟐�𝟐𝟐

= 𝑡𝑡5 ∙ 2

𝑢𝑢(-2)(2) = 𝑡𝑡10




𝑎𝑎𝑛𝑛𝑚𝑚, 1𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛

Page 7-6

Exponents & Rules

• Exponents: review of basic rules

Name Rule Examplezero exponent a0 a0 = 1 (a ≠ 0, 00 is undefined) (15)0 = 1 one exponent a1 a1 = a (But 1n = 1) 71 = 7 , 113 = 1

negative exponent 𝒂𝒂-𝒏𝒏


5-2 = 152

= 125

1𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛 1

6-2 = 62 = 36

�𝑎𝑎𝑏𝑏�

-𝑛𝑛= �

𝑏𝑏𝑎𝑎�𝑛𝑛

�45�

-3= �

54�

3


1. (- 0 .1356)0 = 1 a0 = 1

2. 64-1/3 = 1641/3 = 1

√643 = 1

√433 = 𝟏𝟏𝟒𝟒


, 𝑎𝑎1𝑛𝑛 = √𝑎𝑎𝑛𝑛 , √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

3. –5𝑝𝑝–3/4 𝑞𝑞2/5 𝑟𝑟–1/2 = −𝟓𝟓 𝒒𝒒𝟐𝟐/𝟓𝟓

𝒑𝒑𝟑𝟑/𝟒𝟒 𝒓𝒓𝟏𝟏/𝟐𝟐 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

4. �4𝑐𝑐𝑐𝑐3𝑎𝑎𝑎𝑎

�-4/7

= �𝟑𝟑𝒂𝒂𝟑𝟑𝟒𝟒𝟒𝟒𝟒𝟒

�𝟒𝟒/𝟕𝟕

�𝑎𝑎𝑎𝑎�


𝑎𝑎�𝑛𝑛

• Exponent rules review

Name Ruleproduct of like bases aman = am + n

quotient of like bases 𝑎𝑎𝑚𝑚


power of a power (am)n = am n power of a product (a ∙ b )n = an bn (am ∙ b n)p = amp bnp

power of a quotient �𝑎𝑎𝑎𝑎�𝑛𝑛

= 𝑎𝑎𝑛𝑛

𝑎𝑎𝑛𝑛 �𝑎𝑎

𝑚𝑚



𝑎𝑎𝑛𝑛𝑚𝑚


1. 101/2 10 -2/3 = 101/2 – 2/3 = 103/6 – 4/6 =10-1/6 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏/𝟔𝟔 am an = am + n , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

2. 𝒘𝒘𝟓𝟓/𝟕𝟕

𝒘𝒘𝟐𝟐/𝟕𝟕 = 𝑤𝑤5/7−2/7 = 𝒘𝒘𝟑𝟑/𝟕𝟕 𝑎𝑎𝑚𝑚


3. (y-2/3) 4/5 = y(-2/3) (4/5) = y-8/15 = 𝟏𝟏𝒚𝒚𝟖𝟖/𝟏𝟏𝟓𝟓 (an)m = an m , 𝑎𝑎-𝑛𝑛 = 1

𝑎𝑎𝑛𝑛

4. (u ∙ v )3/5 = u 3/5 v3/5(a ∙ b)n = an bn

5. (a 2 ∙ 𝟑𝟑 -3)2/5 = a 2 ∙ (2/5) ∙ 𝑏𝑏 -3 ∙ (2/5) = a 4/5 ∙ 𝑏𝑏 -6/5 = 𝒂𝒂𝟒𝟒/𝟓𝟓

𝟑𝟑𝟔𝟔/𝟓𝟓 (am ∙bn)p = a m p b n p , 𝑎𝑎-𝑛𝑛 = 1𝑎𝑎𝑛𝑛

6. �𝒙𝒙𝒚𝒚�−𝟑𝟑

= �𝑦𝑦𝑥𝑥�3

= 𝒚𝒚𝟑𝟑

𝒙𝒙𝟑𝟑 �𝑎𝑎𝑎𝑎�


𝑎𝑎�𝑛𝑛

, �𝑎𝑎𝑎𝑎�𝑛𝑛

= 𝑎𝑎𝑛𝑛

𝑎𝑎𝑛𝑛

7. � 𝒕𝒕𝟓𝟓

𝒖𝒖-𝟐𝟐�𝟐𝟐

= 𝑡𝑡5 ∙ 2

𝑢𝑢(-2)(2) = 𝑡𝑡10




𝑎𝑎𝑛𝑛𝑚𝑚, 1𝑎𝑎-𝑛𝑛 = 𝑎𝑎𝑛𝑛

Page 7-6





Page 7-7

Simplifying Radical Expressions

Example

• A radical expression is an algebraic expression containing a radical sign √𝑛𝑛𝑛𝑛 . �7𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦23

• Simplifying radical expressions

Example: Express in simplest radical form.

1. √𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = (𝑥𝑥𝑥𝑥)210 = 𝑥𝑥𝑥𝑥

15 = √𝒙𝒙𝒙𝒙𝟓𝟓𝟓𝟓 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑚𝑚𝑚𝑚

= 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚𝑛𝑛𝑛𝑛

2. �𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐𝟓𝟓𝟓𝟓 = (25 𝑥𝑥𝑥𝑥10𝑦𝑦𝑦𝑦2)15

√𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 = 𝑎𝑎𝑎𝑎1𝑛𝑛𝑛𝑛

= 25 ∙ 15 𝑥𝑥𝑥𝑥10 ∙ 15 𝑦𝑦𝑦𝑦2 ∙ 15 = 𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐 �𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓 (am ∙ bn) p = amp bnp ,

𝑎𝑎𝑎𝑎1𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛

3. √𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 √𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 = (3𝑎𝑎𝑎𝑎)14 (2𝑎𝑎𝑎𝑎)

12 = (3𝑎𝑎𝑎𝑎)

14 (2𝑎𝑎𝑎𝑎)

24

√𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 = 𝑎𝑎𝑎𝑎1𝑛𝑛𝑛𝑛 , LCD = 4

= [(3𝑎𝑎𝑎𝑎)1(2𝑎𝑎𝑎𝑎)2]𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒 = �(3𝑎𝑎𝑎𝑎)(2𝑎𝑎𝑎𝑎)24 amp bnp=(am ∙ bn)p , √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑚𝑚𝑚𝑚


= √𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 am an = am + n

4. �√𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟔𝟔𝟔𝟔 = �𝑎𝑎𝑎𝑎13

6

= (𝑎𝑎𝑎𝑎13)

16


= 𝑎𝑎𝑎𝑎13 ∙ 16 = 𝑎𝑎𝑎𝑎

118 = √𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 (an)m = anm

, 𝑎𝑎𝑎𝑎

1𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛

5. 𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

𝟒𝟒𝟒𝟒𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

= 34𝑦𝑦𝑦𝑦

12−

23 = 3

4𝑦𝑦𝑦𝑦36−

46 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚

𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚−𝑛𝑛𝑛𝑛 , LCD = 6


-16 = 3

4𝑦𝑦𝑦𝑦16

= 𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 �𝒚𝒚𝒚𝒚 𝟔𝟔𝟔𝟔 𝑎𝑎𝑎𝑎-𝑛𝑛𝑛𝑛 = 1

𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛

, 𝑎𝑎𝑎𝑎

1𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 n

6. 𝒙𝒙𝒙𝒙𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 𝒚𝒚𝒚𝒚

𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑 𝒛𝒛𝒛𝒛

𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 = 𝑥𝑥𝑥𝑥

912 𝑦𝑦𝑦𝑦

1612 𝑧𝑧𝑧𝑧

612 LCD = 12

= (𝑥𝑥𝑥𝑥9 𝑦𝑦𝑦𝑦16 𝑧𝑧𝑧𝑧6)𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 = �𝒙𝒙𝒙𝒙𝟗𝟗𝟗𝟗𝒚𝒚𝒚𝒚𝟏𝟏𝟏𝟏𝟔𝟔𝟔𝟔𝒛𝒛𝒛𝒛𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 anm = (an)m

, 𝑎𝑎𝑎𝑎


7. �𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑𝒃𝒃𝒃𝒃𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑

𝒄𝒄𝒄𝒄𝟏𝟏𝟏𝟏/𝟔𝟔𝟔𝟔𝒅𝒅𝒅𝒅𝟏𝟏𝟏𝟏/𝟔𝟔𝟔𝟔�2

= (𝑎𝑎𝑎𝑎1/3)𝟐𝟐𝟐𝟐(𝑏𝑏𝑏𝑏1/3)𝟐𝟐𝟐𝟐

(𝑐𝑐𝑐𝑐1/6𝑑𝑑𝑑𝑑1/6)𝟐𝟐𝟐𝟐 = (𝑎𝑎𝑎𝑎2/𝟑𝟑𝟑𝟑)(𝑏𝑏𝑏𝑏2/𝟑𝟑𝟑𝟑)

𝑐𝑐𝑐𝑐2/6𝑑𝑑𝑑𝑑2/6 �𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚

𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛�𝑝𝑝𝑝𝑝

= 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛𝑚𝑚𝑚𝑚

= (𝑎𝑎𝑎𝑎2/𝟑𝟑𝟑𝟑)(𝑏𝑏𝑏𝑏2/𝟑𝟑𝟑𝟑)(𝑐𝑐𝑐𝑐𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑)(𝑑𝑑𝑑𝑑𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑)

= �(𝑎𝑎𝑎𝑎2)(𝑏𝑏𝑏𝑏2)(𝑐𝑐𝑐𝑐1)(𝑑𝑑𝑑𝑑1)

� 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑 amp bnp=(am ∙bn)p

= �𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒃𝒃𝒃𝒃𝟐𝟐𝟐𝟐

𝒄𝒄𝒄𝒄𝒅𝒅𝒅𝒅

𝟑𝟑𝟑𝟑


Page 7-7

Simplifying Radical Expressions

Example

• A radical expression is an algebraic expression containing a radical sign √𝑛𝑛𝑛𝑛 . �7𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦23

• Simplifying radical expressions

Example: Express in simplest radical form.

1. √𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = (𝑥𝑥𝑥𝑥)210 = 𝑥𝑥𝑥𝑥

15 = √𝒙𝒙𝒙𝒙𝟓𝟓𝟓𝟓 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑚𝑚𝑚𝑚


2. �𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐𝟓𝟓𝟓𝟓 = (25 𝑥𝑥𝑥𝑥10𝑦𝑦𝑦𝑦2)15


= 25 ∙ 15 𝑥𝑥𝑥𝑥10 ∙ 15 𝑦𝑦𝑦𝑦2 ∙ 15 = 𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐 �𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓 (am ∙ bn) p = amp bnp ,


3. √𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 √𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 = (3𝑎𝑎𝑎𝑎)14 (2𝑎𝑎𝑎𝑎)

12 = (3𝑎𝑎𝑎𝑎)

14 (2𝑎𝑎𝑎𝑎)

24

√𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 = 𝑎𝑎𝑎𝑎1𝑛𝑛𝑛𝑛 , LCD = 4

= [(3𝑎𝑎𝑎𝑎)1(2𝑎𝑎𝑎𝑎)2]𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒 = �(3𝑎𝑎𝑎𝑎)(2𝑎𝑎𝑎𝑎)24 amp bnp=(am ∙ bn)p , √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑚𝑚𝑚𝑚


= √𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 am an = am + n

4. �√𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟔𝟔𝟔𝟔 = �𝑎𝑎𝑎𝑎13

6

= (𝑎𝑎𝑎𝑎13)

16


= 𝑎𝑎𝑎𝑎13 ∙ 16 = 𝑎𝑎𝑎𝑎

118 = √𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 (an)m = anm

, 𝑎𝑎𝑎𝑎


5. 𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

𝟒𝟒𝟒𝟒𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑


12−

23 = 3

4𝑦𝑦𝑦𝑦36−

46 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚

𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛= 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚−𝑛𝑛𝑛𝑛 , LCD = 6


-16 = 3

4𝑦𝑦𝑦𝑦16

= 𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 �𝒚𝒚𝒚𝒚 𝟔𝟔𝟔𝟔 𝑎𝑎𝑎𝑎-𝑛𝑛𝑛𝑛 = 1

𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛

, 𝑎𝑎𝑎𝑎

1𝑛𝑛𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 n

6. 𝒙𝒙𝒙𝒙𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 𝒚𝒚𝒚𝒚

𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑 𝒛𝒛𝒛𝒛

𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 = 𝑥𝑥𝑥𝑥

912 𝑦𝑦𝑦𝑦

1612 𝑧𝑧𝑧𝑧

612 LCD = 12

= (𝑥𝑥𝑥𝑥9 𝑦𝑦𝑦𝑦16 𝑧𝑧𝑧𝑧6)𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 = �𝒙𝒙𝒙𝒙𝟗𝟗𝟗𝟗𝒚𝒚𝒚𝒚𝟏𝟏𝟏𝟏𝟔𝟔𝟔𝟔𝒛𝒛𝒛𝒛𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 anm = (an)m

, 𝑎𝑎𝑎𝑎


7. �𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑𝒃𝒃𝒃𝒃𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑

𝒄𝒄𝒄𝒄𝟏𝟏𝟏𝟏/𝟔𝟔𝟔𝟔𝒅𝒅𝒅𝒅𝟏𝟏𝟏𝟏/𝟔𝟔𝟔𝟔�2

= (𝑎𝑎𝑎𝑎1/3)𝟐𝟐𝟐𝟐(𝑏𝑏𝑏𝑏1/3)𝟐𝟐𝟐𝟐

(𝑐𝑐𝑐𝑐1/6𝑑𝑑𝑑𝑑1/6)𝟐𝟐𝟐𝟐 = (𝑎𝑎𝑎𝑎2/𝟑𝟑𝟑𝟑)(𝑏𝑏𝑏𝑏2/𝟑𝟑𝟑𝟑)

𝑐𝑐𝑐𝑐2/6𝑑𝑑𝑑𝑑2/6 �𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚

𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛�𝑝𝑝𝑝𝑝

= 𝑎𝑎𝑎𝑎𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

𝑏𝑏𝑏𝑏𝑛𝑛𝑛𝑛𝑚𝑚𝑚𝑚

= (𝑎𝑎𝑎𝑎2/𝟑𝟑𝟑𝟑)(𝑏𝑏𝑏𝑏2/𝟑𝟑𝟑𝟑)(𝑐𝑐𝑐𝑐𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑)(𝑑𝑑𝑑𝑑𝟏𝟏𝟏𝟏/𝟑𝟑𝟑𝟑)

= �(𝑎𝑎𝑎𝑎2)(𝑏𝑏𝑏𝑏2)(𝑐𝑐𝑐𝑐1)(𝑑𝑑𝑑𝑑1)

� 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑 amp bnp=(am ∙bn)p

= �𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒃𝒃𝒃𝒃𝟐𝟐𝟐𝟐

𝒄𝒄𝒄𝒄𝒅𝒅𝒅𝒅

𝟑𝟑𝟑𝟑






7-3 SIMPLIFY RADICALS USING PRODUCT & QUOTIENT RULES

Product and Quotient Rules

• Product and quotient rule for radicals

Name Rule Exampleproduct rule √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 ∙ √𝑎𝑎𝑛𝑛 a ≥ 0 , b ≥ 0 √12 = √4 ∙ 3 = √4 √3 = �22 √3 = 2√3

quotient rule �𝑎𝑎𝑏𝑏

𝑛𝑛 = √𝑎𝑎𝑛𝑛

√𝑏𝑏𝑛𝑛 a ≥ 0 , b > 0 , b ≠ 0 � 827

3 = √83

√273 =�233

�333 = 23

Read √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 : The nth root of the product of a and b is the product of the nth root of a and the nth root of b .

�𝑎𝑎𝑏𝑏


√𝑏𝑏𝑛𝑛 : The nth root of 𝑎𝑎𝑏𝑏

is the nth root of a over the nth root of b .

• Simplifying radical expressions: a radical expression is in simplest form if it satisfies the

following conditions .

A radical expression is in simplest form when: Simplest Form Not Simplest FormThe exponent (m) of the radical is less than the index (n).

m < n √𝑎𝑎𝑛𝑛 𝑚𝑚 �𝑥𝑥3 5 or √𝑎𝑎5 3

3 < 5

�𝑥𝑥87

8 > 7

No fractions appear within a radical sign. �2𝑦𝑦3 �2

3𝑥𝑥𝑦𝑦4

No radicals appear in the denominator of a fraction. √35

√3√8

• Tips: To use the product and quotient rule for radicals, factor out any perfect square, perfect

cube, and perfect 4th power, perfect 5th power, etc .

Perfect square: a number that is the exact square of a whole number .

Perfect cube: a number that is the exact cube of a whole number .

Perfect nth power: a number that is the exact nth power of a whole number .

Examples

Perfect Square Perfect Cube Perfect 4th Power Perfect 5th Power … …22 = 4 23 = 8 24 = 16 25 = 3232 = 9 33 = 27 34 = 81 35 = 243

42 = 16 43 = 64 44 = 256 45 = 102452 = 25 53 = 125 54 = 625 55 = 312562 = 36 63 = 216 64 = 1296 65 = 777672 = 49 73 = 343 74 = 2401 75 = 1680782 = 64 83 = 512 84 = 4096 85 = 32768

… … … … … … … …

2 y x 5 = 32

or 2 ^ 5 Enter 32

Page 7-8

7-3 SIMPLIFY RADICALS USING PRODUCT & QUOTIENT RULES

Product and Quotient Rules

• Product and quotient rule for radicals

Name Rule Exampleproduct rule √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 ∙ √𝑎𝑎𝑛𝑛 a ≥ 0 , b ≥ 0 √12 = √4 ∙ 3 = √4 √3 = �22 √3 = 2√3



√𝑏𝑏𝑛𝑛 a ≥ 0 , b > 0 , b ≠ 0 � 827

3 = √83

√273 =�233

�333 = 23

Read √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 : The nth root of the product of a and b is the product of the nth root of a and the nth root of b .

�𝑎𝑎𝑏𝑏


√𝑏𝑏𝑛𝑛 : The nth root of 𝑎𝑎𝑏𝑏

is the nth root of a over the nth root of b .

• Simplifying radical expressions: a radical expression is in simplest form if it satisfies the

following conditions .

A radical expression is in simplest form when: Simplest Form Not Simplest FormThe exponent (m) of the radical is less than the index (n).


3 < 5

�𝑥𝑥87

8 > 7


3𝑥𝑥𝑦𝑦4


√3√8

• Tips: To use the product and quotient rule for radicals, factor out any perfect square, perfect

cube, and perfect 4th power, perfect 5th power, etc .

Perfect square: a number that is the exact square of a whole number .

Perfect cube: a number that is the exact cube of a whole number .

Perfect nth power: a number that is the exact nth power of a whole number .

Examples

Perfect Square Perfect Cube Perfect 4th Power Perfect 5th Power … …22 = 4 23 = 8 24 = 16 25 = 3232 = 9 33 = 27 34 = 81 35 = 243

42 = 16 43 = 64 44 = 256 45 = 102452 = 25 53 = 125 54 = 625 55 = 312562 = 36 63 = 216 64 = 1296 65 = 777672 = 49 73 = 343 74 = 2401 75 = 1680782 = 64 83 = 512 84 = 4096 85 = 32768

… … … … … … … …

2 y x 5 = 32

or 2 ^ 5 Enter 32

Page 7-8





Simplifying Radicals

Example: Simplify .

1. √𝟏𝟏𝟏𝟏𝟏𝟏𝟑𝟑 = √2 ∙ 9 𝑥𝑥2 ∙ 𝑥𝑥 Factor out a perfect square: 9 = 32

= �(3𝑥𝑥)2 √2𝑥𝑥 = 𝟑𝟑|𝟏𝟏|√𝟐𝟐𝟏𝟏 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

2. √𝟒𝟒𝟒𝟒 𝟑𝟑 = √5 ∙ 83 = √5 ∙ 233Factor out a perfect cube: 8 = 23

= √53 √233 = 𝟐𝟐√𝟓𝟓 𝟑𝟑√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

3. �𝟐𝟐𝟒𝟒𝟏𝟏𝟓𝟓𝒚𝒚 𝟑𝟑 = �3 ∙ 8𝑥𝑥3𝑥𝑥2𝑦𝑦 3 = �3𝑥𝑥2𝑦𝑦 3 √8𝑥𝑥33√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; regroup

= �3𝑥𝑥2𝑦𝑦 3 �(2𝑥𝑥)33 = 𝟐𝟐𝟏𝟏 �𝟑𝟑𝟏𝟏𝟐𝟐𝒚𝒚 𝟑𝟑 8 = 23 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

4. √𝟏𝟏𝟐𝟐𝟒𝟒 √𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒 = √𝑥𝑥2 ∙ 16𝑥𝑥6 4√𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛 ; am an = am + n

= √24 ∙ 𝑥𝑥8 4Rewrite 16 as a perfect 4th power: 16 = 24

= √244 √𝑥𝑥8 4 = 2 √𝑥𝑥8 4√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

= 2 𝑥𝑥84 = 2x2

√𝑎𝑎 𝑛𝑛 m = 𝑎𝑎

𝑚𝑚𝑛𝑛

5. � 𝟗𝟗𝟒𝟒𝟗𝟗

= √9√49

= √32

√72= 𝟑𝟑

𝟕𝟕�𝑎𝑎𝑏𝑏


√𝑏𝑏𝑛𝑛 ; √𝑎𝑎 𝑛𝑛 𝑛𝑛= 𝑎𝑎

6. � 𝟑𝟑𝟐𝟐𝟐𝟐𝟒𝟒𝟑𝟑

𝟓𝟓= √325

√2435 = √255

�355 = 𝟐𝟐𝟑𝟑 �𝑎𝑎

𝑏𝑏𝑛𝑛 = √𝑎𝑎

𝑛𝑛

√𝑏𝑏𝑛𝑛 ; 25 = 32 ; 35 = 243

5 2nd F √𝑥𝑥 243 = 3

7. � 𝟓𝟓𝟒𝟒𝒂𝒂𝟏𝟏

𝟏𝟏𝒂𝒂𝟒𝟒= � 9𝑎𝑎8

𝑎𝑎4 = √32𝒂𝒂𝟒𝟒 𝑎𝑎𝑚𝑚


= �32(𝒂𝒂𝟐𝟐)𝟐𝟐 = √32 ∙ �(𝑎𝑎2)2 = 3a2√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎 𝑛𝑛 𝑛𝑛

= 𝑎𝑎

8.�𝟏𝟏𝟏𝟏𝒂𝒂𝟓𝟓𝒃𝒃𝟏𝟏𝟑𝟑

�𝟑𝟑𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐𝟑𝟑 = �81𝑎𝑎5𝑏𝑏8

3𝑎𝑎2𝑏𝑏23

= √27𝑎𝑎3𝑎𝑎6 3 = √33𝑎𝑎3𝑎𝑎6 3 √𝑎𝑎𝑛𝑛

√𝑏𝑏𝑛𝑛 = �𝑎𝑎𝑏𝑏

𝑛𝑛 ; 𝑎𝑎𝑚𝑚


= √333 √𝑎𝑎33 √𝑎𝑎63 = 3a𝑎𝑎63 = 3a𝒃𝒃𝟐𝟐 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎 𝑛𝑛 𝑛𝑛

= 𝑎𝑎; √𝑎𝑎 𝑛𝑛 m = 𝑎𝑎

𝑚𝑚𝑛𝑛

9.�𝒂𝒂𝟐𝟐𝒃𝒃𝟒𝟒𝟓𝟓

√𝒂𝒂𝒃𝒃𝟑𝟑 = (𝑎𝑎2𝑏𝑏4)15

(𝑎𝑎𝑏𝑏)13

= 𝑎𝑎25 𝑏𝑏

45

𝑎𝑎13 𝑏𝑏

13

√𝑎𝑎𝑛𝑛 = 𝑎𝑎1𝑛𝑛 ; (am ∙bn) p = amp bnp

= �𝑎𝑎25 – 13� �𝑎𝑎

45 − 13� = �𝑎𝑎

615 – 515� �𝑎𝑎

1215 − 515� 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛 ; LCD = 15

= 𝑎𝑎115 𝑎𝑎

715 = (𝑎𝑎1 𝑎𝑎7)

115 = √𝒂𝒂𝒃𝒃𝟕𝟕𝟏𝟏𝟓𝟓

𝑎𝑎1𝑛𝑛 = √𝑎𝑎𝑛𝑛

9

1n = 2

n = 2

Page 7-9

Simplifying Radicals

Example: Simplify .

1. √𝟏𝟏𝟏𝟏𝟏𝟏𝟑𝟑 = √2 ∙ 9 𝑥𝑥2 ∙ 𝑥𝑥 Factor out a perfect square: 9 = 32

= �(3𝑥𝑥)2 √2𝑥𝑥 = 𝟑𝟑|𝟏𝟏|√𝟐𝟐𝟏𝟏 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

2. √𝟒𝟒𝟒𝟒 𝟑𝟑 = √5 ∙ 83 = √5 ∙ 233Factor out a perfect cube: 8 = 23

= √53 √233 = 𝟐𝟐√𝟓𝟓 𝟑𝟑√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

3. �𝟐𝟐𝟒𝟒𝟏𝟏𝟓𝟓𝒚𝒚 𝟑𝟑 = �3 ∙ 8𝑥𝑥3𝑥𝑥2𝑦𝑦 3 = �3𝑥𝑥2𝑦𝑦 3 √8𝑥𝑥33√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; regroup

= �3𝑥𝑥2𝑦𝑦 3 �(2𝑥𝑥)33 = 𝟐𝟐𝟏𝟏 �𝟑𝟑𝟏𝟏𝟐𝟐𝒚𝒚 𝟑𝟑 8 = 23 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

4. √𝟏𝟏𝟐𝟐𝟒𝟒 √𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒 = √𝑥𝑥2 ∙ 16𝑥𝑥6 4√𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑎𝑎𝑛𝑛 ; am an = am + n

= √24 ∙ 𝑥𝑥8 4Rewrite 16 as a perfect 4th power: 16 = 24

= √244 √𝑥𝑥8 4 = 2 √𝑥𝑥8 4√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

= 2 𝑥𝑥84 = 2x2

√𝑎𝑎 𝑛𝑛 m = 𝑎𝑎

𝑚𝑚𝑛𝑛

5. � 𝟗𝟗𝟒𝟒𝟗𝟗

= √9√49

= √32

√72= 𝟑𝟑

𝟕𝟕�𝑎𝑎𝑏𝑏


√𝑏𝑏𝑛𝑛 ; √𝑎𝑎 𝑛𝑛 𝑛𝑛= 𝑎𝑎

6. � 𝟑𝟑𝟐𝟐𝟐𝟐𝟒𝟒𝟑𝟑

𝟓𝟓= √325

√2435 = √255

�355 = 𝟐𝟐𝟑𝟑 �𝑎𝑎

𝑏𝑏𝑛𝑛 = √𝑎𝑎

𝑛𝑛

√𝑏𝑏𝑛𝑛 ; 25 = 32 ; 35 = 243

5 2nd F √𝑥𝑥 243 = 3

7. � 𝟓𝟓𝟒𝟒𝒂𝒂𝟏𝟏

𝟏𝟏𝒂𝒂𝟒𝟒= � 9𝑎𝑎8

𝑎𝑎4 = √32𝒂𝒂𝟒𝟒 𝑎𝑎𝑚𝑚


= �32(𝒂𝒂𝟐𝟐)𝟐𝟐 = √32 ∙ �(𝑎𝑎2)2 = 3a2√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎 𝑛𝑛 𝑛𝑛

= 𝑎𝑎

8.�𝟏𝟏𝟏𝟏𝒂𝒂𝟓𝟓𝒃𝒃𝟏𝟏𝟑𝟑

�𝟑𝟑𝒂𝒂𝟐𝟐𝒃𝒃𝟐𝟐𝟑𝟑 = �81𝑎𝑎5𝑏𝑏8

3𝑎𝑎2𝑏𝑏23

= √27𝑎𝑎3𝑎𝑎6 3 = √33𝑎𝑎3𝑎𝑎6 3 √𝑎𝑎𝑛𝑛


𝑛𝑛 ; 𝑎𝑎𝑚𝑚


= √333 √𝑎𝑎33 √𝑎𝑎63 = 3a𝑎𝑎63 = 3a𝒃𝒃𝟐𝟐 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; √𝑎𝑎 𝑛𝑛 𝑛𝑛

= 𝑎𝑎; √𝑎𝑎 𝑛𝑛 m = 𝑎𝑎

𝑚𝑚𝑛𝑛

9.�𝒂𝒂𝟐𝟐𝒃𝒃𝟒𝟒𝟓𝟓

√𝒂𝒂𝒃𝒃𝟑𝟑 = (𝑎𝑎2𝑏𝑏4)15

(𝑎𝑎𝑏𝑏)13

= 𝑎𝑎25 𝑏𝑏

45

𝑎𝑎13 𝑏𝑏

13

√𝑎𝑎𝑛𝑛 = 𝑎𝑎1𝑛𝑛 ; (am ∙bn) p = amp bnp

= �𝑎𝑎25 – 13� �𝑎𝑎

45 − 13� = �𝑎𝑎

615 – 515� �𝑎𝑎

1215 − 515� 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛 ; LCD = 15

= 𝑎𝑎115 𝑎𝑎

715 = (𝑎𝑎1 𝑎𝑎7)

115 = √𝒂𝒂𝒃𝒃𝟕𝟕𝟏𝟏𝟓𝟓

𝑎𝑎1𝑛𝑛 = √𝑎𝑎𝑛𝑛

9

1n = 2

n = 2

Page 7-9





7-4 OPERATIONS WITH RADICALS

Adding and Subtracting Radicals

• Add and subtract radical expressions by combining the like radicals (or like terms) .

• Like radicals are radicals with exactly the same index (n) and radicand (a) . √𝑎𝑎𝑛𝑛

Example

Like Radicals 4 √𝟓𝟓𝟓𝟓 𝟑𝟑 and -6 √𝟓𝟓𝟓𝟓 𝟑𝟑 The same index (3) and radicand (5a)

Unlike Radicals 7x √6𝑥𝑥 𝟑𝟑 and 5x √6𝑥𝑥 𝟕𝟕 The same radicand but different index (3 and 7)

Tips: Combine expressions: 7ab – 2ab + 3ab = ab (7 – 2 + 3) = 8ab

Combine radicals: 7√𝟓𝟓𝒂𝒂 – 2√𝟓𝟓𝒂𝒂 + 3√𝟓𝟓𝒂𝒂 = √𝟓𝟓𝒂𝒂 (7 – 2 + 3) = 8√𝑎𝑎𝑎𝑎


1. 5√𝟑𝟑 + 4√𝟑𝟑 – 2√𝟑𝟑 = √𝟑𝟑 (5 + 4 – 2) = 7√𝟑𝟑 Factor out √3.

2. 3√𝟐𝟐𝟓𝟓𝟐𝟐 + √𝟐𝟐𝟔𝟔 = 3√64 ∙ 4 + √64 Rewrite: 256 = 64 ∙ 4

= 3√𝟐𝟐𝟔𝟔√4 + √𝟐𝟐𝟔𝟔 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛

= √𝟐𝟐𝟔𝟔 (3√4 + 1) Factor out √64.

= √82 (3∙2 + 1) = 8 ∙ 7 = 56 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

3. 9√𝟑𝟑𝟓𝟓𝒙𝒙𝟑𝟑 − 𝟐𝟐√𝟕𝟕𝒙𝒙𝟑𝟑 = 9√5 ∙ 7𝑥𝑥3 – 2√7𝑥𝑥3 Rewrite: 35 = 5∙7

= 9√5 √𝟕𝟕𝒙𝒙𝟑𝟑 − 2√𝟕𝟕𝒙𝒙𝟑𝟑 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛

= √7𝑥𝑥3 (9√5 − 2) Factor out √7𝑥𝑥3.

= √7𝑥𝑥 ∙ 𝑥𝑥2 (9√5 − 2) am an = am + n

= 𝒙𝒙√𝟕𝟕𝒙𝒙 (𝟗𝟗√𝟓𝟓 − 𝟐𝟐) √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

4. 2 �𝟏𝟏𝟐𝟐𝟏𝟏 𝟑𝟑 – 3 �𝟓𝟓𝟔𝟔𝟏𝟏𝟔𝟔 𝟑𝟑 = 2 �2 ∙ 8𝑦𝑦 3 − 3 �2 ∙ 27𝑦𝑦 ∙ 𝑦𝑦3 3Rewrite: 𝑦𝑦4 = 𝑦𝑦1𝑦𝑦3; 16 = 2∙8 ; 54 = 2∙27

= 2 �2𝑦𝑦 3 √23 3 − 3 �2𝑦𝑦 3 √33 3 �𝑦𝑦3 3√𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛 ; regroup

= 2 �2𝑦𝑦 3 ∙ 2 − 3 ∙ �2𝑦𝑦 3 ∙ 3 ∙ 𝑦𝑦 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

= 4 �𝟐𝟐𝟏𝟏 𝟑𝟑 − 9 𝑦𝑦 �𝟐𝟐𝟏𝟏 𝟑𝟑

= �𝟐𝟐𝟏𝟏 𝟑𝟑 (𝟔𝟔 − 𝟗𝟗𝟏𝟏) Factor out �2𝑦𝑦 3.

Page 7-10





Example






1. 5√𝟑𝟑 + 4√𝟑𝟑 – 2√𝟑𝟑 = √𝟑𝟑 (5 + 4 – 2) = 7√𝟑𝟑 Factor out √3.

2. 3√𝟐𝟐𝟓𝟓𝟐𝟐 + √𝟐𝟐𝟔𝟔 = 3√64 ∙ 4 + √64 Rewrite: 256 = 64 ∙ 4

= 3√𝟐𝟐𝟔𝟔√4 + √𝟐𝟐𝟔𝟔 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛

= √𝟐𝟐𝟔𝟔 (3√4 + 1) Factor out √64.

= √82 (3∙2 + 1) = 8 ∙ 7 = 56 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎



= √7𝑥𝑥3 (9√5 − 2) Factor out √7𝑥𝑥3.

= √7𝑥𝑥 ∙ 𝑥𝑥2 (9√5 − 2) am an = am + n

= 𝒙𝒙√𝟕𝟕𝒙𝒙 (𝟗𝟗√𝟓𝟓 − 𝟐𝟐) √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎



= 2 �2𝑦𝑦 3 ∙ 2 − 3 ∙ �2𝑦𝑦 3 ∙ 3 ∙ 𝑦𝑦 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

= 4 �𝟐𝟐𝟏𝟏 𝟑𝟑 − 9 𝑦𝑦 �𝟐𝟐𝟏𝟏 𝟑𝟑

= �𝟐𝟐𝟏𝟏 𝟑𝟑 (𝟔𝟔 − 𝟗𝟗𝟏𝟏) Factor out �2𝑦𝑦 3.

Page 7-10





Example






1. 5√𝟑𝟑 + 4√𝟑𝟑 – 2√𝟑𝟑 = √𝟑𝟑 (5 + 4 – 2) = 7√𝟑𝟑 Factor out √3.

2. 3√𝟐𝟐𝟓𝟓𝟐𝟐 + √𝟐𝟐𝟔𝟔 = 3√64 ∙ 4 + √64 Rewrite: 256 = 64 ∙ 4

= 3√𝟐𝟐𝟔𝟔√4 + √𝟐𝟐𝟔𝟔 √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑎𝑎𝑛𝑛

= √𝟐𝟐𝟔𝟔 (3√4 + 1) Factor out √64.

= √82 (3∙2 + 1) = 8 ∙ 7 = 56 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎



= √7𝑥𝑥3 (9√5 − 2) Factor out √7𝑥𝑥3.

= √7𝑥𝑥 ∙ 𝑥𝑥2 (9√5 − 2) am an = am + n

= 𝒙𝒙√𝟕𝟕𝒙𝒙 (𝟗𝟗√𝟓𝟓 − 𝟐𝟐) √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎



= 2 �2𝑦𝑦 3 ∙ 2 − 3 ∙ �2𝑦𝑦 3 ∙ 3 ∙ 𝑦𝑦 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

= 4 �𝟐𝟐𝟏𝟏 𝟑𝟑 − 9 𝑦𝑦 �𝟐𝟐𝟏𝟏 𝟑𝟑

= �𝟐𝟐𝟏𝟏 𝟑𝟑 (𝟔𝟔 − 𝟗𝟗𝟏𝟏) Factor out �2𝑦𝑦 3.

Page 7-10





Multiplying Radicals

• Multiplying radical expressions is based on the product rule and distributive property . Multiply

• Review: Product rule: √𝑎𝑎 𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏 𝑛𝑛Note: √𝑎𝑎 𝑛𝑛 √𝑏𝑏 𝑛𝑛 = √𝑎𝑎𝑏𝑏 𝑛𝑛

Distributive property: a (b + c) = a b + a c Simplify


1. 2√𝟑𝟑 (3√𝟐𝟐 +√𝟑𝟑) = 2√3 ∙ 3√2 + 2√3√3 = 6√3 ∙ 2 + 2√32 Distributive property

= 6√6 + 2 ∙ 3 = 𝟔𝟔(√𝟔𝟔 + 1) √𝑎𝑎𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

2. √𝟐𝟐 𝟒𝟒 �√𝟖𝟖𝟒𝟒 − 𝟑𝟑 √𝟑𝟑 𝟒𝟒 � = √2 4 √84 − 3√24 √3 4 = √164 − 3√64Distribute, √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= √244 − 3√64 = 𝟐𝟐 − 𝟑𝟑√𝟔𝟔𝟒𝟒√𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

3. 𝟐𝟐 √𝒙𝒙 𝟑𝟑 �√𝒙𝒙𝟐𝟐𝟑𝟑 + �𝒙𝒙𝟐𝟐𝒚𝒚𝟑𝟑 𝟑𝟑 − 𝟑𝟑√𝒙𝒙𝟐𝟐𝟑𝟑 � = 2√𝑥𝑥3 √𝑥𝑥23 + 2√𝑥𝑥3 �𝑥𝑥2𝑦𝑦3 3 + 2 √𝑥𝑥 3 (-3 √𝑥𝑥2 3 ) Distribute

= 2√𝑥𝑥 𝑥𝑥23 + 2 �𝑥𝑥 𝑥𝑥2𝑦𝑦3 3 − 6 √𝑥𝑥 𝑥𝑥2 3√𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= 2√𝑥𝑥3𝟑𝟑 + 2 �𝑥𝑥3𝑦𝑦3 3 − 6√𝑥𝑥3𝟑𝟑am an = am + n

= 2𝑥𝑥 + 2 √𝑥𝑥3 3 �𝑦𝑦3 3 − 6𝑥𝑥 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

= 2𝑥𝑥𝑦𝑦 − 4𝑥𝑥 = 𝟐𝟐𝒙𝒙 (𝒚𝒚 − 𝟐𝟐) Factor out 2x.

4. (√𝟑𝟑 − √𝟐𝟐) (3√𝟐𝟐 +√𝟑𝟑) = 3√3 √2 + √3√3 − 3√2√2 − √2√3 FOIL

= 3√6 + √32 − 3√22 − √6 √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

= 3√𝟔𝟔 + 3 − 3 ∙ 2 − √𝟔𝟔 = 2√𝟔𝟔 − 𝟑𝟑 Combine like radicals .

5. (√𝟐𝟐 − √𝟓𝟓)2 = √22− 2√2√5 + √5

2= 2 − 2√10 + 5 = 7 − 𝟐𝟐√𝟏𝟏𝟏𝟏 (𝑎𝑎 − 𝑏𝑏)2 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2

• Conjugates are two binomials (2 terms) whose only difference is the sign of one term . (Switch the middle sign of a pair of binomials, then conjugate to (a + b) is (a – b) .)

Conjugates Switch the middle sign3 − 4 3 + 4

2x + √5 2x − √5√𝑎𝑎 − √𝑏𝑏 √𝑎𝑎 + √𝑏𝑏

5 + 7i 5 − 7i


1. (√5 + √3) (√5 − √3) = √52− √3

2= 5 – 3 = 2 (𝑎𝑎 + 𝑏𝑏) (𝑎𝑎 – 𝑏𝑏) = 𝑎𝑎2– 𝑏𝑏2

or = √52− √5√3 + √3√5 −√3

2= √5

2− √3

2FOIL

2. (√𝑎𝑎 − √3) (√𝑎𝑎 + √3) = √𝑎𝑎2− √3

2= a − 3 (𝑎𝑎 + 𝑏𝑏) (𝑎𝑎 – 𝑏𝑏) = 𝑎𝑎2– 𝑏𝑏2

Tip: The radicals will disappear if a pair of conjugates are mulitplied .

Example:

Page 7-11

Multiplying Radicals

• Multiplying radical expressions is based on the product rule and distributive property . Multiply

• Review: Product rule: √𝑎𝑎 𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏 𝑛𝑛Note: √𝑎𝑎 𝑛𝑛 √𝑏𝑏 𝑛𝑛 = √𝑎𝑎𝑏𝑏 𝑛𝑛

Distributive property: a (b + c) = a b + a c Simplify


1. 2√𝟑𝟑 (3√𝟐𝟐 +√𝟑𝟑) = 2√3 ∙ 3√2 + 2√3√3 = 6√3 ∙ 2 + 2√32 Distributive property

= 6√6 + 2 ∙ 3 = 𝟔𝟔(√𝟔𝟔 + 1) √𝑎𝑎𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

2. √𝟐𝟐 𝟒𝟒 �√𝟖𝟖𝟒𝟒 − 𝟑𝟑 √𝟑𝟑 𝟒𝟒 � = √2 4 √84 − 3√24 √3 4 = √164 − 3√64Distribute, √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= √244 − 3√64 = 𝟐𝟐 − 𝟑𝟑√𝟔𝟔𝟒𝟒√𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

3. 𝟐𝟐 √𝒙𝒙 𝟑𝟑 �√𝒙𝒙𝟐𝟐𝟑𝟑 + �𝒙𝒙𝟐𝟐𝒚𝒚𝟑𝟑 𝟑𝟑 − 𝟑𝟑√𝒙𝒙𝟐𝟐𝟑𝟑 � = 2√𝑥𝑥3 √𝑥𝑥23 + 2√𝑥𝑥3 �𝑥𝑥2𝑦𝑦3 3 + 2 √𝑥𝑥 3 (-3 √𝑥𝑥2 3 ) Distribute

= 2√𝑥𝑥 𝑥𝑥23 + 2 �𝑥𝑥 𝑥𝑥2𝑦𝑦3 3 − 6 √𝑥𝑥 𝑥𝑥2 3√𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= 2√𝑥𝑥3𝟑𝟑 + 2 �𝑥𝑥3𝑦𝑦3 3 − 6√𝑥𝑥3𝟑𝟑am an = am + n

= 2𝑥𝑥 + 2 √𝑥𝑥3 3 �𝑦𝑦3 3 − 6𝑥𝑥 √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

= 2𝑥𝑥𝑦𝑦 − 4𝑥𝑥 = 𝟐𝟐𝒙𝒙 (𝒚𝒚 − 𝟐𝟐) Factor out 2x.

4. (√𝟑𝟑 − √𝟐𝟐) (3√𝟐𝟐 +√𝟑𝟑) = 3√3 √2 + √3√3 − 3√2√2 − √2√3 FOIL

= 3√6 + √32 − 3√22 − √6 √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎

= 3√𝟔𝟔 + 3 − 3 ∙ 2 − √𝟔𝟔 = 2√𝟔𝟔 − 𝟑𝟑 Combine like radicals .

5. (√𝟐𝟐 − √𝟓𝟓)2 = √22− 2√2√5 + √5

2= 2 − 2√10 + 5 = 7 − 𝟐𝟐√𝟏𝟏𝟏𝟏 (𝑎𝑎 − 𝑏𝑏)2 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2


Conjugates Switch the middle sign3 − 4 3 + 4

2x + √5 2x − √5√𝑎𝑎 − √𝑏𝑏 √𝑎𝑎 + √𝑏𝑏

5 + 7i 5 − 7i


1. (√5 + √3) (√5 − √3) = √52− √3

2= 5 – 3 = 2 (𝑎𝑎 + 𝑏𝑏) (𝑎𝑎 – 𝑏𝑏) = 𝑎𝑎2– 𝑏𝑏2

or = √52− √5√3 + √3√5 −√3

2= √5

2− √3

2FOIL

2. (√𝑎𝑎 − √3) (√𝑎𝑎 + √3) = √𝑎𝑎2− √3

2= a − 3 (𝑎𝑎 + 𝑏𝑏) (𝑎𝑎 – 𝑏𝑏) = 𝑎𝑎2– 𝑏𝑏2

Tip: The radicals will disappear if a pair of conjugates are mulitplied .

Example:

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7-5 DIVIDING RADICALS

Rationalizing Denominators

• Rationalize the denominator by getting rid of the radicals in the denominator to satisfy the

simplest condition – no radical appears in the denominator .

• Rationalize a monomial by multiplying both denominator (bottom) and numerator (top) by

the root in the denominator .

In General Example

𝑎𝑎√𝑏𝑏

= 𝑎𝑎√𝒃𝒃√𝑏𝑏√𝒃𝒃

= 𝑎𝑎√𝑏𝑏√𝑏𝑏2

= 𝑎𝑎√𝑏𝑏𝑏𝑏

Multiply by √𝑏𝑏 to get a perfect square. 𝟐𝟐√𝟔𝟔

= 2√𝟔𝟔√6√𝟔𝟔

= 2√6√62

= 2√66

= √𝟔𝟔𝟑𝟑

√𝑏𝑏2

= √𝑏𝑏2 2 = 𝑏𝑏

Example: � 𝟑𝟑𝒂𝒂𝟑𝟑

𝟕𝟕𝒂𝒂𝟖𝟖=� 3𝑎𝑎3

7𝑎𝑎3𝑎𝑎5=� 3

7𝑎𝑎5 am an = a m + n ; √𝑎𝑎𝑛𝑛


𝑛𝑛

= √3�7𝑎𝑎5

= √3�7𝑎𝑎5

�𝟕𝟕𝒂𝒂𝟓𝟓

�𝟕𝟕𝒂𝒂𝟓𝟓 =

�3∙7𝑎𝑎5

�(7𝑎𝑎5)(7𝑎𝑎5) Multiply by √7𝑎𝑎5 ; √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= √21𝑎𝑎∙𝑎𝑎4

�(7𝑎𝑎5)2 = √21𝑎𝑎�(𝑎𝑎2)2

�(7𝑎𝑎5)2 = 𝑎𝑎2√21𝑎𝑎

7𝑎𝑎5 = √𝟐𝟐𝟐𝟐𝒂𝒂

𝟕𝟕𝒂𝒂𝟑𝟑 am an = a m + n ; anm = (an)m ; √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

• Rationalize a binomial (two terms) in denominator by multiplying both denominator and

numerator by the conjugate of the denominator .

In General Example

1. 𝒂𝒂√𝒃𝒃 +√𝒄𝒄

= 𝑎𝑎�√𝒃𝒃 –√𝒄𝒄��√𝑏𝑏 +√𝑐𝑐��√𝒃𝒃 –√𝒄𝒄�

Multiply by �√𝑏𝑏 –√𝑐𝑐�. 1. 𝟒𝟒√𝟔𝟔+√𝟐𝟐

= 4�√𝟔𝟔 −√𝟐𝟐��√6 +√2��√𝟔𝟔 −√𝟐𝟐�

Multiply by �√6 –√2�.

= 𝑎𝑎(√𝑏𝑏−√𝑐𝑐)(√𝑏𝑏)2 −(√𝑐𝑐)2

(a + b)(a – b) = a2 – b2 : = 4(√6−√2)(√6)2 −(√2)2

(a + b)(a – b) = a2 – b2 :

= 𝑎𝑎(√𝑏𝑏−√𝑐𝑐)𝑏𝑏−𝑐𝑐

= 4�√6−√2�6−2

= √𝟔𝟔 − √𝟐𝟐

2. 𝒂𝒂√𝒃𝒃−√𝒄𝒄

= 𝑎𝑎(√𝒃𝒃+√𝒄𝒄)(√𝑏𝑏 −√𝑐𝑐)(√𝒃𝒃+√𝒄𝒄)

Multiply by �√𝑏𝑏 + √𝑐𝑐�. 2. 𝟓𝟓√𝟐𝟐𝟏𝟏−√𝟓𝟓

= 5(√10+√5)(√10 −√5)(√10+√5)

Multiply by �√10 + √5�.

= 𝑎𝑎(√𝑏𝑏+√𝑐𝑐)(√𝑏𝑏)2−(√𝑐𝑐)2

a2 – b2 = (a + b)(a – b) : = 5(√10+√5)(√10)2−(√5)2

(a + b)(a – b) = a2 – b2:

= 𝑎𝑎(√𝑏𝑏+√𝑐𝑐)𝑏𝑏−𝑐𝑐

= 5(√10+√5)10−5

= √𝟐𝟐𝟏𝟏 + √𝟓𝟓

1

3

𝑎𝑎 = √𝑏𝑏 , 𝑏𝑏 = √𝑐𝑐√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎 √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

𝑎𝑎 = √6 , 𝑏𝑏 = √2

𝑎𝑎 = √𝑏𝑏 , 𝑏𝑏 = √𝑐𝑐√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

𝑎𝑎 = √10 , 𝑏𝑏 = √5√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

Page 7-12

7-5 DIVIDING RADICALS

Rationalizing Denominators


simplest condition – no radical appears in the denominator .

• Rationalize a monomial by multiplying both denominator (bottom) and numerator (top) by

the root in the denominator .

In General Example

𝑎𝑎√𝑏𝑏

= 𝑎𝑎√𝒃𝒃√𝑏𝑏√𝒃𝒃

= 𝑎𝑎√𝑏𝑏√𝑏𝑏2

= 𝑎𝑎√𝑏𝑏𝑏𝑏

Multiply by √𝑏𝑏 to get a perfect square. 𝟐𝟐√𝟔𝟔

= 2√𝟔𝟔√6√𝟔𝟔

= 2√6√62

= 2√66

= √𝟔𝟔𝟑𝟑

√𝑏𝑏2

= √𝑏𝑏2 2 = 𝑏𝑏

Example: � 𝟑𝟑𝒂𝒂𝟑𝟑

𝟕𝟕𝒂𝒂𝟖𝟖=� 3𝑎𝑎3

7𝑎𝑎3𝑎𝑎5=� 3

7𝑎𝑎5 am an = a m + n ; √𝑎𝑎𝑛𝑛


𝑛𝑛

= √3�7𝑎𝑎5

= √3�7𝑎𝑎5

�𝟕𝟕𝒂𝒂𝟓𝟓

�𝟕𝟕𝒂𝒂𝟓𝟓 =

�3∙7𝑎𝑎5

�(7𝑎𝑎5)(7𝑎𝑎5) Multiply by √7𝑎𝑎5 ; √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= √21𝑎𝑎∙𝑎𝑎4

�(7𝑎𝑎5)2 = √21𝑎𝑎�(𝑎𝑎2)2

�(7𝑎𝑎5)2 = 𝑎𝑎2√21𝑎𝑎

7𝑎𝑎5 = √𝟐𝟐𝟐𝟐𝒂𝒂

𝟕𝟕𝒂𝒂𝟑𝟑 am an = a m + n ; anm = (an)m ; √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

• Rationalize a binomial (two terms) in denominator by multiplying both denominator and

numerator by the conjugate of the denominator .

In General Example

1. 𝒂𝒂√𝒃𝒃 +√𝒄𝒄

= 𝑎𝑎�√𝒃𝒃 –√𝒄𝒄��√𝑏𝑏 +√𝑐𝑐��√𝒃𝒃 –√𝒄𝒄�

Multiply by �√𝑏𝑏 –√𝑐𝑐�. 1. 𝟒𝟒√𝟔𝟔+√𝟐𝟐

= 4�√𝟔𝟔 −√𝟐𝟐��√6 +√2��√𝟔𝟔 −√𝟐𝟐�

Multiply by �√6 –√2�.

= 𝑎𝑎(√𝑏𝑏−√𝑐𝑐)(√𝑏𝑏)2 −(√𝑐𝑐)2

(a + b)(a – b) = a2 – b2 : = 4(√6−√2)(√6)2 −(√2)2

(a + b)(a – b) = a2 – b2 :

= 𝑎𝑎(√𝑏𝑏−√𝑐𝑐)𝑏𝑏−𝑐𝑐

= 4�√6−√2�6−2

= √𝟔𝟔 − √𝟐𝟐

2. 𝒂𝒂√𝒃𝒃−√𝒄𝒄

= 𝑎𝑎(√𝒃𝒃+√𝒄𝒄)(√𝑏𝑏 −√𝑐𝑐)(√𝒃𝒃+√𝒄𝒄)

Multiply by �√𝑏𝑏 + √𝑐𝑐�. 2. 𝟓𝟓√𝟐𝟐𝟏𝟏−√𝟓𝟓

= 5(√10+√5)(√10 −√5)(√10+√5)

Multiply by �√10 + √5�.

= 𝑎𝑎(√𝑏𝑏+√𝑐𝑐)(√𝑏𝑏)2−(√𝑐𝑐)2

a2 – b2 = (a + b)(a – b) : = 5(√10+√5)(√10)2−(√5)2

(a + b)(a – b) = a2 – b2:

= 𝑎𝑎(√𝑏𝑏+√𝑐𝑐)𝑏𝑏−𝑐𝑐

= 5(√10+√5)10−5

= √𝟐𝟐𝟏𝟏 + √𝟓𝟓

1

3

𝑎𝑎 = √𝑏𝑏 , 𝑏𝑏 = √𝑐𝑐√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎 √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

𝑎𝑎 = √6 , 𝑏𝑏 = √2

𝑎𝑎 = √𝑏𝑏 , 𝑏𝑏 = √𝑐𝑐√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

𝑎𝑎 = √10 , 𝑏𝑏 = √5√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

Page 7-12





Dividing Radicals

• Dividing radical expressions is based on the quotient rule . Dividing

• Recall quotient rule: Note:

Simplify

• Tip: Multiply both denominator and numerator by a radical to get a perfect square, perfect

cube, perfect 4th power, perfect 5th power, etc . in the denominator .

Example: Perform the indicated operations (rationalize each denominator) .

1. �𝟓𝟓𝟑𝟑

= √5√3

= √5√𝟑𝟑√3√𝟑𝟑

�𝑎𝑎𝑏𝑏


√𝑏𝑏𝑛𝑛 ;

= √5 ∙ 3

√32 = √𝟏𝟏𝟓𝟓

𝟑𝟑 √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

2. √𝟐𝟐𝟐𝟐 𝟑𝟑

�𝒚𝒚 𝟑𝟑 = √2𝑥𝑥 3 �𝒚𝒚𝟐𝟐 𝟑𝟑

�𝑦𝑦1 3 �𝒚𝒚𝟐𝟐 𝟑𝟑 = �2𝑥𝑥𝑦𝑦2 3

�𝑦𝑦1 𝑦𝑦2 3

= �2𝑥𝑥𝑦𝑦2 3

�𝑦𝑦1+2 3 = �2𝑥𝑥𝑦𝑦2 3

�𝑦𝑦3 3 = �𝟐𝟐𝟐𝟐𝒚𝒚𝟐𝟐 𝟑𝟑

𝒚𝒚am an = am + n ; √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

3. 𝟑𝟑𝟐𝟐√𝟐𝟐 − √𝟐𝟐

= 3𝑥𝑥 �√𝟐𝟐 + √𝟐𝟐��√2 – √𝑥𝑥��√𝟐𝟐 + √𝟐𝟐�

Multiply by √2 + √𝑥𝑥.

= 3𝑥𝑥 �√2 + √𝑥𝑥�

�√2�2− �√𝑥𝑥�

2 Distribute; (a + b)(a – b) = a2 – b2

= 𝟑𝟑𝟐𝟐√𝟐𝟐 + 𝟑𝟑𝟐𝟐√𝟐𝟐𝟐𝟐 − 𝟐𝟐

√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

4. 𝟐𝟐√𝟑𝟑 + √𝟐𝟐√𝟐𝟐 − √𝟑𝟑

= �2√3 + √2��√𝟐𝟐 +√𝟑𝟑��√2 − √3��√𝟐𝟐 + √𝟑𝟑�

Multiply by √2 + √3.

= 2√3√2 + 2√3√3 + √2√2 + √2√3

�√2�2− �√3�

2 FOIL; (a + b)(a – b) = a2 – b2

= 2√3∙2 + 2√3∙3 + √2∙2 + √2∙32 − 3

= 2√𝟔𝟔 + 6 + 2 + √𝟔𝟔2 − 3

√𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= 3√𝟔𝟔 + 8-1

= -(𝟑𝟑√𝟔𝟔 + 𝟖𝟖) Combine like radicals .

5. 𝟑𝟑𝟑𝟑�𝟐𝟐𝟑𝟑𝟑𝟑𝒃𝒃𝟐𝟐

𝟒𝟒 = 3𝑎𝑎 �𝟐𝟐𝟑𝟑𝟑𝟑𝟏𝟏𝒃𝒃𝟐𝟐 𝟒𝟒

�𝟐𝟐𝟑𝟑𝟑𝟑𝒃𝒃𝟐𝟐 𝟒𝟒 �𝟐𝟐𝟑𝟑𝟑𝟑𝟏𝟏𝒃𝒃𝟐𝟐

𝟒𝟒 Multiply by √23𝑎𝑎1𝑏𝑏2 4 to get a perfect 4th power .

= 3𝑎𝑎 √23𝑎𝑎1𝑏𝑏2 4

�(2∙23)(𝑎𝑎3𝑎𝑎1)(𝑏𝑏2𝑏𝑏2) 4 = 3𝑎𝑎 √23𝑎𝑎𝑏𝑏2 4

√24𝑎𝑎4𝑏𝑏4 4 am an = am + n

= 3𝑎𝑎 √8𝑎𝑎𝑏𝑏2 4

√24 4 √𝑎𝑎4 4 √𝑏𝑏4 4 = 3𝑎𝑎 √8𝑎𝑎𝑏𝑏2 4

2𝑎𝑎𝑏𝑏 = 𝟑𝟑 �𝟖𝟖𝟑𝟑𝒃𝒃𝟐𝟐

𝟒𝟒

𝟐𝟐𝒃𝒃 √𝑎𝑎𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

Multiply by �𝑦𝑦2 3 to get a perfect cube; √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

√𝑎𝑎𝑛𝑛


𝑛𝑛 √𝑎𝑎𝑛𝑛


𝑛𝑛

multiply by √3 to get a perfect square .

Page 7-13

Dividing Radicals

• Dividing radical expressions is based on the quotient rule . Dividing

• Recall quotient rule: Note:

Simplify

• Tip: Multiply both denominator and numerator by a radical to get a perfect square, perfect

cube, perfect 4th power, perfect 5th power, etc . in the denominator .

Example: Perform the indicated operations (rationalize each denominator) .

1. �𝟓𝟓𝟑𝟑

= √5√3

= √5√𝟑𝟑√3√𝟑𝟑

�𝑎𝑎𝑏𝑏


√𝑏𝑏𝑛𝑛 ;

= √5 ∙ 3

√32 = √𝟏𝟏𝟓𝟓

𝟑𝟑 √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

2. √𝟐𝟐𝟐𝟐 𝟑𝟑

�𝒚𝒚 𝟑𝟑 = √2𝑥𝑥 3 �𝒚𝒚𝟐𝟐 𝟑𝟑

�𝑦𝑦1 3 �𝒚𝒚𝟐𝟐 𝟑𝟑 = �2𝑥𝑥𝑦𝑦2 3

�𝑦𝑦1 𝑦𝑦2 3

= �2𝑥𝑥𝑦𝑦2 3

�𝑦𝑦1+2 3 = �2𝑥𝑥𝑦𝑦2 3

�𝑦𝑦3 3 = �𝟐𝟐𝟐𝟐𝒚𝒚𝟐𝟐 𝟑𝟑

𝒚𝒚am an = am + n ; √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

3. 𝟑𝟑𝟐𝟐√𝟐𝟐 − √𝟐𝟐

= 3𝑥𝑥 �√𝟐𝟐 + √𝟐𝟐��√2 – √𝑥𝑥��√𝟐𝟐 + √𝟐𝟐�

Multiply by √2 + √𝑥𝑥.

= 3𝑥𝑥 �√2 + √𝑥𝑥�

�√2�2− �√𝑥𝑥�

2 Distribute; (a + b)(a – b) = a2 – b2

= 𝟑𝟑𝟐𝟐√𝟐𝟐 + 𝟑𝟑𝟐𝟐√𝟐𝟐𝟐𝟐 − 𝟐𝟐

√𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

4. 𝟐𝟐√𝟑𝟑 + √𝟐𝟐√𝟐𝟐 − √𝟑𝟑

= �2√3 + √2��√𝟐𝟐 +√𝟑𝟑��√2 − √3��√𝟐𝟐 + √𝟑𝟑�

Multiply by √2 + √3.

= 2√3√2 + 2√3√3 + √2√2 + √2√3

�√2�2− �√3�

2 FOIL; (a + b)(a – b) = a2 – b2

= 2√3∙2 + 2√3∙3 + √2∙2 + √2∙32 − 3

= 2√𝟔𝟔 + 6 + 2 + √𝟔𝟔2 − 3

√𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

= 3√𝟔𝟔 + 8-1

= -(𝟑𝟑√𝟔𝟔 + 𝟖𝟖) Combine like radicals .

5. 𝟑𝟑𝟑𝟑�𝟐𝟐𝟑𝟑𝟑𝟑𝒃𝒃𝟐𝟐

𝟒𝟒 = 3𝑎𝑎 �𝟐𝟐𝟑𝟑𝟑𝟑𝟏𝟏𝒃𝒃𝟐𝟐 𝟒𝟒

�𝟐𝟐𝟑𝟑𝟑𝟑𝒃𝒃𝟐𝟐 𝟒𝟒 �𝟐𝟐𝟑𝟑𝟑𝟑𝟏𝟏𝒃𝒃𝟐𝟐

𝟒𝟒 Multiply by √23𝑎𝑎1𝑏𝑏2 4 to get a perfect 4th power .

= 3𝑎𝑎 √23𝑎𝑎1𝑏𝑏2 4

�(2∙23)(𝑎𝑎3𝑎𝑎1)(𝑏𝑏2𝑏𝑏2) 4 = 3𝑎𝑎 √23𝑎𝑎𝑏𝑏2 4

√24𝑎𝑎4𝑏𝑏4 4 am an = am + n

= 3𝑎𝑎 √8𝑎𝑎𝑏𝑏2 4

√24 4 √𝑎𝑎4 4 √𝑏𝑏4 4 = 3𝑎𝑎 √8𝑎𝑎𝑏𝑏2 4

2𝑎𝑎𝑏𝑏 = 𝟑𝟑 �𝟖𝟖𝟑𝟑𝒃𝒃𝟐𝟐

𝟒𝟒

𝟐𝟐𝒃𝒃 √𝑎𝑎𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 ; √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

Multiply by �𝑦𝑦2 3 to get a perfect cube; √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛

√𝑎𝑎𝑛𝑛


𝑛𝑛 √𝑎𝑎𝑛𝑛


𝑛𝑛

multiply by √3 to get a perfect square .

Page 7-13





7-6 SOLVING EQUATIONS WITH RADICALS

Square Root Equations

• A square root equation is an equation containing a square root . Example

√𝑥𝑥 − 5 = 3• To solve a square root equation

Steps Example: Solve √𝒙𝒙 − 𝟐𝟐 = 𝟑𝟑 .

- Isolate the square root term (on one side of the √𝑥𝑥 = 3 + 2 Add 2 .

equation). √𝑥𝑥 = 5

- Get rid of the square root by squaring both sides . √𝑥𝑥2

= 52 Square both sides .

- Solve for x : x = 25 √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎, √𝑥𝑥2

= √𝑥𝑥2 2

- Check . ?

√25 − 2 = 3 Replace x by 25 in the equation . ?

√52 − 2 = 3 √

5 – 2 = 3 Correct!

Example: Solve for x .5

√4−3𝑥𝑥= 1

5√4−3𝑥𝑥

∙ √4 − 3𝑥𝑥 = 1 ∙ √4 − 3𝑥𝑥 Multiply by √4 − 3𝑥𝑥.

√4 − 3𝑥𝑥 = 5

√4 − 3𝑥𝑥2

= 52 Square both sides ; √𝑎𝑎𝑛𝑛 𝑛𝑛 = 𝑎𝑎

4 – 3x = 25 Solve for x ; subtract 4 .

-3x = 21 Divide by -3 .

x = -7 ?

Check: 5

�4−3(-𝟕𝟕) = 1 Replace x by -7 in the equation .

?

5

√25 = 1

√55

= 1 Correct!

Page 7-14











= √𝑥𝑥2 2

- Check . ?


√52 − 2 = 3 √

5 – 2 = 3 Correct!


√4−3𝑥𝑥= 1

5√4−3𝑥𝑥


√4 − 3𝑥𝑥 = 5

√4 − 3𝑥𝑥2



-3x = 21 Divide by -3 .

x = -7 ?

Check: 5


?

5

√25 = 1

√55

= 1 Correct!

Page 7-14











= √𝑥𝑥2 2

- Check . ?


√52 − 2 = 3 √

5 – 2 = 3 Correct!


√4−3𝑥𝑥= 1

5√4−3𝑥𝑥


√4 − 3𝑥𝑥 = 5

√4 − 3𝑥𝑥2



-3x = 21 Divide by -3 .

x = -7 ?

Check: 5


?

5

√25 = 1

√55

= 1 Correct!

Page 7-14





Page 7-15

Square Root Equations & Extraneous Solutions

Example: Solve the following equation. √𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 + 𝟕𝟕𝟕𝟕 − 𝟑𝟑𝟑𝟑 = 𝟑𝟑𝟑𝟑

√3𝑥𝑥𝑥𝑥 + 7 = 3 + 𝑥𝑥𝑥𝑥 Isolate the √ term: add x

√3𝑥𝑥𝑥𝑥 + 72

= (3 + 𝑥𝑥𝑥𝑥)2 Square both sides ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛 = 𝑎𝑎𝑎𝑎

3x + 7 = 9 + 6x + x2 (𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 + 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2

x2 + 3x + 2 = 0 Combine like terms: subtract 3x ; subtract 7.

(x + 1)(x + 2) = 0 Factor.

(x + 1) = 0 , (x + 2) = 0 Zero product property

x = -1 or x = -2 Solve for x.

Check: x = -1 x = -2 √3𝑥𝑥𝑥𝑥 + 7− 𝑥𝑥𝑥𝑥 = 3 ? ?

�3(−𝟏𝟏𝟏𝟏) + 7 − �-𝟏𝟏𝟏𝟏� = 3 �3(−𝟐𝟐𝟐𝟐) + 7 − �-𝟐𝟐𝟐𝟐� = 3 Replace x by -1 and -2 in the equation. ? ?

√4 + 1 = 3 √1 + 2 = 3 √ √ 2 + 1 = 3 1 + 2 = 3 Correct!

• The squaring, cubing, etc. process can sometimes create extraneous solutions that do not

satisfy the original equation. So always check solutions. Checking is necessary, not optional.

• An extraneous solution (false solution) is a solution that does not satisfy the original equation.

Example: x = -3 Original equation

x2 = (-3)2 Square both sides.

x2 = 9 Take the square root of both sides. √𝑥𝑥𝑥𝑥2 = ±√9 If x2 = A , then √𝑥𝑥𝑥𝑥2 = ±√𝐴𝐴𝐴𝐴 .

x = ±3 Solutions � 𝑥𝑥𝑥𝑥 = 3 ,

𝟑𝟑𝟑𝟑 = −𝟑𝟑𝟑𝟑 ,

Example: 𝟏𝟏𝟏𝟏 + 𝟑𝟑𝟑𝟑 − √𝟓𝟓𝟓𝟓 − 𝟑𝟑𝟑𝟑 = 𝟎𝟎𝟎𝟎 1 + 𝑥𝑥𝑥𝑥 = √5 − 𝑥𝑥𝑥𝑥 Isolate the √ term: add √5 − 𝑥𝑥𝑥𝑥.

(1 + 𝑥𝑥𝑥𝑥)2 = √5 − 𝑥𝑥𝑥𝑥2 Square both sides.

1+ 2x + x2 = 5 − 𝑥𝑥𝑥𝑥 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛 = 𝑎𝑎𝑎𝑎 , (𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 + 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2 x2 + 3x – 4 = 0 , (x –1) (x + 4) = 0 Factor.

𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏 or 𝑥𝑥𝑥𝑥 = -4 Solve for x (zero product property).

Check: 𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 = -𝟒𝟒𝟒𝟒 1 + 𝑥𝑥𝑥𝑥 − √5 − 𝑥𝑥𝑥𝑥 = 0

? ?

1 + 𝟏𝟏𝟏𝟏 − √5 − 𝟏𝟏𝟏𝟏 = 0 1 + (−𝟒𝟒𝟒𝟒) −�5 − (−𝟒𝟒𝟒𝟒) = 0 Replace x by 1 and -4 in the equation. √

2 – 2 = 0 True -3 – 3 ≠ 0 False (an extraneous solution)

an extraneous solution (discard it, 3 ≠ -3) the solution of the original equation

Page 7-15

Square Root Equations & Extraneous Solutions

Example: Solve the following equation. √𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 + 𝟕𝟕𝟕𝟕 − 𝟑𝟑𝟑𝟑 = 𝟑𝟑𝟑𝟑

√3𝑥𝑥𝑥𝑥 + 7 = 3 + 𝑥𝑥𝑥𝑥 Isolate the √ term: add x

√3𝑥𝑥𝑥𝑥 + 72

= (3 + 𝑥𝑥𝑥𝑥)2 Square both sides ; √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛 = 𝑎𝑎𝑎𝑎

3x + 7 = 9 + 6x + x2 (𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 + 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2

x2 + 3x + 2 = 0 Combine like terms: subtract 3x ; subtract 7.

(x + 1)(x + 2) = 0 Factor.

(x + 1) = 0 , (x + 2) = 0 Zero product property

x = -1 or x = -2 Solve for x.

Check: x = -1 x = -2 √3𝑥𝑥𝑥𝑥 + 7− 𝑥𝑥𝑥𝑥 = 3 ? ?

�3(−𝟏𝟏𝟏𝟏) + 7 − �-𝟏𝟏𝟏𝟏� = 3 �3(−𝟐𝟐𝟐𝟐) + 7 − �-𝟐𝟐𝟐𝟐� = 3 Replace x by -1 and -2 in the equation. ? ?

√4 + 1 = 3 √1 + 2 = 3 √ √ 2 + 1 = 3 1 + 2 = 3 Correct!


satisfy the original equation. So always check solutions. Checking is necessary, not optional.

• An extraneous solution (false solution) is a solution that does not satisfy the original equation.

Example: x = -3 Original equation

x2 = (-3)2 Square both sides.

x2 = 9 Take the square root of both sides. √𝑥𝑥𝑥𝑥2 = ±√9 If x2 = A , then √𝑥𝑥𝑥𝑥2 = ±√𝐴𝐴𝐴𝐴 .

x = ±3 Solutions � 𝑥𝑥𝑥𝑥 = 3 ,

𝟑𝟑𝟑𝟑 = −𝟑𝟑𝟑𝟑 ,

Example: 𝟏𝟏𝟏𝟏 + 𝟑𝟑𝟑𝟑 − √𝟓𝟓𝟓𝟓 − 𝟑𝟑𝟑𝟑 = 𝟎𝟎𝟎𝟎 1 + 𝑥𝑥𝑥𝑥 = √5 − 𝑥𝑥𝑥𝑥 Isolate the √ term: add √5 − 𝑥𝑥𝑥𝑥.

(1 + 𝑥𝑥𝑥𝑥)2 = √5 − 𝑥𝑥𝑥𝑥2 Square both sides.

1+ 2x + x2 = 5 − 𝑥𝑥𝑥𝑥 √𝑎𝑎𝑎𝑎𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛 = 𝑎𝑎𝑎𝑎 , (𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 + 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2 x2 + 3x – 4 = 0 , (x –1) (x + 4) = 0 Factor.

𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏 or 𝑥𝑥𝑥𝑥 = -4 Solve for x (zero product property).

Check: 𝟑𝟑𝟑𝟑 = 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 = -𝟒𝟒𝟒𝟒 1 + 𝑥𝑥𝑥𝑥 − √5 − 𝑥𝑥𝑥𝑥 = 0

? ?

1 + 𝟏𝟏𝟏𝟏 − √5 − 𝟏𝟏𝟏𝟏 = 0 1 + (−𝟒𝟒𝟒𝟒) −�5 − (−𝟒𝟒𝟒𝟒) = 0 Replace x by 1 and -4 in the equation. √

2 – 2 = 0 True -3 – 3 ≠ 0 False (an extraneous solution)

an extraneous solution (discard it, 3 ≠ -3) the solution of the original equation





Radical Equations

• A radical equation is an equation containing radical expressions. Example (Expressions containing radical signs √𝑛𝑛 ) √𝑥𝑥3 − 2𝑥𝑥 = 7

• Solve a radical equation by generalizing the squaring property to other powers .

Tips: - Get rid of a square root by squaring . - Get rid of a cube root by cubing . - Get rid of the 4th root by raising to the 4th power . … … - Get rid of the nth root by raising to the nth power . √a𝑛𝑛 𝑛𝑛

= a

Radical Equation Do Example Solutionsquare root equation squaring √𝑥𝑥 = 2, √𝑥𝑥

2= 22 x = 4

cube root equation cubing √𝑥𝑥3 = 2, √𝑥𝑥3 3= 23 x = 8

4th root equation raising to the 4th power √𝑥𝑥4 = 2, √𝑥𝑥4 4= 24 x = 16


nth root equation raising to the nth power √𝑥𝑥𝑛𝑛 = 2, √𝑥𝑥𝑛𝑛 𝑛𝑛= 2𝑛𝑛 𝑥𝑥 = 2𝑛𝑛

• Procedure to solve a radical equationSteps Example Solve √𝒂𝒂𝟑𝟑 − 𝟏𝟏 = 𝟐𝟐

- Isolate the radical term (on one side of the equation) . √𝑎𝑎3 = 2 + 1 Isolate √a3 .

√𝑎𝑎3 = 3

- Get rid of the radical by raising the power of both √𝑎𝑎3 3= 33 Cube both sides .

sides to n .

- Solve for variable. a = 27 √an n= a

- Check . ?

√𝟐𝟐𝟐𝟐3 − 1 = 2 Replace a by 27. ? √

√333 − 1 = 2 , 3 – 1 = 2 Correct!

Example: 𝟑𝟑 + √𝒂𝒂 + 𝟐𝟐𝟒𝟒 = 𝟓𝟓 Isolate √𝑎𝑎 + 24 .

√𝑎𝑎 + 24 = 2 Raise to the 4th power .

√𝑎𝑎 + 24 4= 24 √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

a + 2 = 16 , a = 14 Solve for a. ?Check: 3 + √14 + 24 = 5

?3 + √164 = 5

? √

3 + √244 = 5 , 3 + 2 = 5 Correct!

Page 7-16

Radical Equations

• A radical equation is an equation containing radical expressions. Example (Expressions containing radical signs √𝑛𝑛 ) √𝑥𝑥3 − 2𝑥𝑥 = 7


Tips: - Get rid of a square root by squaring . - Get rid of a cube root by cubing . - Get rid of the 4th root by raising to the 4th power . … … - Get rid of the nth root by raising to the nth power . √a𝑛𝑛 𝑛𝑛

= a


2= 22 x = 4





• Procedure to solve a radical equationSteps Example Solve √𝒂𝒂𝟑𝟑 − 𝟏𝟏 = 𝟐𝟐

- Isolate the radical term (on one side of the equation) . √𝑎𝑎3 = 2 + 1 Isolate √a3 .

√𝑎𝑎3 = 3

- Get rid of the radical by raising the power of both √𝑎𝑎3 3= 33 Cube both sides .

sides to n .

- Solve for variable. a = 27 √an n= a

- Check . ?

√𝟐𝟐𝟐𝟐3 − 1 = 2 Replace a by 27. ? √

√333 − 1 = 2 , 3 – 1 = 2 Correct!

Example: 𝟑𝟑 + √𝒂𝒂 + 𝟐𝟐𝟒𝟒 = 𝟓𝟓 Isolate √𝑎𝑎 + 24 .

√𝑎𝑎 + 24 = 2 Raise to the 4th power .

√𝑎𝑎 + 24 4= 24 √𝑎𝑎𝑛𝑛 𝑛𝑛

= 𝑎𝑎

a + 2 = 16 , a = 14 Solve for a. ?Check: 3 + √14 + 24 = 5

?3 + √164 = 5

? √

3 + √244 = 5 , 3 + 2 = 5 Correct!

Page 7-16





Page 7-17

Equations With Two Radicals

Equations with two radicals: perform the same steps as equations with one radical term.

Example: Solve the following equations.

√𝟕𝟕𝟕𝟕 + 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 − √𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟏𝟏𝟏𝟏 = 𝟎𝟎𝟎𝟎 Isolate one √ term.

√7 + 2𝑎𝑎𝑎𝑎 = √𝑎𝑎𝑎𝑎 + 15 − 1 Add √𝑎𝑎𝑎𝑎 + 15 ; subtract 1.

√7 + 2𝑎𝑎𝑎𝑎𝟐𝟐𝟐𝟐 = (√𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏)2 Square both sides.

7 + 2𝑎𝑎𝑎𝑎 = √𝑎𝑎𝑎𝑎 + 152 − 2√𝑎𝑎𝑎𝑎 + 15 ∙ 1 + 12 (𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 − 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2:

a = √𝑎𝑎𝑎𝑎 + 15 ; b = 1

7 + 2𝑎𝑎𝑎𝑎 = a + 15 − 2√𝑎𝑎𝑎𝑎 + 15 + 1

7 + 2𝑎𝑎𝑎𝑎 = a + 16 − 2√𝑎𝑎𝑎𝑎 + 15 15 + 1 = 16

2√𝑎𝑎𝑎𝑎 + 15 = 9 − a Isolate √ again: add 2√𝑎𝑎𝑎𝑎 + 15 ; subtract 2a & 7.

22 √𝑎𝑎𝑎𝑎 + 152 = (9 − a)2

Square both sides.

4(a + 15) = 92 − 18a + a2 (𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 − 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2

4a + 60 = 81 − 18a + a2 Combine like terms: subtract 4a & 60.

𝑎𝑎𝑎𝑎2 − 22𝑎𝑎𝑎𝑎 + 21 = 0 Factor.

(a – 1)(a – 21) = 0 , (a – 1) = 0 (a – 21) = 0 Zero product rule

a = 1 a = 21 Solve for a.

Check: √7 + 2𝑎𝑎𝑎𝑎 − √𝑎𝑎𝑎𝑎 + 15 + 1 = 0 Original equation

a = 1 a = 21 ? ?

√7 + 2 ∙ 1 − √1 + 15 + 1 = 0 √7 + 2 ∙ 21 − √21 + 15 + 1 = 0 ? ?

√9 −√16 + 1 = 0 √49 − √36 + 1 = 0 √ ?

3 − 4 + 1 = 0 7 − 6 + 1 = 0 √

−1 + 1 = 0 2 ≠ 0

Solution: a = 1 a = 21 is an extraneous solution.

Write in descending order.

Page 7-17

Equations With Two Radicals

Equations with two radicals: perform the same steps as equations with one radical term.

Example: Solve the following equations.

√𝟕𝟕𝟕𝟕 + 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 − √𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟏𝟏𝟏𝟏 = 𝟎𝟎𝟎𝟎 Isolate one √ term.

√7 + 2𝑎𝑎𝑎𝑎 = √𝑎𝑎𝑎𝑎 + 15 − 1 Add √𝑎𝑎𝑎𝑎 + 15 ; subtract 1.

√7 + 2𝑎𝑎𝑎𝑎𝟐𝟐𝟐𝟐 = (√𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏)2 Square both sides.

7 + 2𝑎𝑎𝑎𝑎 = √𝑎𝑎𝑎𝑎 + 152 − 2√𝑎𝑎𝑎𝑎 + 15 ∙ 1 + 12 (𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 − 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2:

a = √𝑎𝑎𝑎𝑎 + 15 ; b = 1

7 + 2𝑎𝑎𝑎𝑎 = a + 15 − 2√𝑎𝑎𝑎𝑎 + 15 + 1

7 + 2𝑎𝑎𝑎𝑎 = a + 16 − 2√𝑎𝑎𝑎𝑎 + 15 15 + 1 = 16

2√𝑎𝑎𝑎𝑎 + 15 = 9 − a Isolate √ again: add 2√𝑎𝑎𝑎𝑎 + 15 ; subtract 2a & 7.

22 √𝑎𝑎𝑎𝑎 + 152 = (9 − a)2

Square both sides.

4(a + 15) = 92 − 18a + a2 (𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 − 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2

4a + 60 = 81 − 18a + a2 Combine like terms: subtract 4a & 60.

𝑎𝑎𝑎𝑎2 − 22𝑎𝑎𝑎𝑎 + 21 = 0 Factor.

(a – 1)(a – 21) = 0 , (a – 1) = 0 (a – 21) = 0 Zero product rule

a = 1 a = 21 Solve for a.

Check: √7 + 2𝑎𝑎𝑎𝑎 − √𝑎𝑎𝑎𝑎 + 15 + 1 = 0 Original equation

a = 1 a = 21 ? ?

√7 + 2 ∙ 1 − √1 + 15 + 1 = 0 √7 + 2 ∙ 21 − √21 + 15 + 1 = 0 ? ?

√9 −√16 + 1 = 0 √49 − √36 + 1 = 0 √ ?

3 − 4 + 1 = 0 7 − 6 + 1 = 0 √

−1 + 1 = 0 2 ≠ 0

Solution: a = 1 a = 21 is an extraneous solution.

Write in descending order.





7-7 COMPLEX NUMBERS

Complex Number System

• Recall an even root of a negative number is not a real number .Such as 𝑥𝑥 = √−1 is not real, ∵ x2 = -1 has no real solution (no real number squared gives -1.)

• The complex number system is an expanded number system that is larger than the real

number system and includes an even root of a negative number such as �-1 .

• A complex number is an expression of the form A + iB, which is the sum of a real number

A and an imaginary number Bi .Complex Number Example

A – real partA + iB iB – imaginary part

A and B are real numbers

3 + 7i 3 – real part

7i – imaginary part

• Imaginary unit (i): the square root of negative one . Imaginary Unit

𝑖𝑖 = �-1 , i2 = -1 (𝑖𝑖2 = �-12

= -1)

• Examples of complex numbersComplex Number Real Part Imaginary Part

-4 + √3 𝑖𝑖 -4 √3 𝑖𝑖6 − 𝜋𝜋 𝑖𝑖 6 - 𝜋𝜋 𝑖𝑖

√23

−12

𝑖𝑖 √23

-12

𝑖𝑖

-7 i 0 -7 𝑖𝑖Note: Either part can be 0 .

• Extended number system

Complex Numbers (A + Bi) 𝑖𝑖 = �-1

3 + 5i, 4i, 7- …Real Numbers

3, 0, -7, 4/9, π, 5 …Rational Numbers Irrational Numbers6, 4/5 , -4 .5 … 5 , π … .

Page 7-18

7-7 COMPLEX NUMBERS

Complex Number System

• Recall an even root of a negative number is not a real number .Such as 𝑥𝑥 = √−1 is not real, ∵ x2 = -1 has no real solution (no real number squared gives -1.)



• A complex number is an expression of the form A + iB, which is the sum of a real number

A and an imaginary number Bi .Complex Number Example

A – real partA + iB iB – imaginary part

A and B are real numbers




𝑖𝑖 = �-1 , i2 = -1 (𝑖𝑖2 = �-12

= -1)

• Examples of complex numbersComplex Number Real Part Imaginary Part

-4 + √3 𝑖𝑖 -4 √3 𝑖𝑖6 − 𝜋𝜋 𝑖𝑖 6 - 𝜋𝜋 𝑖𝑖

√23

−12

𝑖𝑖 √23

-12

𝑖𝑖

-7 i 0 -7 𝑖𝑖Note: Either part can be 0 .





Page 7-18





Imaginary Unit i

• Powers of imaginary unit i Powers of i

i = �-1 i

i2 = ��-1� 2 = -1 i2=-1i3 = i2 ∙ i = (-1) i = -i i3= -ii4 = i2 ∙ i2 = (-1)(-1) = 1 i4= 1i5 = i4 ∙ i = (1) i i5 = ii6 = i4 ∙ i2 = (1)(-1) = -1 i6= -1i7 = i6 ∙ i1 = (-1) i i7= -ii8 = i4 ∙ i4 = 1 ∙ 1 i8 = 1

Note: i , -1, -i and 1 keep repeating .

Example: Write in terms of i .

1. �-13 = �(-1)(13) = √𝟏𝟏𝟏𝟏 i 𝑖𝑖 = �-1

2. -�-20 = -��-1�(4 ∙ 5)

= -�-1√4√5 = -𝟐𝟐√𝟓𝟓 𝒊𝒊 𝑖𝑖 = �-1


1. -5 i 4 = -5 ∙ 1 = -5 i 4 = 1

2. 3 + 4 ∙ i 3 = 3 + 4 (-i) = 3 – 4i i 3 = - i

3. 7 i 13 = 7 i 5 ∙ i 8 = 7 ∙ i 5 ∙ 1 am an = am+ n ; i 8 = 1

= 7 ∙ i ∙ 1 = 7 i i 5 = i

• A shortcut for powers of imaginary unit iPower of i Example

i n = i R

R = the remainder of n ÷ 4

5i 23 = i 3 = -i 4 23

203

Proof: i 23 = (i4)5 ∙ i3 = 15 ∙ i3 = i3 ( i 4 = 1)


1. i 85 = i 1 = i 85 ÷ 4 = 21 R 1 214 85

8 5 4

2. i 91 = i 3 = -i 91 ÷ 4 = 22 R 3 1 i 3 = -i

… …

One cycle

Another cycle

anm = (an)m ; 1 n = 1

Page 7-19

Imaginary Unit i

• Powers of imaginary unit i Powers of i

i = �-1 i

i2 = ��-1� 2 = -1 i2=-1i3 = i2 ∙ i = (-1) i = -i i3= -ii4 = i2 ∙ i2 = (-1)(-1) = 1 i4= 1i5 = i4 ∙ i = (1) i i5 = ii6 = i4 ∙ i2 = (1)(-1) = -1 i6= -1i7 = i6 ∙ i1 = (-1) i i7= -ii8 = i4 ∙ i4 = 1 ∙ 1 i8 = 1

Note: i , -1, -i and 1 keep repeating .

Example: Write in terms of i .

1. �-13 = �(-1)(13) = √𝟏𝟏𝟏𝟏 i 𝑖𝑖 = �-1

2. -�-20 = -��-1�(4 ∙ 5)

= -�-1√4√5 = -𝟐𝟐√𝟓𝟓 𝒊𝒊 𝑖𝑖 = �-1


1. -5 i 4 = -5 ∙ 1 = -5 i 4 = 1

2. 3 + 4 ∙ i 3 = 3 + 4 (-i) = 3 – 4i i 3 = - i

3. 7 i 13 = 7 i 5 ∙ i 8 = 7 ∙ i 5 ∙ 1 am an = am+ n ; i 8 = 1

= 7 ∙ i ∙ 1 = 7 i i 5 = i

• A shortcut for powers of imaginary unit iPower of i Example

i n = i R


5i 23 = i 3 = -i 4 23

203

Proof: i 23 = (i4)5 ∙ i3 = 15 ∙ i3 = i3 ( i 4 = 1)


1. i 85 = i 1 = i 85 ÷ 4 = 21 R 1 214 85

8 5 4

2. i 91 = i 3 = -i 91 ÷ 4 = 22 R 3 1 i 3 = -i

… …

One cycle

Another cycle

anm = (an)m ; 1 n = 1

Page 7-19





Operations With Complex Numbers

• Add and subtract complex numbers by combining real numbers together and imaginary

numbers together (combine like terms) .

Adding/Subtracting Complex Numbers Example(A + Bi) + (C + Di) = (A + C) + (B + D) i (2 + 5i) + (1 – 3i) = (2 + 1) + (5 – 3) i = 3 + 2i (A + Bi) – (C + Di) = (A – C) + (B – D) i

Or remove parentheses and combine like terms .(7 + 6i) – (2 + 3i) = (7 – 2) + (6 – 3) i = 5 + 3ior (7 + 6i) – (2 + 3i) = 7 + 6i – 2 – 3i = 5 + 3i


1. (-2 – 3i ) + (6 – 7i) = (-2 + 6) + [-3 + (-7)]i = 4 – 10i = 2(2 – 5i)

2. (4 + 9i) – (5 – 6i) = (4 – 5) + [9 – (-6)]i = -1 + 15i

or (4 + 9i) – (5 – 6i) = 4 + 9i – 5 + 6i = -1 + 15i Treat i as a variable .

• Multiply complex numbers using the FOIL method .

Multiplying Complex Numbers Example(A + Bi) (C + Di) = AC + ADi + BCi + BDi2

F O I L = AC + (AD + BC)i + BD(-1)

= (AC – BD) + (AD + BC) i

(4 + 3i) (3 + 7i) = 12 + 28i + 9i + 21i2

F O I L = 12 + 37i – 21 (i 2 = -1) = -9 + 37i


1. (3 – 4i) (4 + 5i) = 12 + 15i – 16i – 20i2FOIL

= 12 – i – 20(-1) i 2 = -1

= 32 – i

2. 4i (3 – 2i) = 12 i – 8 i2 = 12 i – 8(-1) = 8 + 12i Distribute , i 2 = -1

3. 2i – (3 – 2i)2 = 2 i – [32 – 2 ∙ 3 (2i) + (2i)2] (𝑎𝑎 − 𝑏𝑏)2 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2

= 2 i – (9 – 12 i + 4 i2)

= 2 i – 9 + 12 i – 4 i2

= 14 i – 9 – 4(-1) i 2 = -1

= – 5 + 14 i

4. √𝒕𝒕𝟖𝟖𝟖𝟖 ∙ √ 𝟒𝟒 = �( 1)(81) ∙ �( 1)(4) = √ 𝟖𝟖 √92 ∙ √ 𝟖𝟖√22

= (𝒊𝒊 9) ∙ (𝒊𝒊 2) = i2 18 = (-1) 18 = -18 i 2 = -1

(i 2 = -1)

𝑎𝑎 = 3, 𝑏𝑏 = 2𝑖𝑖

Note: √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛 This rule applies only when both a and b are non-negative .

√𝑡𝑡81√𝑡𝑡4 = �(𝑡𝑡81)(𝑡𝑡4) = √92 ∙ 22 = 9 ∙ 2 = 18 Incorrect!

Page 7-20

Operations With Complex Numbers

• Add and subtract complex numbers by combining real numbers together and imaginary

numbers together (combine like terms) .

Adding/Subtracting Complex Numbers Example(A + Bi) + (C + Di) = (A + C) + (B + D) i (2 + 5i) + (1 – 3i) = (2 + 1) + (5 – 3) i = 3 + 2i (A + Bi) – (C + Di) = (A – C) + (B – D) i

Or remove parentheses and combine like terms .(7 + 6i) – (2 + 3i) = (7 – 2) + (6 – 3) i = 5 + 3ior (7 + 6i) – (2 + 3i) = 7 + 6i – 2 – 3i = 5 + 3i


1. (-2 – 3i ) + (6 – 7i) = (-2 + 6) + [-3 + (-7)]i = 4 – 10i = 2(2 – 5i)

2. (4 + 9i) – (5 – 6i) = (4 – 5) + [9 – (-6)]i = -1 + 15i

or (4 + 9i) – (5 – 6i) = 4 + 9i – 5 + 6i = -1 + 15i Treat i as a variable .

• Multiply complex numbers using the FOIL method .

Multiplying Complex Numbers Example(A + Bi) (C + Di) = AC + ADi + BCi + BDi2

F O I L = AC + (AD + BC)i + BD(-1)

= (AC – BD) + (AD + BC) i

(4 + 3i) (3 + 7i) = 12 + 28i + 9i + 21i2

F O I L = 12 + 37i – 21 (i 2 = -1) = -9 + 37i


1. (3 – 4i) (4 + 5i) = 12 + 15i – 16i – 20i2FOIL

= 12 – i – 20(-1) i 2 = -1

= 32 – i

2. 4i (3 – 2i) = 12 i – 8 i2 = 12 i – 8(-1) = 8 + 12i Distribute , i 2 = -1

3. 2i – (3 – 2i)2 = 2 i – [32 – 2 ∙ 3 (2i) + (2i)2] (𝑎𝑎 − 𝑏𝑏)2 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2

= 2 i – (9 – 12 i + 4 i2)

= 2 i – 9 + 12 i – 4 i2

= 14 i – 9 – 4(-1) i 2 = -1

= – 5 + 14 i

4. √𝒕𝒕𝟖𝟖𝟖𝟖 ∙ √ 𝟒𝟒 = �( 1)(81) ∙ �( 1)(4) = √ 𝟖𝟖 √92 ∙ √ 𝟖𝟖√22

= (𝒊𝒊 9) ∙ (𝒊𝒊 2) = i2 18 = (-1) 18 = -18 i 2 = -1

(i 2 = -1)

𝑎𝑎 = 3, 𝑏𝑏 = 2𝑖𝑖

Note: √𝑎𝑎𝑛𝑛 √𝑏𝑏𝑛𝑛 = √𝑎𝑎𝑏𝑏𝑛𝑛 This rule applies only when both a and b are non-negative .

√𝑡𝑡81√𝑡𝑡4 = �(𝑡𝑡81)(𝑡𝑡4) = √92 ∙ 22 = 9 ∙ 2 = 18 Incorrect!

Page 7-20

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- - --

- - -





Complex Conjugates and Division

Example

• Review conjugates: A + B and A – B 3 + √2 and 3 – √2

• Complex conjugates: each complex number has a complex conjugate .

Complex Conjugates Example Conjugates

A + Bi and A – Bi

Conjugates

4 + 3i and 4 – 3i

Tip: Change the sign of the imaginary part .

Example

Complex Number Complex Conjugate 11 + 7i 11 – 7i -2 – 5i -2 + 5i

-21i (0 – 21i) 21i (0 + 21i)

• To divide complex numbers, rationalize the denominator to get rid of the imaginary

number in the denominator .

Steps Example: -𝟑𝟑𝟐𝟐+𝒊𝒊

- Multiply the nominator and denominator - 32+𝑖𝑖

= - 3(𝟐𝟐−𝒊𝒊)(2+𝑖𝑖)(𝟐𝟐−𝒊𝒊)

Multiply by (2 – i) .

by the conjugate of the denominator .

- Apply (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏) = 𝑎𝑎2 − 𝑏𝑏2 . = - 6+3𝑖𝑖22−𝒊𝒊2

a = 2 , b = i

- Simplify . = - 6+3𝑖𝑖 4−(-𝟏𝟏)

i 2 = -1

- Write in the A + Bi form . = - 6 + 3𝑖𝑖 5

= - 𝟔𝟔 𝟓𝟓

+ 𝟑𝟑 𝟓𝟓𝒊𝒊

Page 7-21

Complex Conjugates and Division

Example

• Review conjugates: A + B and A – B 3 + √2 and 3 – √2

• Complex conjugates: each complex number has a complex conjugate .

Complex Conjugates Example Conjugates

A + Bi and A – Bi

Conjugates

4 + 3i and 4 – 3i

Tip: Change the sign of the imaginary part .

Example

Complex Number Complex Conjugate 11 + 7i 11 – 7i -2 – 5i -2 + 5i

-21i (0 – 21i) 21i (0 + 21i)

• To divide complex numbers, rationalize the denominator to get rid of the imaginary

number in the denominator .

Steps Example: -𝟑𝟑𝟐𝟐+𝒊𝒊

- Multiply the nominator and denominator - 32+𝑖𝑖

= - 3(𝟐𝟐−𝒊𝒊)(2+𝑖𝑖)(𝟐𝟐−𝒊𝒊)

Multiply by (2 – i) .

by the conjugate of the denominator .

- Apply (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏) = 𝑎𝑎2 − 𝑏𝑏2 . = - 6+3𝑖𝑖22−𝒊𝒊2

a = 2 , b = i

- Simplify . = - 6+3𝑖𝑖 4−(-𝟏𝟏)

i 2 = -1

- Write in the A + Bi form . = - 6 + 3𝑖𝑖 5

= - 𝟔𝟔 𝟓𝟓

+ 𝟑𝟑 𝟓𝟓𝒊𝒊

Page 7-21





Page 7-22

Complex Division and Solution

• Complex division

Example: Simplify and write the answer in the form A + Bi.

1. 𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 − 𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑

= (2 − 3𝑖𝑖𝑖𝑖)(𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑)(3 − 4𝑖𝑖𝑖𝑖)(𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑)

Multiply by (3 + 4i) (the conjugate of (3 − 4i)).

= 6 + 8𝑖𝑖𝑖𝑖 − 9𝑖𝑖𝑖𝑖 − 12 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

32 − (4𝑖𝑖𝑖𝑖)2 FOIL ; (𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏)(𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏) = 𝑎𝑎𝑎𝑎2 − 𝑏𝑏𝑏𝑏2

= 6 − 𝑖𝑖𝑖𝑖 − 12(-𝟏𝟏𝟏𝟏)9 − 42 𝑖𝑖𝑖𝑖2

i2 = -1

= 6 − 𝑖𝑖𝑖𝑖 + 129 − 16 (−1)

= 18 − 𝑖𝑖𝑖𝑖 25

= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐− 𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 A + Bi

2. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

= 2𝟑𝟑𝟑𝟑5𝑖𝑖𝑖𝑖 ∙ 𝟑𝟑𝟑𝟑

= 2𝑖𝑖𝑖𝑖5𝑖𝑖𝑖𝑖2

Multiply by i.

= 2𝑖𝑖𝑖𝑖5(-1)

i 2 = -1

= - 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 A + Bi (0 − 𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑)

3. 𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑

= (2 − 3𝑖𝑖𝑖𝑖)𝟑𝟑𝟑𝟑4𝑖𝑖𝑖𝑖 ∙ 𝟑𝟑𝟑𝟑

Multiply by i.

= 2𝑖𝑖𝑖𝑖 − 3𝑖𝑖𝑖𝑖2

4(-1) i 2 = -1

=2𝑖𝑖𝑖𝑖 − 3(-1)

-4

=2𝑖𝑖𝑖𝑖 + 3

-4

= -𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒− 𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 A + B i

• Complex solution

Example: Determine whether the complex number (2 – i) is a solution of the equation.

y2 – 4y + 5 = 0. ? (2 – i)2 – 4(2 – i) + 5 = 0 Replace y with (2 – i). ? (22 – 2 + 2i + i 2) – 8 + 4i + 5 = 0 (𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 − 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2; combine like terms. ? i 2 + 1 = 0 i 2 = -1 √

-1 + 1 = 0 Yes, (2 – i) is a solution. Page 7-22

Complex Division and Solution

• Complex division

Example: Simplify and write the answer in the form A + Bi.

1. 𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 − 𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑

= (2 − 3𝑖𝑖𝑖𝑖)(𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑)(3 − 4𝑖𝑖𝑖𝑖)(𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑)

Multiply by (3 + 4i) (the conjugate of (3 − 4i)).

= 6 + 8𝑖𝑖𝑖𝑖 − 9𝑖𝑖𝑖𝑖 − 12 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

32 − (4𝑖𝑖𝑖𝑖)2 FOIL ; (𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏)(𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏) = 𝑎𝑎𝑎𝑎2 − 𝑏𝑏𝑏𝑏2

= 6 − 𝑖𝑖𝑖𝑖 − 12(-𝟏𝟏𝟏𝟏)9 − 42 𝑖𝑖𝑖𝑖2

i2 = -1

= 6 − 𝑖𝑖𝑖𝑖 + 129 − 16 (−1)

= 18 − 𝑖𝑖𝑖𝑖 25

= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐− 𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 A + Bi

2. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

= 2𝟑𝟑𝟑𝟑5𝑖𝑖𝑖𝑖 ∙ 𝟑𝟑𝟑𝟑

= 2𝑖𝑖𝑖𝑖5𝑖𝑖𝑖𝑖2

Multiply by i.

= 2𝑖𝑖𝑖𝑖5(-1)

i 2 = -1

= - 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 A + Bi (0 − 𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑)

3. 𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑

= (2 − 3𝑖𝑖𝑖𝑖)𝟑𝟑𝟑𝟑4𝑖𝑖𝑖𝑖 ∙ 𝟑𝟑𝟑𝟑

Multiply by i.

= 2𝑖𝑖𝑖𝑖 − 3𝑖𝑖𝑖𝑖2

4(-1) i 2 = -1

=2𝑖𝑖𝑖𝑖 − 3(-1)

-4

=2𝑖𝑖𝑖𝑖 + 3

-4

= -𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒− 𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 A + B i

• Complex solution

Example: Determine whether the complex number (2 – i) is a solution of the equation.

y2 – 4y + 5 = 0. ? (2 – i)2 – 4(2 – i) + 5 = 0 Replace y with (2 – i). ? (22 – 2 + 2i + i 2) – 8 + 4i + 5 = 0 (𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏)2 = 𝑎𝑎𝑎𝑎2 − 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 𝑏𝑏𝑏𝑏2; combine like terms. ? i 2 + 1 = 0 i 2 = -1 √

-1 + 1 = 0 Yes, (2 – i) is a solution.





Unit 7 Summary • Square roots




If x2 = 9,

Then �𝑥𝑥 = √9 = 3 𝑥𝑥 = -√9 = -3


• Radical (root) is an expression that uses a root, such as square root, cube root, etc .

• Radical notation for the nth root √𝒏𝒏

√𝑎𝑎𝑛𝑛 �√ − the radical sign 𝑎𝑎 − the radicand (a real number) 𝑛𝑛 − the index (a positive integer > 1)

√𝑎𝑎𝑛𝑛 − radical or radical expression

• Rational (fractional) exponent notation a is a fractional power or a number is raised to a fraction .

nth Root Example


Note: if n = 2 , write √𝑎𝑎 rather than √𝑎𝑎2 . Omit 2 in √52 , write √5 . • nth root to the nth power Example

√𝒂𝒂𝒏𝒏 𝒏𝒏= 𝒂𝒂 Note: √𝒂𝒂𝒏𝒏 𝒏𝒏


= (𝟕𝟕𝟐𝟐)𝟏𝟏𝟐𝟐 = 𝟕𝟕

𝟐𝟐𝟐𝟐 = 𝟕𝟕𝟏𝟏 = 𝟕𝟕

• nth root Example


Example• If the index n is an even natural number: √𝑎𝑎𝑛𝑛 𝑛𝑛

= |𝑎𝑎| √ 3𝟒𝟒 𝟒𝟒= �-3� = 3

• If the index n is an odd natural number: √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎


• Index of a radical (n)Index n Read Example Read








111 = √211 the 11th root of 2





radical notation

√6𝟓𝟓 𝟓𝟓= 6 a = 6

�-6𝟓𝟓

𝟓𝟓

= - 6 a = - 6

√0 𝟓𝟓 𝟓𝟓= 0 a = 0


Negative root

1n

– a

Page 7-23

Unit 7 Summary • Square roots




If x2 = 9,

Then �𝑥𝑥 = √9 = 3 𝑥𝑥 = -√9 = -3


• Radical (root) is an expression that uses a root, such as square root, cube root, etc .

• Radical notation for the nth root √𝒏𝒏

√𝑎𝑎𝑛𝑛 �√ − the radical sign 𝑎𝑎 − the radicand (a real number) 𝑛𝑛 − the index (a positive integer > 1)

√𝑎𝑎𝑛𝑛 − radical or radical expression

• Rational (fractional) exponent notation a is a fractional power or a number is raised to a fraction .

nth Root Example


Note: if n = 2 , write √𝑎𝑎 rather than √𝑎𝑎2 . Omit 2 in √52 , write √5 . • nth root to the nth power Example

√𝒂𝒂𝒏𝒏 𝒏𝒏= 𝒂𝒂 Note: √𝒂𝒂𝒏𝒏 𝒏𝒏


= (𝟕𝟕𝟐𝟐)𝟏𝟏𝟐𝟐 = 𝟕𝟕

𝟐𝟐𝟐𝟐 = 𝟕𝟕𝟏𝟏 = 𝟕𝟕

• nth root Example


Example• If the index n is an even natural number: √𝑎𝑎𝑛𝑛 𝑛𝑛

= |𝑎𝑎| √ 3𝟒𝟒 𝟒𝟒= �-3� = 3

• If the index n is an odd natural number: √𝑎𝑎𝑛𝑛 𝑛𝑛= 𝑎𝑎


• Index of a radical (n)Index n Read Example Read








111 = √211 the 11th root of 2





radical notation

√6𝟓𝟓 𝟓𝟓= 6 a = 6

�-6𝟓𝟓

𝟓𝟓

= - 6 a = - 6

√0 𝟓𝟓 𝟓𝟓= 0 a = 0


Negative root

1n

– a

Page 7-23





• Powers of roots The nth Root to the mth Power Example

𝑎𝑎𝑚𝑚𝑛𝑛 = (√𝑎𝑎𝑛𝑛 )𝑚𝑚 = √𝑎𝑎𝑛𝑛 𝑚𝑚

= √𝑎𝑎𝑚𝑚𝑛𝑛 m – power 723 = (√73 )2 = √73 2

= √723

• A radical expression is an algebraic expression containing a radical sign √𝑛𝑛 .

• Product and quotient rule for radicalsName Rule Example

product rule √𝑎𝑎𝑎𝑎𝑛𝑛 = √𝑎𝑎𝑛𝑛 ∙ √𝑎𝑎𝑛𝑛 a ≥ 0 , b ≥ 0 √12 = √4 ∙ 3 = √4 √3 = �22 √3 = 2√3



√𝑏𝑏𝑛𝑛 a ≥ 0 , b > 0 , b ≠ 0 � 827

3 = √83

√273 =�233

�333 = 23

• Simplifying radical expressionsA radical expression is in simplest form when: Simplest Form Not Simplest FormThe exponent (m) of the radical is less than the index (n).


3 < 5

�𝑥𝑥87

8 > 7


3𝑥𝑥𝑦𝑦4


√3√8

• To add and subtract radical expressions by combining the like radicals (or like terms) .




simplest condition — no radical appears in the denominator .

• A square root equation is an equation containing a square root . Example √𝑥𝑥 − 5 = 3


satisfy the original equation . So always check solutions . Checking is necessary, not optional .

• A radical equation contains radical expressions. Example (Expressions containing radical signs √𝑛𝑛 ) √𝑥𝑥3 − 2𝑥𝑥 = 7



2= 22 x = 4





Page 7-24







Complex Number Example A – real part

A + iB iB – imaginary part A and B are real numbers




𝑖𝑖 = �-1 , i2 = -1 (𝑖𝑖2 = �-12

= -1)





• Shortcut for powers of imaginary unit iPower of i Example

i n = i R


5i 23 = i 3 = -i 4 23

203

• Adding and subtracting complex numbers Adding/Subtracting Complex Numbers Example

(A + Bi) + (C + Di) = (A + C) + (B + D) i (2 + 5i) + (1 – 3i) = (2 + 1) + (5 – 3) i = 3 + 2i (A + Bi) – (C + Di) = (A – C) + (B – D) i

Or remove parentheses and combine like terms(7 + 6i) – (2 + 3i) = (7 – 2) + (6 – 3) i = 5 + 3ior (7 + 6i) – (2 + 3i) = 7 + 6i – 2 – 3i = 5 + 3i

• Multiplying complex numbersMultiplying Complex Numbers Example

(A + Bi) (C + Di) = AC + ADi + BCi + BDi2

F O I L = AC + (AD + BC)i + BD(-1)

= (AC – BD) + (AD + BC) i

(4 + 3i) (3 + 7i) = 12 + 28i + 9i + 21i2

F O I L = 12 + 37i – 21 (i 2 = -1) = -9 + 37i

• Complex conjugates: each complex number has a complex conjugate .Complex Conjugates Example

Conjugates

A + Bi and A – Bi

Conjugates

4 + 3i and 4 – 3i

• To divide complex numbers: Rationalize the denominator to get rid of the imaginary number in the denominator .

(i 2 = -1)

Page 7-25





PRACTICE QUIZ

Unit 7 Radicals

1. Given the function f (x) = √5𝑥𝑥 + 2 ,

a. determine the function values f (3) and f (0) .

b. identify the domain .

2. Find each root .

a. � 181

4

b . √8𝑢𝑢33

3. Given the function f (x) = - √7𝑥𝑥 − 1 3 ,

determine the function value f (4) .

4. Express each of the following in positive exponential form .

a . -3𝑥𝑥-2/3 𝑦𝑦3/4 𝑧𝑧-1/5

b . (u -3 ∙ 𝑣𝑣 2)3/4

c . �𝑥𝑥3

𝑦𝑦-4�3

d . �𝑎𝑎𝑏𝑏�

-4

e . (-37.56891)0

5. Express in simplest radical form .

a . �16𝑥𝑥8𝑦𝑦34

b . �√𝑏𝑏45

c . 𝑢𝑢23 𝑣𝑣

12 𝑤𝑤

34

d . � 𝑎𝑎1/4𝑏𝑏1/4

𝑐𝑐1/12𝑑𝑑1/12�3

Page 11

PRACTICE QUIZ

Unit 7 Radicals

1. Given the function f (x) = √5𝑥𝑥 + 2 ,

a. determine the function values f (3) and f (0) .

b. identify the domain .

2. Find each root .

a. � 181

4

b . √8𝑢𝑢33

3. Given the function f (x) = - √7𝑥𝑥 − 1 3 ,

determine the function value f (4) .

4. Express each of the following in positive exponential form .

a . -3𝑥𝑥-2/3 𝑦𝑦3/4 𝑧𝑧-1/5

b . (u -3 ∙ 𝑣𝑣 2)3/4

c . �𝑥𝑥3

𝑦𝑦-4�3

d . �𝑎𝑎𝑏𝑏�

-4

e . (-37.56891)0

5. Express in simplest radical form .

a . �16𝑥𝑥8𝑦𝑦34

b . �√𝑏𝑏45

c . 𝑢𝑢23 𝑣𝑣

12 𝑤𝑤

34

d . � 𝑎𝑎1/4𝑏𝑏1/4

𝑐𝑐1/12𝑑𝑑1/12�3

Page 11





6. Simplify the following .

a . �56𝑥𝑥4𝑦𝑦 3

b . √32𝑎𝑎7𝑏𝑏94

√2𝑎𝑎3𝑏𝑏44


a . 4�15𝑦𝑦3 − 3�5𝑦𝑦3

b . 3 √𝑎𝑎 3 �√𝑎𝑎23 + √𝑎𝑎2𝑏𝑏3 3 − 5√𝑎𝑎23 �

c . �5𝑦𝑦 3

√𝑥𝑥 3

8. Solve the following equations:

a . 2 + √7𝑥𝑥+ 13 4 = 4

b . √𝑥𝑥 − 6 − √𝑥𝑥 + 9 + 3 = 0

9. Simplify and write the answer in the form A + Bi . 3 − 4𝑖𝑖5 − 2𝑖𝑖

Page 12




Algebra I & II Key Concepts, Practice, and Quizzes Unit 8 – Quadratic Equations and Inequalities

UNIT 8 QUADRATIC EQUATIONS AND INEQUALITIES

8-1 SOLVING QUADRATIC EQUATIONS

Incomplete Quadratic Equations

• A quadratic equation: an equation that has a squared term, such as 2x2 + 7x – 3 = 0 .

Quadratic Equations in Standard Formax2 + bx + c = 0 a ≠ 0

• Incomplete quadratic equation

Incomplete Quadratic Equation Example a b cax2 + bx = 0 (c = 0) 7x2 – 4x = 0 7 -4 0

ax2 + c = 0 (b = 0) 3x2 + 16 = 0 3 0 16

• Zero-product propertyZero-Product Property

If A · B = 0, then either A = 0 or B = 0 (or both)(A and B are algebraic expressions .)

Note: “or” means possibility of both .

• Solving incomplete quadratic equationsIncomplete

Quadratic Equation Steps Example

Use the zero-product

property to solve

ax2 + bx = 0.

- Express in ax2 + bx = 0 - Factor: x (ax + b) = 0 - Apply the zero-product property:

x = 0 ax + b = 0

- Solve for x : x = 0 abx -=

Solve 9x2 = -5x 9x2 + 5x = 0

x (9x + 5) = 0

x = 0 9x + 5 = 0

x = 0 95-=x

Use the square root

method to solve

ax2 – c = 0(or ax2 = c).

- Express in ax2 = c

- Divide both sides by a: acx =2

- Take the square root of both sides:acx ±=

Solve 7x2 – 4 = 0 7x2 = 4

742 =x

76 .074

±≈±=x

Exact solutions Approximate solutions

Add 5x

Page 8-1

UNIT 8 QUADRATIC EQUATIONS AND INEQUALITIES

8-1 SOLVING QUADRATIC EQUATIONS

Incomplete Quadratic Equations

• A quadratic equation: an equation that has a squared term, such as 2x2 + 7x – 3 = 0 .


• Incomplete quadratic equation

Incomplete Quadratic Equation Example a b cax2 + bx = 0 (c = 0) 7x2 – 4x = 0 7 -4 0

ax2 + c = 0 (b = 0) 3x2 + 16 = 0 3 0 16



Note: “or” means possibility of both .




property to solve

ax2 + bx = 0.


x = 0 ax + b = 0


Solve 9x2 = -5x 9x2 + 5x = 0

x (9x + 5) = 0

x = 0 9x + 5 = 0

x = 0 95-=x

Use the square root

method to solve

ax2 – c = 0(or ax2 = c).




Solve 7x2 – 4 = 0 7x2 = 4

742 =x

76 .074

±≈±=x


Add 5x

Page 8-1





Quadratic Equations

• Solve a quadratic equation: a quadratic equation can be written as:

(x + a)(x + b) = 0 Factor .

Set each term equal to zero: x + a = 0 x + b = 0 Zero-product property

Solutions： x = -a x = -b Solve for a and b.

Example: Solve for x. (x + 5)(x –7) = 0

x + 5 = 0 x – 7 = 0 Zero-product property

x = -5 x = 7 Solve for x.

The solution set is {-5, 7} .

• The x-intercepts of a quadratic equation: the solutions of a quadratic equation .Recall: The x-intercept is the point at which the graph crosses the x-axis .

• To find the x-intercept (x, 0): solve for x in the quadratic equation x2 + bx + c = 0 ∵ All points on the x-axis (the x-intercepts) have a y-coordinate that is zero, x2 + bx + c = 0 is x2 + bx + c = y with y = 0 .

Example: 1. Solve the quadratic equation x2 – 2x – 8 = 0.

2. Identify the x-intercepts of f (x) = x2 – 2x – 8 .

Solution: 1. x2 – 2 x – 8 = 0

(x + 2)(x – 4) = 0 Factor .


x = -2 x = 4

2. f (x) = x2 – 2x – 8

x y = x2 – 2x – 8 (x, y) 0 02 – 2 · 0 – 8 = -8 (0, -8)1 12 – 2 · 1 – 8 = -9 (1, -9)

-1 (-1)2 – 2 (-1) – 8 = -5 (-1, -5)2 22 – 2 ∙ 2 – 8 = -8 (2, -8)

-2 (-2)2 – 2(-2) – 8 = 0 (-2, 0) 4 42 – 2 · 4 – 8 = 0 (4, 0)

x-intercepts: (-2, 0) , (4, 0)Example: Solve (y + 3)2 = 2

�(𝑦𝑦 + 3)2 = ±√2 y + 3 = √2 y + 3 = -√2

y = √2 – 3 ≈ -1 .586 y = -√2 − 3 ≈ -4 .414 Subtract 3 .

Solutions: y = - 𝟑𝟑 ± √𝟐𝟐 or y ≈ �-𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓-𝟒𝟒. 𝟒𝟒𝟏𝟏𝟒𝟒


x-intercepts

Tips: - Choose points on both sides of the axis .- The solutions of f (x) = x2 – 2x – 8 are the first coordinates of the x-intercepts .

∙ (4, 0)(-2, 0) ∙

(-1, -5) ∙

∙ (2, -8)

x

∙ (1, -9) (0, -8) ∙

0

Take the square root of both sides .

Page 8-2

Quadratic Equations

• Solve a quadratic equation: a quadratic equation can be written as:

(x + a)(x + b) = 0 Factor .

Set each term equal to zero: x + a = 0 x + b = 0 Zero-product property

Solutions： x = -a x = -b Solve for a and b.

Example: Solve for x. (x + 5)(x –7) = 0


x = -5 x = 7 Solve for x.

The solution set is {-5, 7} .

• The x-intercepts of a quadratic equation: the solutions of a quadratic equation .Recall: The x-intercept is the point at which the graph crosses the x-axis .

• To find the x-intercept (x, 0): solve for x in the quadratic equation x2 + bx + c = 0 ∵ All points on the x-axis (the x-intercepts) have a y-coordinate that is zero, x2 + bx + c = 0 is x2 + bx + c = y with y = 0 .

Example: 1. Solve the quadratic equation x2 – 2x – 8 = 0.

2. Identify the x-intercepts of f (x) = x2 – 2x – 8 .

Solution: 1. x2 – 2 x – 8 = 0

(x + 2)(x – 4) = 0 Factor .


x = -2 x = 4

2. f (x) = x2 – 2x – 8

x y = x2 – 2x – 8 (x, y) 0 02 – 2 · 0 – 8 = -8 (0, -8)1 12 – 2 · 1 – 8 = -9 (1, -9)

-1 (-1)2 – 2 (-1) – 8 = -5 (-1, -5)2 22 – 2 ∙ 2 – 8 = -8 (2, -8)

-2 (-2)2 – 2(-2) – 8 = 0 (-2, 0) 4 42 – 2 · 4 – 8 = 0 (4, 0)

x-intercepts: (-2, 0) , (4, 0)Example: Solve (y + 3)2 = 2

�(𝑦𝑦 + 3)2 = ±√2 y + 3 = √2 y + 3 = -√2

y = √2 – 3 ≈ -1 .586 y = -√2 − 3 ≈ -4 .414 Subtract 3 .

Solutions: y = - 𝟑𝟑 ± √𝟐𝟐 or y ≈ �-𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓-𝟒𝟒. 𝟒𝟒𝟏𝟏𝟒𝟒


x-intercepts

Tips: - Choose points on both sides of the axis .- The solutions of f (x) = x2 – 2x – 8 are the first coordinates of the x-intercepts .

∙ (4, 0)(-2, 0) ∙

(-1, -5) ∙

∙ (2, -8)

x

∙ (1, -9) (0, -8) ∙

0


Page 8-2





8-2 COMPLETING THE SQUARE

Completing the Square

• Completing the square can be used to solve quadratic equations that are not factorable .

• Completing the square: the process of finding a number to add to a quadratic equation and to form a perfect square, such as: x2 + 10x + ? = (x + 5)2

• Procedure to complete the square – Case I: x2 + bx + c = 0 Steps Example: Solve x2 + 10 x – 1 = 0 .

- Express in the form x2 + b x = -c . x2 + 10 x = 1 Add 1 .

- Add to both sides of the equation . x2 + 10 x + = 1 +

- Determine �𝑏𝑏2�2

(Take half of the coefficient of x and square it .) �𝑏𝑏2�2

= �102�2

= 25

- Add �𝑏𝑏2�2to both sides of the equation .

𝑥𝑥2+bx + �𝑏𝑏2�

2= −𝑐𝑐 + �𝑏𝑏

2�

2x2 + 10x + 25 = 1 + 25

- Factor the left side . (x + 5)2 = 26- Take the square root of both sides . x 26)5( 2 ±=+

- Solve for x . 𝑥𝑥 + 5 = ±√26 𝑥𝑥 + 5 = √26 𝑥𝑥 + 5 = -√26 x = -5 + √26 x = -5 – √26

Solutions: x = -5 ± √𝟐𝟐𝟐𝟐

• Procedure to complete the square – Case II: Ax2 + Bx + C = 0

Steps Example: Solve 2x2 + 4x – 70 = 0 .

Method 1 Method 2

- Express in the form Ax2 + Bx= - C . 2x2 + 4x = 70 2x2 + 4x = 70 Add 70 .

- Make the coefficient of x2 equal to 1 . 2

702

42

2 2

=+xx Divide by 2 . 2(x2 + 2x) = 70 Factor out 2 .

- Add to both sides of the equation . x2 + 2x + = 35 + 2(x2 + 2x + ) = 70 + 2 ∙

- Determine �𝑏𝑏2�2 . �𝑏𝑏

2�2

= �22�2

= 𝟏𝟏 �𝑏𝑏2�2

= �22�2

= 𝟏𝟏

- Add �𝑏𝑏2�2

to both sides of the equation . x2 + 2x + 1 = 35 + 1 2(x2 + 2x + 1 )= 70 + 2 ∙ 1

- Factor the left side . (x + 1)2 = 36 2(x + 1)2 = 72 Divide by 2 .

- Take the square root of both sides . x 36)1( 2 ±=+ x 36)1( 2 ±=+ - Solve for x . x 61 ±=+

x 61=+ x 6-1=+ Solutions: x = 5 x = -7

Add 2 ∙ to the right side .

Page 8-3

8-2 COMPLETING THE SQUARE

Completing the Square

• Completing the square can be used to solve quadratic equations that are not factorable .

• Completing the square: the process of finding a number to add to a quadratic equation and to form a perfect square, such as: x2 + 10x + ? = (x + 5)2

• Procedure to complete the square – Case I: x2 + bx + c = 0 Steps Example: Solve x2 + 10 x – 1 = 0 .

- Express in the form x2 + b x = -c . x2 + 10 x = 1 Add 1 .

- Add to both sides of the equation . x2 + 10 x + = 1 +

- Determine �𝑏𝑏2�2

(Take half of the coefficient of x and square it .) �𝑏𝑏2�2

= �102�2

= 25

- Add �𝑏𝑏2�2to both sides of the equation .

𝑥𝑥2+bx + �𝑏𝑏2�

2= −𝑐𝑐 + �𝑏𝑏

2�

2x2 + 10x + 25 = 1 + 25

- Factor the left side . (x + 5)2 = 26- Take the square root of both sides . x 26)5( 2 ±=+

- Solve for x . 𝑥𝑥 + 5 = ±√26 𝑥𝑥 + 5 = √26 𝑥𝑥 + 5 = -√26 x = -5 + √26 x = -5 – √26

Solutions: x = -5 ± √𝟐𝟐𝟐𝟐

• Procedure to complete the square – Case II: Ax2 + Bx + C = 0

Steps Example: Solve 2x2 + 4x – 70 = 0 .

Method 1 Method 2

- Express in the form Ax2 + Bx= - C . 2x2 + 4x = 70 2x2 + 4x = 70 Add 70 .

- Make the coefficient of x2 equal to 1 . 2

702

42

2 2

=+xx Divide by 2 . 2(x2 + 2x) = 70 Factor out 2 .

- Add to both sides of the equation . x2 + 2x + = 35 + 2(x2 + 2x + ) = 70 + 2 ∙

- Determine �𝑏𝑏2�2 . �𝑏𝑏

2�2

= �22�2

= 𝟏𝟏 �𝑏𝑏2�2

= �22�2

= 𝟏𝟏


to both sides of the equation . x2 + 2x + 1 = 35 + 1 2(x2 + 2x + 1 )= 70 + 2 ∙ 1

- Factor the left side . (x + 1)2 = 36 2(x + 1)2 = 72 Divide by 2 .

- Take the square root of both sides . x 36)1( 2 ±=+ x 36)1( 2 ±=+ - Solve for x . x 61 ±=+

x 61=+ x 6-1=+ Solutions: x = 5 x = -7

Add 2 ∙ to the right side .

Page 8-3





Example: 1. Solve x2 – 4x + 9 = 0 by completing the square .

2. Identify the x-intercepts of f (x) = x2 – 4x + 9 .

Steps Solution

1. x2 – 4x + 9 = 0

- Express in the form x2 + bx= - c . x2 – 4x = -9 Subtract 9 .

- Determine �𝑏𝑏2�

2 . �-4

2�2

= 4 b = -4


to both sides of the equation . (x2 – 4x + 4) = -9 + 4- Factor the left side . (x – 2)2 = -5

- Take the square root of both sides . x 5-)2( 2 ±=−

- Solve for x . x 5-2 ±=−

52

5)1-(2

5-2

i

x

±=

⋅±=

±=

�-1 = 𝑖𝑖𝑖𝑖

Solutions: i x 52 += or i x 52 −=x = 𝟐𝟐 ± 𝒊𝒊√𝟓𝟓

2. f (x) = x2 – 4x + 9 . f (x) = x2 – 4x + 9

x y = x2 – 4x + 9 (x, y)0 02 – 4 · 0 + 9 = 9 (0, 9) 1 12 – 4 · 1 + 9 = 6 (1, 6)

-1 (-1)2 – 4 (-1) + 9 = 14 (-1, 14) 2 22 – 4 ∙ 2 + 9 = 5 (2, 5) 3 32 – 4 ∙ 3 + 9 = 6 (3, 6) 4 42 – 4 · 4 + 9 = 9 (4, 9)5 52 – 4 · 5 + 9 = 14 (5, 14)

Tip: Choose points on both sides of the y-axis .

No x-intercepts (x are non-real complex numbers) .

(-1, 14) ∙ (5, 14) ∙

y

Add 2 .

∙ (4, 9)

∙ (3, 6)

x

(1, 6) ∙

(0 9) ∙

∙ (2, 5)

0

The graph does not cross the x-axis anywhere .

Page 8-4





Applications

• Simple interest: interest computed on the original principal .

• Compound interest: interest computed on both the principal and the past interest earned .

• Compound interest formulaFormula Component

A = P(1 + r) tA – new valueP – starting principalr – interest ratet – time (years)

Example: $100 grows to $121 in 2 years . Calculate interest rate r .

A = P(1 + r) t � A = $121

P = $100

𝑡𝑡 = 2 years121= 100(1 + r) 2

121100

= (1 + r) 2Divide by 100 .

±�121100

= �(1 + 𝑟𝑟)2 Take the square root of both sides .

±�121100

= 1 + 𝑟𝑟

-1 ±�121100

= 𝑟𝑟 Subtract 1

r = -1 +�121100

r = -1 –�121100

Solve for r.

= -1 +�112

102= -1 –�112

102

= -1 + 1110

= -1 –1110

= - 1010

+ 1110

= - 1010− 11

10

= 110

= 0.1 √ = - 2110

×

r = 0.1 or 10 % Interest rate is 10% .

Interest rate cannot be negative, eliminate it .

Page 8-5





8-3 THE QUADRATIC FORMULA

Methods to Solve Quadratic Equations

• Methods for solving quadratic equations

Quadratic Equation Methodax2 + c = 0 (no x term) Square root method

ax2 + bx = 0 (c = 0) Zero-product property

ax2 + bx + c = 0 Try factoring first .ax2 + bx + c = 0

Not factorable (or does not factor easily)Completing the square or quadratic formula

Factoring: fast and easy to use, but is limited in scope . (Some quadratic equations are not factorable .)

Completing the square: can be used to solve all quadratic equations, but is tedious .

The quadratic formula: a general formula that can be used to solve any quadratic equation .

The Quadratic Formula

quadratic equation: ax2 + bx + c = 0

solutions: a

acb-bx2

42 −±= (a ≠ 0)

Note: The plus or minus sign (±) shows that the quadratic formula gives two solutions .

aacb-bx

242 −+

= anda

acb-bx2

42 −−=

Page 8-6

8-3 THE QUADRATIC FORMULA

Methods to Solve Quadratic Equations


Quadratic Equation Methodax2 + c = 0 (no x term) Square root method

ax2 + bx = 0 (c = 0) Zero-product property

ax2 + bx + c = 0 Try factoring first .ax2 + bx + c = 0

Not factorable (or does not factor easily)Completing the square or quadratic formula

Factoring: fast and easy to use, but is limited in scope . (Some quadratic equations are not factorable .)

Completing the square: can be used to solve all quadratic equations, but is tedious .

The quadratic formula: a general formula that can be used to solve any quadratic equation .


quadratic equation: ax2 + bx + c = 0

solutions: a

acb-bx2

42 −±= (a ≠ 0)

Note: The plus or minus sign (±) shows that the quadratic formula gives two solutions .

aacb-bx

242 −+

= anda

acb-bx2

42 −−=

Page 8-6






Solving quadratic equations using the quadratic formula

Example: Use the quadratic formula to solve x2 + 5x = - 4 .

Steps Solution

x2 + 5x = - 4

- Write in standard form (ax2 + bx + c = 0) . 1x2 + 5x + 4 = 0 Add 4 .

- Identify a, b, and c . a = 1, b = 5, c = 4

- Substitute the values of a, b, and c into 12

41455-2

4 22

⋅⋅⋅−±

=−±

=a

acb-bx

the formula and calculate . 2

35-2

95-2

16255- ±=

±=

−±=x

𝑥𝑥 = -5±32

- Solve for x . x 1-2

35-=

+= x 4-

235-=

−=

Solutions: x = -4 or -1

Example: Use the quadratic formula to solve - 6x = 3 – 4x2 .

Steps Solution

- 6x = 3 – 4x2

- Write in standard form . 4x2 – 6x – 3 = 0 Add 4x2; subtract 3 .

- Identify a, b and c . a = 4, b = -6, c = -3

- Substitute a, b, and c into the formula .

42)3-(44)6-()6-(-

24

2

2

⋅⋅⋅−±

=

−±=

aacb-bx

- Solve for x . = 6 ± √36 + 488

= 6 ± √848

= 6 ± √21 ∙ 48

= 6 ± 2√218

Divide by 2 .

Solutions: 𝒙𝒙 = 𝟑𝟑 ± √𝟐𝟐𝟐𝟐𝟒𝟒

Page 8-7


Solving quadratic equations using the quadratic formula

Example: Use the quadratic formula to solve x2 + 5x = - 4 .

Steps Solution

x2 + 5x = - 4

- Write in standard form (ax2 + bx + c = 0) . 1x2 + 5x + 4 = 0 Add 4 .

- Identify a, b, and c . a = 1, b = 5, c = 4

- Substitute the values of a, b, and c into 12

41455-2

4 22

⋅⋅⋅−±

=−±

=a

acb-bx

the formula and calculate . 2

35-2

95-2

16255- ±=

±=

−±=x

𝑥𝑥 = -5±32

- Solve for x . x 1-2

35-=

+= x 4-

235-=

−=

Solutions: x = -4 or -1

Example: Use the quadratic formula to solve - 6x = 3 – 4x2 .

Steps Solution

- 6x = 3 – 4x2

- Write in standard form . 4x2 – 6x – 3 = 0 Add 4x2; subtract 3 .

- Identify a, b and c . a = 4, b = -6, c = -3

- Substitute a, b, and c into the formula .

42)3-(44)6-()6-(-

24

2

2

⋅⋅⋅−±

=

−±=

aacb-bx

- Solve for x . = 6 ± √36 + 488

= 6 ± √848

= 6 ± √21 ∙ 48

= 6 ± 2√218

Divide by 2 .

Solutions: 𝒙𝒙 = 𝟑𝟑 ± √𝟐𝟐𝟐𝟐𝟒𝟒

Page 8-7





Example: 1. Use the quadratic formula to solve (x – 3)(x – 4) – 8 = 0 .

2. Identify the x-intercepts of f (x) = (x – 3)(x – 4) – 8 .

Solution: 1. x2 – 4x – 3x + 12 – 8 = 0 FOIL

x2 – 7x + 4 = 0 Standard form

12414)7-()7-(-

24

2

2

⋅⋅⋅−±

=

−±=

aacb-bx

a = 1, b = -7, c = 4

= 𝟕𝟕 ± √𝟒𝟒𝟒𝟒− 𝟏𝟏𝟏𝟏𝟐𝟐

= 𝟕𝟕 ± √𝟑𝟑𝟑𝟑𝟐𝟐

2. The x-intercepts: �𝟕𝟕+ √𝟑𝟑𝟑𝟑𝟐𝟐

, 𝟎𝟎� or �𝟕𝟕− √𝟑𝟑𝟑𝟑𝟐𝟐

, 𝟎𝟎�Example: Use the quadratic formula to solve the following .

1. 2x (x – 1) + (x + 3) = x2 + x Remove parentheses .

2x2 – 2x + x + 3 = x2 + x Write in standard form .

x2 – 2x + 3 = 0 Subtract x2 and x.

12314)2-()2-(-

24 22

⋅⋅⋅−±

=−±

=a

acb-bx a = 1, b = -2, c = 3

21242 −±

=2

8-2 ±=

242)1-(2 ⋅⋅±

=2

222 i±= �-𝟏𝟏 = 𝒊𝒊 ; divide by 2 .

𝒙𝒙 = 𝟏𝟏 ± 𝒊𝒊√𝟐𝟐

2. 𝟏𝟏𝒕𝒕

+ 𝟏𝟏𝒕𝒕−𝟏𝟏

= 𝟏𝟏𝟓𝟓

LCD = 5t (t - 1)

𝟏𝟏𝒕𝒕𝟓𝟓𝒕𝒕(𝒕𝒕 − 𝟏𝟏) + 𝟏𝟏

𝒕𝒕−𝟏𝟏𝟓𝟓𝒕𝒕(𝒕𝒕 − 𝟏𝟏) = 𝟏𝟏

𝟓𝟓𝟓𝟓𝒕𝒕(𝒕𝒕 − 𝟏𝟏) Multiply by the LCD .

𝟓𝟓(𝒕𝒕 − 𝟏𝟏) + 𝟓𝟓𝒕𝒕 = 𝒕𝒕(𝒕𝒕 − 𝟏𝟏) Distribute

𝟓𝟓𝒕𝒕 − 𝟓𝟓 + 𝟓𝟓𝒕𝒕 = 𝒕𝒕𝟐𝟐 − 𝒕𝒕

𝒕𝒕𝟐𝟐 − 𝟏𝟏𝟏𝟏𝒕𝒕 + 𝟓𝟓 = 𝟎𝟎 Write in standard form .

12514)11-()11-(-

24 22

⋅⋅⋅−±

=−±

=a

acb-bt a = 1, b = -11, c = 5

22012111 −±

=

𝒕𝒕 = 𝟏𝟏𝟏𝟏±√𝟏𝟏𝟎𝟎𝟏𝟏𝟐𝟐

Exact solutions . t ≈ �𝟏𝟏𝟎𝟎.𝟓𝟓𝟐𝟐𝟓𝟓𝟎𝟎.𝟒𝟒𝟕𝟕𝟓𝟓 Approximate solutions .

Page 8-8





8-4 APPLICATIONS OF QUADRATIC EQUATIONS

Quadratic Applications

Recall steps for solving word problems

Procedure for Solving Word Problems• Organize the facts given from the problem .• Identify and label the unknown quantity (let x = unknown) .• Draw a diagram if it will make the problem clearer . • Convert the wording into a mathematical equation . • Solve the equation .• Check and answer the question.

Example: Evan is going to replace old carpet in his bedroom, which is a rectangle and has a

length 2 meters greater than its width . If the area of his bedroom is 48 square meters,

what will be the dimensions (length and width) of the carpet?

Steps Solution

- List the facts and label the unknown .

FactsArea A = 48m2

Length = Width + 2mUnknowns Width = x , Length = x + 2m

- Draw a diagram .Width = x

Length = x + 2

- Write an equation . x (x + 2) = 48 (Area: A = lw )

- Solve the equation . x2 + 2x = 48 Distribute

- Standard form: x2 + 2x – 48 = 0

- Factor: (x + 8)(x – 6) = 0

- Zero-product property: x + 8 = 0 x – 6 = 0

- Solutions: x = -8 x = 6 (Since the width of a rectangle cannot be negative, eliminate x = -8 .)

? √- Check . 6 (6 + 2) = 48 , 48 = 48

- Answer (the size of the carpet): Width = x = 6mLength = x + 2 = 6 + 2 = 8m

Page 8-9

8-4 APPLICATIONS OF QUADRATIC EQUATIONS

Quadratic Applications

Recall steps for solving word problems

Procedure for Solving Word Problems• Organize the facts given from the problem .• Identify and label the unknown quantity (let x = unknown) .• Draw a diagram if it will make the problem clearer . • Convert the wording into a mathematical equation . • Solve the equation .• Check and answer the question.

Example: Evan is going to replace old carpet in his bedroom, which is a rectangle and has a

length 2 meters greater than its width . If the area of his bedroom is 48 square meters,

what will be the dimensions (length and width) of the carpet?

Steps Solution


FactsArea A = 48m2

Length = Width + 2mUnknowns Width = x , Length = x + 2m

- Draw a diagram .Width = x

Length = x + 2

- Write an equation . x (x + 2) = 48 (Area: A = lw )

- Solve the equation . x2 + 2x = 48 Distribute

- Standard form: x2 + 2x – 48 = 0

- Factor: (x + 8)(x – 6) = 0

- Zero-product property: x + 8 = 0 x – 6 = 0

- Solutions: x = -8 x = 6 (Since the width of a rectangle cannot be negative, eliminate x = -8 .)

? √- Check . 6 (6 + 2) = 48 , 48 = 48

- Answer (the size of the carpet): Width = x = 6mLength = x + 2 = 6 + 2 = 8m

Page 8-9





More Examples

Example: Alice plans to make a circular flower garden in her yard . If the area is 36 square

meters and she is going to put a statue in the middle of the circle, what will be the

distance from the edge of the circle to the statue (radius r)?

Steps Solution

- List the facts and label the unknown .Fact Area A = 36m2

Unknown Radius r = ?- Diagram .

●

- Equation: A = π r2 The area of a circle: A = πr2

- Solve for r.πAr =2 Divide both sides by π .

𝑟𝑟 = ±�𝐴𝐴𝜋𝜋


= ±�36𝜋𝜋

≈ ±3.385 m Area A = 36 m 2

- Solution: r ≈ 3.385 m The radius cannot be negative, eliminate - 3 .385m .

- Answer: the distance from the edge of the circle to the statue is 3 .385 m .

Example: Tom is going to make a small square table for his kids . If the diagonal of the square

is 3 inches, what is the side of this square table?

Steps Solution

- List the facts and label the unknown .Fact The diagonal of a square = 3″.

Unknown The side x = ?- Diagram .

3″ x

- Equation: x2 + x2 = 32 Pythagorean theorem

r

x

Page 8-10

More Examples

Example: Alice plans to make a circular flower garden in her yard . If the area is 36 square

meters and she is going to put a statue in the middle of the circle, what will be the

distance from the edge of the circle to the statue (radius r)?

Steps Solution

- List the facts and label the unknown .Fact Area A = 36m2

Unknown Radius r = ?- Diagram .

●

- Equation: A = π r2 The area of a circle: A = πr2

- Solve for r.πAr =2 Divide both sides by π .

𝑟𝑟 = ±�𝐴𝐴𝜋𝜋


= ±�36𝜋𝜋

≈ ±3.385 m Area A = 36 m 2

- Solution: r ≈ 3.385 m The radius cannot be negative, eliminate - 3 .385m .

- Answer: the distance from the edge of the circle to the statue is 3 .385 m .

Example: Tom is going to make a small square table for his kids . If the diagonal of the square

is 3 inches, what is the side of this square table?

Steps Solution

- List the facts and label the unknown .Fact The diagonal of a square = 3″.

Unknown The side x = ?- Diagram .

3″ x

- Equation: x2 + x2 = 32 Pythagorean theorem

r

x

Page 8-10





- Solve for x: x2 + x2 = 32 , 2x2 = 9

292 =x

12 .25 .429

±≈±=±=x


- Solution: x = 2.12″

- Answer: the side of the table is about 2 .12 inches .

Example: The product of two consecutive integers is 156 . Find the two integers .

Steps Solution


Unknowns Let 1st integer = x , 2nd integer = x + 1

Organize Facts x ∙ (x + 1) = 156 product is

- Equation: x (x + 1) = 156

- Solve for x . x2 + x = 156

x2 + x – 156 = 0

aacbbx

2251-2

6251-12

)156-(1411-2

4-

2

2

±=

±=

⋅⋅⋅−±

=

−±=

a = 1, b = 1, c = -156

122

251-=

+=x 13-

2251-

=−

=x

Solutions: x = 12 or -13

Answer: the first integer x is 12 or -13 .

Tips: - If the first integer is x = 12, the second consecutive integer is x + 1 = 12 + 1 = 13 .

- If the first integer is x = -13, the second consecutive integer is x + 1 = -13 + 1 = -12 .

The side of the table cannot be negative, eliminate -2.12″.

Does not factor easily .

Standard form: ax2 + bx + c = 0

Quadratic formula

Page 8-11





8-5 DISCRIMINANT OF QUADRATIC EQUATIONS

The Discriminant & Solutions

• Discriminant: The expression that appears under the square root sign in the quadratic

formula . It can predict the type of quadratic solution without solving it .

Quadratic Formula Discriminant

𝑥𝑥 =- 𝑏𝑏 ± √𝒃𝒃𝟐𝟐 − 𝟒𝟒𝟒𝟒𝟒𝟒

2𝑎𝑎b2− 4ac

• Three cases: A quadratic equation may have one or two real solutions, or two complex

solutions .

Recall the number system

Complex Numbers (a + i b) 3 + 5i, 4i, 7- …

Real Numbers

Rational Numbers Irrational Numbers6, 4/5 , -4 .5 5 , π … .

If (b2 – 4ac) = 0, there is one real solution

In general Example: x2 − 4x + 4 = 0

ab

ab

aacbbx

2-

20-

24- 2

=±

=−±

=12

414)4-()4-(- 2

⋅⋅⋅−±

=x ax2 + bx + c = 0

224

204

==±

=One real solution

If (b2 – 4ac) > 0, there are two unequal real solutions

In general Example: x2 + 5x + 4 = 0

aacbbx

24- 2 −±

=

235-

295-

1241455- 2

±=

±=

⋅⋅⋅−±

=x

aacb-bx

242 −+

=a

acb-bx2

42 −−=

𝑥𝑥 = -5+3

2 = -1 𝑥𝑥 = -5−3

2= -4

Two real solutionsTwo real solutions

b2−4ac = 0One real solution

b2−4ac = 9 > 0

√−1 = 𝑖𝑖𝑖𝑖

Page 8-12

8-5 DISCRIMINANT OF QUADRATIC EQUATIONS

The Discriminant & Solutions

• Discriminant: The expression that appears under the square root sign in the quadratic

formula . It can predict the type of quadratic solution without solving it .

Quadratic Formula Discriminant

𝑥𝑥 =- 𝑏𝑏 ± √𝒃𝒃𝟐𝟐 − 𝟒𝟒𝟒𝟒𝟒𝟒

2𝑎𝑎b2− 4ac

• Three cases: A quadratic equation may have one or two real solutions, or two complex

solutions .

Recall the number system

Complex Numbers (a + i b) 3 + 5i, 4i, 7- …

Real Numbers

Rational Numbers Irrational Numbers6, 4/5 , -4 .5 5 , π … .

If (b2 – 4ac) = 0, there is one real solution


ab

ab

aacbbx

2-

20-

24- 2

=±

=−±

=12

414)4-()4-(- 2

⋅⋅⋅−±

=x ax2 + bx + c = 0

224

204

==±

=One real solution

If (b2 – 4ac) > 0, there are two unequal real solutions

In general Example: x2 + 5x + 4 = 0

aacbbx

24- 2 −±

=

235-

295-

1241455- 2

±=

±=

⋅⋅⋅−±

=x

aacb-bx

242 −+

=a

acb-bx2

42 −−=

𝑥𝑥 = -5+3

2 = -1 𝑥𝑥 = -5−3

2= -4

Two real solutionsTwo real solutions

b2−4ac = 0One real solution

b2−4ac = 9 > 0

√−1 = 𝑖𝑖𝑖𝑖

Page 8-12





Page 8-13

If (b2 – 4ac) < 0, there are two unequal complex solutions


aacbib

aacbb

aacbbx

24-

241--

2)4(--

2

2

2

-±=

-±=

-±=

ii

x

522

544

2201-4

220-4

12914)4-((-4)-

2

±=⋅±

=

±=

±=

⋅⋅⋅-±

=

Two non-real

Two non-real

• Discriminant and solutions

Discriminant: b2 – 4ac Solution Example b2 – 4ac

(b2 – 4ac ) = 0 one real solution x2 − 4x + 4 = 0

x = 2 (-4)2− 4(1)(4) = 0

(b2 – 4ac) > 0 two real solutions x2 + 5x + 4 = 0 x = -4 or -1 52− 4(1)(4) = 9 > 0

(b2 – 4ac ) < 0 two non-real (complex solutions)

x2 – 4x + 9 = 0 x = 2 ± i √5 (-4)2− 4(1)(9) = -20 < 0

Example: Use the discriminant to determine the nature of the solutions to the equations.

1. 2y2 = 5

- Write in standard form. 2y2 + 0y − 5 = 0 ax2 + bx + c = 0

- Identify a, b, and c. a = 2, b = 0, c = -5

- Calculate the discriminant. b2 – 4ac = 02 − 4∙2∙(-5) = 40

- It has two real solutions. b2 – 4ac = 40 > 0

2. n2 − 3√𝟐𝟐𝟐𝟐𝒏𝒏𝒏𝒏 + 7 = 0

- Identify a, b, and c. a = 1, b = - 3√2, c = 7

- Calculate the discriminant. b2 – 4ac = (-3√2)2 − 4∙1∙7

= (-3)2√22 4∙1∙7 −

= 18 28 −

= -10

- It has two non-real solutions. b2 – 4ac = -10 < 0

b2 – 4ac < 0

b2− 4ac = -20 < 0

√-1 = 𝑖𝑖𝑖𝑖

Divide by 2.

b2 − 4ac

(Complex solutions)





Writing Equation From Solutions

• Writing equation from solutions: applying the zero-product property in reverse .

• Recall zero-product property: If A ∙ B = 0, then either A = 0 or B = 0

• Zero-product property in reverse: If A = 0 or B = 0 , then A ∙ B = 0

• Steps for writing equation from solutions

Steps Example: -5 or 7 are solutions - Let x = two solutions . x = -5 x = 7

- Make one side zero . x + 5 = 0 (Add 5 .) x – 7 = 0 Subtract 7 .

- Apply the zero-product property in reverse . (x + 5)(x – 7) = 0

- Use FOIL . x2 − 7x + 5x − 35 = 0 FOIL

Equation: x2 – 2x – 35 = 0

Example: Write a quadratic equation having the given numbers as solutions .1. - √𝟐𝟐 and 3√𝟐𝟐

x = - √2 x = 3√2 Let x = two solutions .

x + √2 = 0 x – 3√2 = 0 Make one side zero .

(x + √2)(x – 3√2) = 0 Apply the zero-product property in reverse .

x2 − 3√2 x +√2 x − 3√2 √2 = 0 FOIL

Equation: x2 − 𝟐𝟐√𝟐𝟐 x − 6 = 02. 5 - only solution (A double solution)

x = 5 x = 5 Let x = 5 .

x – 5 = 0 x – 5 = 0 Make one side zero .

(x - 5)(x – 5) = 0 Apply the zero-product property in reverse .

x2 – 5x – 5x + 25 = 0 FOIL

Equation: x2 −𝟏𝟏𝟏𝟏x + 25 = 0

3. - 𝒂𝒂𝟐𝟐

and 𝒃𝒃𝟑𝟑

x = - 𝑎𝑎2

x = 𝑏𝑏3

Let x = two solutions .

x + 𝑎𝑎2

= 0 x – 𝑏𝑏3

= 0 Make one side zero .

2x + a = 0 3x – b = 0 Multiply 2 or 3 .

(2x + a)(3x – b) = 0 Apply the zero-product property in reverse .

6x2 – 2bx + 3ax – ab = 0 FOIL

Equation: 6x2 + (3a – 2b)x – ab = 0

(A and B are algebraic expressions .)

Page 8-14

Writing Equation From Solutions


• Recall zero-product property: If A ∙ B = 0, then either A = 0 or B = 0


• Steps for writing equation from solutions

Steps Example: -5 or 7 are solutions - Let x = two solutions . x = -5 x = 7

- Make one side zero . x + 5 = 0 (Add 5 .) x – 7 = 0 Subtract 7 .

- Apply the zero-product property in reverse . (x + 5)(x – 7) = 0

- Use FOIL . x2 − 7x + 5x − 35 = 0 FOIL

Equation: x2 – 2x – 35 = 0

Example: Write a quadratic equation having the given numbers as solutions .1. - √𝟐𝟐 and 3√𝟐𝟐

x = - √2 x = 3√2 Let x = two solutions .

x + √2 = 0 x – 3√2 = 0 Make one side zero .

(x + √2)(x – 3√2) = 0 Apply the zero-product property in reverse .

x2 − 3√2 x +√2 x − 3√2 √2 = 0 FOIL

Equation: x2 − 𝟐𝟐√𝟐𝟐 x − 6 = 02. 5 - only solution (A double solution)

x = 5 x = 5 Let x = 5 .

x – 5 = 0 x – 5 = 0 Make one side zero .

(x - 5)(x – 5) = 0 Apply the zero-product property in reverse .

x2 – 5x – 5x + 25 = 0 FOIL

Equation: x2 −𝟏𝟏𝟏𝟏x + 25 = 0

3. - 𝒂𝒂𝟐𝟐

and 𝒃𝒃𝟑𝟑

x = - 𝑎𝑎2

x = 𝑏𝑏3

Let x = two solutions .

x + 𝑎𝑎2

= 0 x – 𝑏𝑏3

= 0 Make one side zero .

2x + a = 0 3x – b = 0 Multiply 2 or 3 .

(2x + a)(3x – b) = 0 Apply the zero-product property in reverse .

6x2 – 2bx + 3ax – ab = 0 FOIL

Equation: 6x2 + (3a – 2b)x – ab = 0


Page 8-14





8-6 SOLVING EQUATIONS IN QUADRATIC FORM

Equations in Quadratic Form

• Recall quadratic equation: ax2 + bx + c = 0

• Equations in quadratic form are equations that are not really quadratic but can be reduced

to the quadratic form by using proper substitution .

Example: Although x4 + bx2 + c = 0 is a fourth-degree equation (in one variable), it has

a form similar to a quadratic equation .

• Substitution: x4 + bx2 + c = 0(x2)2 + bx2 + c = 0 Replace x4 with (x2)2 .

Equation in quadratic form: u2 + bu + c = 0 Let u = x2 (let u = middle term’s variable) .

• Solving equations in quadratic form Steps Example: Solve x4 – 5x2 + 6 = 0 .

- Rewrite: x4 = (x2)2 . (x2)2 – 5x2 + 6 = 0

- Let u = x2. (Let u = middle term’s variable .) u2 – 5u + 6 = 0

- Factor . (u –2)(u – 3) = 0

- Apply the zero-product property: u – 2 = 0 u – 3 = 0u = 2 u = 3

- Substitute x2 back for u (to find x) . x2 = 2 x2 = 3- Solve for x . 𝒙𝒙 = ±√𝟐𝟐 𝒙𝒙 = ±√𝟑𝟑

Steps Example: Solve y + 2�𝒚𝒚 – 3 = 0 .- Let u =�𝒚𝒚 . (Let u = middle term’s variable .) u2 + 2u – 3 = 0 u =�𝑦𝑦 , u2 =�𝑦𝑦 2 = y

- Factor . (u –1)(u + 3) = 0

- Apply the zero-product property . u – 1 = 0 u + 3 = 0u = 1 u = -3

- Substitute �𝒚𝒚 back for u (to find y). �𝒚𝒚 = 1 �𝒚𝒚 = -3 u =�𝑦𝑦

- Solve for y . y = 1 y = 9 Square both sides .

- Check . y + 2�𝒚𝒚 – 3 = 0 Original equation . .

1 + 2√1 – 3 = 0 9 + 2√9 – 3 = 0

1 + 2 – 3 = 0 9 + 2 ∙ 3 – 3 = 0

0 = 0 12 ≠ 0 An extraneous solution .

Solution: y = 1

Equation in quadratic form . (Solve it like normal .)

?

?

?

?

√


Page 8-15

8-6 SOLVING EQUATIONS IN QUADRATIC FORM

Equations in Quadratic Form

• Recall quadratic equation: ax2 + bx + c = 0

• Equations in quadratic form are equations that are not really quadratic but can be reduced

to the quadratic form by using proper substitution .

Example: Although x4 + bx2 + c = 0 is a fourth-degree equation (in one variable), it has

a form similar to a quadratic equation .

• Substitution: x4 + bx2 + c = 0(x2)2 + bx2 + c = 0 Replace x4 with (x2)2 .

Equation in quadratic form: u2 + bu + c = 0 Let u = x2 (let u = middle term’s variable) .

• Solving equations in quadratic form Steps Example: Solve x4 – 5x2 + 6 = 0 .

- Rewrite: x4 = (x2)2 . (x2)2 – 5x2 + 6 = 0

- Let u = x2. (Let u = middle term’s variable .) u2 – 5u + 6 = 0

- Factor . (u –2)(u – 3) = 0

- Apply the zero-product property: u – 2 = 0 u – 3 = 0u = 2 u = 3

- Substitute x2 back for u (to find x) . x2 = 2 x2 = 3- Solve for x . 𝒙𝒙 = ±√𝟐𝟐 𝒙𝒙 = ±√𝟑𝟑

Steps Example: Solve y + 2�𝒚𝒚 – 3 = 0 .- Let u =�𝒚𝒚 . (Let u = middle term’s variable .) u2 + 2u – 3 = 0 u =�𝑦𝑦 , u2 =�𝑦𝑦 2 = y

- Factor . (u –1)(u + 3) = 0

- Apply the zero-product property . u – 1 = 0 u + 3 = 0u = 1 u = -3

- Substitute �𝒚𝒚 back for u (to find y). �𝒚𝒚 = 1 �𝒚𝒚 = -3 u =�𝑦𝑦

- Solve for y . y = 1 y = 9 Square both sides .

- Check . y + 2�𝒚𝒚 – 3 = 0 Original equation . .

1 + 2√1 – 3 = 0 9 + 2√9 – 3 = 0

1 + 2 – 3 = 0 9 + 2 ∙ 3 – 3 = 0

0 = 0 12 ≠ 0 An extraneous solution .

Solution: y = 1

Equation in quadratic form . (Solve it like normal .)

?

?

?

?

√


Page 8-15





Solving Equations in Quadratic Form

Example: Solve t -2 – 7 t -1 – 8 = 0 by factoring .

- Let u = t -1 . (Let u = middle term’s variable .) u2 – 7u – 8 = 0 u = t -1, u2 = (t -1) 2 = t -2

- Factor . (u + 1)(u – 8) = 0

- Apply the zero-product property . (u + 1) = 0 (u – 8) = 0

u = -1 u = 8

- Substitute t -1 =𝟏𝟏𝒕𝒕 back for u . 𝟏𝟏

𝒕𝒕 = -1 𝟏𝟏

𝒕𝒕= 8

- Solve for t. t = -𝟏𝟏 𝒕𝒕 = 𝟏𝟏𝟖𝟖

Example: Determine the x-intercepts of the function .

f (x) = (x2 – 2)2 – (x2 – 2) – 6

- Let f (x) = 0 . (x2 – 2)2 – (x2 – 2) – 6 = 0

- Let u = x2 – 2 . (Let u = middle term’s variable .) u2 – u – 6 = 0

- Factor . (u + 2)(u – 3) = 0


u = -2 u = 3

- Substitute x2 – 2 back for u . x2 – 2 = -2 x2 – 2 = 3 u = x2 – 2

- Solve for x . x2 = 0 x2 = 5

x = 0 x = ±√5 Take the square root .

- The x-intercepts of the function: (0, 0), (-√𝟓𝟓, 0), (√𝟓𝟓, 0)

Summary: Substitution for variable

Equation in Quadratic Form Substitution Quadratic Form5t 4 – 2 t 2 + 7 = 0 Let u = t 2 5u 2 – 2 u + 7 = 0a 6 – 5 a 3 + 4 = 0 Let u = a3 u2 – 5 u + 4 = 02w-2 – 7 w -1 + 5 = 0 Let u = w-1

u2= w-2 2u2 – 7u + 5 = 07x + 4 √𝒙𝒙 = 3 Let u = √𝑥𝑥 u2= x 7u 2 + 4 u – 3 = 0(x 2 + 3x) 2 – 5(x 2 + 3x) + 4 = 0 Let u = x2 + 3x u 2 – 5 u + 4 = 03b½ – b ¼ = 2 Let u = b¼

u2= (b1/4)2 = b1/2 3u 2 – u = 22t 2/3 + 3t 1/3 – 5 = 0 Let u = t 1/3 u2= (t1/3)2 = t2/3 2u 2 + 3u – 5 = 0

t -2 – 7 t -1 – 8 = 0

(Let u = middle term’s variable .)

Page 8-16

Solving Equations in Quadratic Form

Example: Solve t -2 – 7 t -1 – 8 = 0 by factoring .

- Let u = t -1 . (Let u = middle term’s variable .) u2 – 7u – 8 = 0 u = t -1, u2 = (t -1) 2 = t -2

- Factor . (u + 1)(u – 8) = 0


u = -1 u = 8

- Substitute t -1 =𝟏𝟏𝒕𝒕 back for u . 𝟏𝟏

𝒕𝒕 = -1 𝟏𝟏

𝒕𝒕= 8

- Solve for t. t = -𝟏𝟏 𝒕𝒕 = 𝟏𝟏𝟖𝟖

Example: Determine the x-intercepts of the function .

f (x) = (x2 – 2)2 – (x2 – 2) – 6

- Let f (x) = 0 . (x2 – 2)2 – (x2 – 2) – 6 = 0

- Let u = x2 – 2 . (Let u = middle term’s variable .) u2 – u – 6 = 0

- Factor . (u + 2)(u – 3) = 0


u = -2 u = 3

- Substitute x2 – 2 back for u . x2 – 2 = -2 x2 – 2 = 3 u = x2 – 2

- Solve for x . x2 = 0 x2 = 5

x = 0 x = ±√5 Take the square root .

- The x-intercepts of the function: (0, 0), (-√𝟓𝟓, 0), (√𝟓𝟓, 0)

Summary: Substitution for variable

Equation in Quadratic Form Substitution Quadratic Form5t 4 – 2 t 2 + 7 = 0 Let u = t 2 5u 2 – 2 u + 7 = 0a 6 – 5 a 3 + 4 = 0 Let u = a3 u2 – 5 u + 4 = 02w-2 – 7 w -1 + 5 = 0 Let u = w-1



t -2 – 7 t -1 – 8 = 0

(Let u = middle term’s variable .)

Page 8-16





8-7 QUADRATIC AND RATIONAL INEQUALITIES

Quadratic Inequalities

• Review inequality symbols

Symbol Meaning> is greater than< is less than≥ is greater than or equal to≤ is less than or equal to

• Quadratic inequality: an inequality written in one of the following forms .

Standard Quadratic Inequality ExampleAx2 + Bx + C > 0 3x2 + 5x + 7 > 0Ax2 + Bx + C < 0 7x2 – 4x + 3 < 0Ax2 + Bx + C ≥ 0 4x2 + 11x - 6 ≥ 0Ax2 + Bx + C ≤ 0 2x2 – 3x – 2 ≤ 0

Procedure for Solving Quadratic Inequalities

Convert the given inequality to standard form .

Solve the related quadratic equation (Ax2 + Bx + C = 0) and find the cut

points (x-intercepts) .

Use cut points to divide the number line into intervals (create a sign chart) .

Test each interval and determine the solution set (pick test values within

each interval) .

Graph and write the solution(s) .

• Graphing real-number inequalities (review)

The empty circle ○ or open interval ( ) : the endpoints are excluded .


Use a heavy line (shade) and open or closed interval, or use an empty circle versus

filled-in circle to graph the intervals .

Page 8-17

8-7 QUADRATIC AND RATIONAL INEQUALITIES

Quadratic Inequalities

• Review inequality symbols

Symbol Meaning> is greater than< is less than≥ is greater than or equal to≤ is less than or equal to

• Quadratic inequality: an inequality written in one of the following forms .

Standard Quadratic Inequality ExampleAx2 + Bx + C > 0 3x2 + 5x + 7 > 0Ax2 + Bx + C < 0 7x2 – 4x + 3 < 0Ax2 + Bx + C ≥ 0 4x2 + 11x - 6 ≥ 0Ax2 + Bx + C ≤ 0 2x2 – 3x – 2 ≤ 0


Convert the given inequality to standard form .

Solve the related quadratic equation (Ax2 + Bx + C = 0) and find the cut

points (x-intercepts) .

Use cut points to divide the number line into intervals (create a sign chart) .

Test each interval and determine the solution set (pick test values within

each interval) .

Graph and write the solution(s) .

• Graphing real-number inequalities (review)

The empty circle ○ or open interval ( ) : the endpoints are excluded .


Use a heavy line (shade) and open or closed interval, or use an empty circle versus

filled-in circle to graph the intervals .

Page 8-17





Page 8-18

Solving Quadratic Inequalities

Example: Solve x2 – 3x < 4 and graph the solution set.

Steps Solution

x2 – 3x < 4

- Convert to standard form. (Ax2 + Bx + C < 0) x2 – 3x – 4 < 0 Subtract 4.

- Solve the related equation and find the x2 – 3x – 4 = 0

cut points (x-intercepts). (x – 4)(x + 1) = 0 Factor.

x – 4 = 0 x + 1 = 0 Zero-product property

Cut points: x = 4 x = -1

- Use cut points to divide the number line into three distinct intervals (create a sign chart).

x < -1 -1 < x < 4 x > 4 x = 4 and -1 are where the graph crosses the x-axis.

-1 4 Note: The open circles ○ are used because the sign of the inequality is “less than” (< ).

- Test each interval and determine the solution sets (pick a number within each interval).

Test x < -1: pick any value Test x > 4: pick any value Test -1 < x < 4: pick any value less than -1, say x = -2. great than 4, say x = 5. between -1 and 4, say x = 1.

x2 – 3x < 4 x2 – 3x < 4 x2 – 3x < 4 ? ? ?

(-2)2 – 3(-2) < 4 52 – 3 ∙ 𝟓𝟓𝟓𝟓 < 4 12 – 3 ∙ 𝟏𝟏𝟏𝟏 < 4 ? ? ? 4 + 6 < 4 25 – 15 < 4 1 – 3 < 4

× × √ 10 < 4, false 10 < 4 , false -2 < 4 , true

- Graph (shade) the solution set on the number line (on the sign chart).

× √ × -1 < x < 4

-1 4

- Solution set: { x | -1 < x < 4 } or (-1, 4)

x

x

Page 8-18

Solving Quadratic Inequalities

Example: Solve x2 – 3x < 4 and graph the solution set.

Steps Solution

x2 – 3x < 4

- Convert to standard form. (Ax2 + Bx + C < 0) x2 – 3x – 4 < 0 Subtract 4.

- Solve the related equation and find the x2 – 3x – 4 = 0

cut points (x-intercepts). (x – 4)(x + 1) = 0 Factor.

x – 4 = 0 x + 1 = 0 Zero-product property

Cut points: x = 4 x = -1

- Use cut points to divide the number line into three distinct intervals (create a sign chart).

x < -1 -1 < x < 4 x > 4 x = 4 and -1 are where the graph crosses the x-axis.

-1 4 Note: The open circles ○ are used because the sign of the inequality is “less than” (< ).

- Test each interval and determine the solution sets (pick a number within each interval).

Test x < -1: pick any value Test x > 4: pick any value Test -1 < x < 4: pick any value less than -1, say x = -2. great than 4, say x = 5. between -1 and 4, say x = 1.

x2 – 3x < 4 x2 – 3x < 4 x2 – 3x < 4 ? ? ?

(-2)2 – 3(-2) < 4 52 – 3 ∙ 𝟓𝟓𝟓𝟓 < 4 12 – 3 ∙ 𝟏𝟏𝟏𝟏 < 4 ? ? ? 4 + 6 < 4 25 – 15 < 4 1 – 3 < 4

× × √ 10 < 4, false 10 < 4 , false -2 < 4 , true

- Graph (shade) the solution set on the number line (on the sign chart).

× √ × -1 < x < 4

-1 4

- Solution set: { x | -1 < x < 4 } or (-1, 4)

x

x





Example: Solve 2x2 – 7x ≥ -3 and graph the solution set .

Steps Solution

2x2 – 7x ≥ -3

- Convert to standard form . 2x2 – 7x + 3 ≥ 0 Add 3 to both sides .

- Solve the related equation . 2x2 – 7x + 3 = 0

(2x – 1)(x – 3) = 0 Factor .

2x – 1 = 0 x – 3 = 0 Zero-product property

- Cut points: x =21 x = 3

- Intervals: x ≤

21

21 ≤ x ≤ 3 x ≥ 3

21 3

Note: The closed circles ● are used because the sign of the inequality is “great than or equal to” (≥ ) .

- Test .

Test x ≤ 21 : pick any value Test x ≥ 3: pick any value Test

21 ≤ x ≤ 3: pick any value

less than21 , say x = 0. great than 3, say x = 4. between

21 and 3, say x = 1.

2x2 – 7x ≥ -3 2x2 – 7x ≥ -3 2x2 – 7x ≥ -3 ? ? ?

2 ∙ 02 – 7 ∙ 0 ≥ -3 2 ∙ 42 – 7 ∙ 4 ≥ -3 2 ∙ 12 – 7 ∙ 1 ≥ -3

? ? ?

0 – 0 ≥ -3 32 – 28 ≥ -3 2 – 7 ≥ -3

√ √ × 0 > -3 , true 4 > -3 , true -5 > -3 , false

- Graph .√ × √

x ≤21

21 3 x ≥ 3

Solution set: { x | x ≤ 𝟏𝟏𝟐𝟐

or x ≥ 3 } or (-∞, 𝟏𝟏𝟐𝟐� ∪ [3, ∞)

x

Page 8-19





Rational Inequalities

• Rational inequality: an inequality involving fractional expression(s) .

Examples: 02≥

−xx , 0

45≥

−+

xx

• Analyzing and graphing a rational inequality/function

Example: 𝒙𝒙𝒙𝒙−𝟐𝟐

≥ 𝟎𝟎

𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝒙𝒙−𝟐𝟐

x y = f (x) = 𝒙𝒙𝒙𝒙−𝟐𝟐

0 01 -1-1 0 .332 ∞3 34 25 1 .67

Vertical asymptote

- Where is this � 𝒙𝒙𝒙𝒙−𝟐𝟐

≥ 𝟎𝟎� true?

- Or when is 𝒙𝒙𝒙𝒙−𝟐𝟐

positive? (≥ 𝟎𝟎)

- Or, graphically, for what x is 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝒙𝒙−𝟐𝟐

above the x–axis?

- Or where can 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝒙𝒙−𝟐𝟐

change its sign (from negative to positive, or vice versa)?

- 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝒙𝒙−𝟐𝟐

changes sign (crosses the x-axis) when x = 0 (the numerator = 0) and when

x = 2 (the denominator = 0, or f (x) = ∞ or undefined) .

- 𝐓𝐓𝐓𝐓𝐓𝐓 𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐓𝐓 𝐨𝐨𝐨𝐨 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝒙𝒙−𝟐𝟐

has a vertical asymptote at points where the denominator is

0 . The sign of 𝒇𝒇(𝒙𝒙) might change from -∞ to +∞ (it is undefined for 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝒙𝒙−𝟐𝟐

) .

• Cut points (or critical points) of rational inequalities: the points where the rational

inequality changes sign or is 0 (the numerator) and undefined (the denominator) .

Example: The cut points for 𝒙𝒙𝒙𝒙−𝟐𝟐

≥ 𝟎𝟎 are x = 0 and x = 2 .

y

x0 2∙ ∙

∙ ∙ ∙

∙

-∞

+∞

∙

Page 8-20





Solving Rational Inequalities

Example: Solve 045≥

−+

xx and graph the solution set .

Steps Solution

- Find the cut points: 045≥

−+

xx

Set the numerator = 0 x + 5 = 0, x = -5 Subtract 5 .

Set the denominator = 0 x – 4 = 0, x = 4 Add 4 .

- Use cut points to divide the number line into intervals (create a sign chart) .

x ≤ -5 -5 ≤ x < 4 x > 4

-5 4 Note: -5 is included . 4 is not included (since x = 4 is undefined) .

- Test each interval and determine the solution sets .

Test x ≤ -5: pick x = -6 Test x > 4: pick x = 5 Test -5 ≤ x < 4: pick x = 1

45

−+

xx ≥ 0

45

−+

xx ≥ 0

45

−+

xx ≥ 0

? ? ?

46-56-

−+ ≥ 0

4555

−+ ≥ 0

4151

−+ ≥ 0

√ √ ×

101

10-1-= > 0, true 10

110

= > 0, true 2-3-

6= > 0, false

- Graph: √ × √

x ≤ -5 -5 4 x > 4

- Solution sets: { x | x ≤ -5 or x > 4 } or (-∞, -5 ] ∪ (4, ∞)

Page 8-21

Solving Rational Inequalities

Example: Solve 045≥

−+

xx and graph the solution set .

Steps Solution

- Find the cut points: 045≥

−+

xx

Set the numerator = 0 x + 5 = 0, x = -5 Subtract 5 .

Set the denominator = 0 x – 4 = 0, x = 4 Add 4 .

- Use cut points to divide the number line into intervals (create a sign chart) .

x ≤ -5 -5 ≤ x < 4 x > 4

-5 4 Note: -5 is included . 4 is not included (since x = 4 is undefined) .

- Test each interval and determine the solution sets .

Test x ≤ -5: pick x = -6 Test x > 4: pick x = 5 Test -5 ≤ x < 4: pick x = 1

45

−+

xx ≥ 0

45

−+

xx ≥ 0

45

−+

xx ≥ 0

? ? ?

46-56-

−+ ≥ 0

4555

−+ ≥ 0

4151

−+ ≥ 0

√ √ ×

101

10-1-= > 0, true 10

110

= > 0, true 2-3-

6= > 0, false

- Graph: √ × √

x ≤ -5 -5 4 x > 4

- Solution sets: { x | x ≤ -5 or x > 4 } or (-∞, -5 ] ∪ (4, ∞)

Page 8-21





Unit 8 Summary

• A quadratic equation: an equation that has a squared term .







property to solve

ax2 + bx = 0.


x = 0 ax + b = 0


Solve 9x2 = -5x 9x2 + 5x = 0

x (9x + 5) = 0

x = 0 9x + 5 = 0

x = 0 95-=x

Use the square root

method to solve

ax2 – c = 0(or ax2 = c).




Solve 7x2 – 4 = 0 7x2 = 4

742 =x

76 .074

±≈±=x


• The x-intercepts of a quadratic equation are the solutions of a quadratic equation .

• Completing the square is the process of finding a number to add to a quadratic equation and

to form a perfect square, such as: x2 + 10x + ? = (x + 5)2

• Procedure to complete the square – Case I: x2 + bx + c = 0 - Express in the form x2 + b x = -c .- Add to both sides of the equation .

- Determine �b2�

2 (Take half of the coefficient of x and square it .)

- Add �b2�

2to both sides of the equation . 𝑥𝑥2 + bx + �b

2�

2= -𝑐𝑐 + �b

2�

2

- Factor the left side . - Take the square root of both sides . - Solve for x .

Add 5x.

Page 8-22

Unit 8 Summary

• A quadratic equation: an equation that has a squared term .







property to solve

ax2 + bx = 0.


x = 0 ax + b = 0


Solve 9x2 = -5x 9x2 + 5x = 0

x (9x + 5) = 0

x = 0 9x + 5 = 0

x = 0 95-=x

Use the square root

method to solve

ax2 – c = 0(or ax2 = c).




Solve 7x2 – 4 = 0 7x2 = 4

742 =x

76 .074

±≈±=x


• The x-intercepts of a quadratic equation are the solutions of a quadratic equation .

• Completing the square is the process of finding a number to add to a quadratic equation and

to form a perfect square, such as: x2 + 10x + ? = (x + 5)2

• Procedure to complete the square – Case I: x2 + bx + c = 0 - Express in the form x2 + b x = -c .- Add to both sides of the equation .


2 (Take half of the coefficient of x and square it .)

- Add �b2�

2to both sides of the equation . 𝑥𝑥2 + bx + �b

2�

2= -𝑐𝑐 + �b

2�

2


Add 5x.

Page 8-22





• Procedure to complete the square - Case II: Ax2 + Bx + C = 0 - Express in the form Ax2 + Bx = - C . - Make the coefficient of x2 equal to 1 . - Add to both sides of the equation .


2 .

- Add �b2�

2to both sides of the equation .


• Simple interest: interest computed on the original principal .

• Compound interest: interest computed on both the principal and the past interest earned .

• Compound interest formula Formula Component

A = P(1 + r) tA – new valueP – starting principalr – interest ratet – time (year)


Quadratic Equation Methodax2 + c = 0 (no x term) square root methodax2 + bx = 0 (c = 0) zero-product propertyax2 + bx + c = 0 try factoring first

ax2 + bx + c = 0Not factorable (or does not factor easily)

completing the squareor quadratic formula

• The quadratic formula: a general formula that can be used to solve any quadratic equation .

The Quadratic Formula Quadratic equation: ax2 + bx + c = 0

The solutions:a

acb-bx2

42 −±= (a ≠ 0)

• DiscriminantQuadratic Formula Discriminant

𝒙𝒙 = - 𝒃𝒃 ± �𝒃𝒃𝟐𝟐−𝟒𝟒𝟒𝟒𝟒𝟒𝟐𝟐𝒂𝒂

• Discriminant and solutionsDiscriminant: b2 – 4ac Nature of Solution Example b2 – 4ac

(b2 – 4ac ) = 0 One real solution x2 – 4x + 4 = 0 x = 2 (-4)2− 4(1)(4) = 0

(b2 – 4ac) > 0 Two real solutions x2 + 5x + 4 = 0 x = -4 or -1 52− 4(1)(4) = 9 > 0

(b2 – 4ac ) < 0 Two non-real(complex solutions)

x2 – 4x + 9 = 0x = 2 ± i √5

(-4)2− 4(1)(9) = -20 < 0

b2 – 4ac

Page 8-23







• Equations in quadratic form: equations that are not really quadratic but can be reduced to

the quadratic form by using proper substitution .

• Substitution for variableEquation in Quadratic Form Substitution Quadratic Form

5t 4 – 2 t 2 + 7 = 0 Let u = t 2 5u 2 – 2 u + 7 = 0a 6 – 5 a 3 + 4 = 0 Let u = a3 u2 – 5 u + 4 = 02w-2 – 7 w -1 + 5 = 0 Let u = w-1



• Quadratic inequality: an inequality written in one of the following forms .Standard Quadratic Inequality Example

Ax2 + Bx + C > 0 3x2 + 5x + 7 > 0Ax2 + Bx + C < 0 7x2 – 4x + 3 < 0Ax2 + Bx + C ≥ 0 4x2 + 11x - 6 ≥ 0Ax2 + Bx + C ≤ 0 2x2 – 3x – 2 ≤ 0


Convert the given inequality to standard form . Solve the related quadratic equation (Ax2 + Bx + C = 0) and find the cut

points (x-intercepts) . Use cut points to divide the number line into intervals (create a sign chart) . Test each interval and determine the solution set (pick test values within

each interval) . Graph and write the solution(s) .

• Rational inequality: an inequality involving fractional expression(s) .

• Cut points (or critical points) of rational inequalities: the points where the rational

inequality is 0 (the numerator) or undefined (the denominator) .


(Let u = middle term’s variable)

Page 8-24





PRACTICE QUIZ

Unit 8 Quadratic Equations and Inequalities

1. Solve the following equation: (x + 5)2 = 3

2. Solve the following by completing the square. x2 + 4 x – 3 = 0

3. 5,000 grows to $5,500 in 2 years . Calculate interest rate r .

4. Use the quadratic formula to solve the following .

a . x2 – 4x + 1 = 0

b . 3x (x + 2) – (x – 5) = x2 – x

5. How much fencing would be required for a rectangular field of area

4,000 m2 if the length is 30m more than the width?

6. Find the length of the side of a square with diagonal equal to 10m .

7. The product of two consecutive integers is 132 . Find the two integers .

8. Use the discriminant to determine the nature of the solutions of x2 + 5√3𝑥𝑥 – 4 = 0

9. Solve m -2 – 6m -1 – 7 = 0 .

10 . Solve x2 + 2x – 8 ≥ 0 and graph the solution set .

Page 13

PRACTICE QUIZ

Unit 8 Quadratic Equations and Inequalities

1. Solve the following equation: (x + 5)2 = 3

2. Solve the following by completing the square. x2 + 4 x – 3 = 0

3. 5,000 grows to $5,500 in 2 years . Calculate interest rate r .

4. Use the quadratic formula to solve the following .

a . x2 – 4x + 1 = 0

b . 3x (x + 2) – (x – 5) = x2 – x

5. How much fencing would be required for a rectangular field of area

4,000 m2 if the length is 30m more than the width?

6. Find the length of the side of a square with diagonal equal to 10m .

7. The product of two consecutive integers is 132 . Find the two integers .

8. Use the discriminant to determine the nature of the solutions of x2 + 5√3𝑥𝑥 – 4 = 0

9. Solve m -2 – 6m -1 – 7 = 0 .

10 . Solve x2 + 2x – 8 ≥ 0 and graph the solution set .

Page 13




Algebra I & II Key Concepts, Practice, and Quizzes Unit 9 – Conics

UNIT 9 CONICS

• A conic section: a curve obtained as the intersection of a plane with a cone .

• Slicing a cone (or a double cone) can generate a circle, a parabola, an ellipse or a hyperbola .

Cone Circle Ellipse Parabola Hyperbola

9-1 CIRCLES

The Distance Formula

• Distance formula can determine the distance between any two points on a coordinate plane .

• The distance (d) between two points (x1, y1) and (x2, y2)Distance Formula Example

𝑑𝑑 = �(𝑥𝑥2 − 𝑥𝑥1)2+(𝑦𝑦𝑦𝑦2 − 𝑦𝑦𝑦𝑦1)2(x1, y1) = (1, 4) , (x2, y2) = (5, 1)𝑑𝑑 = �(5 − 1)2+(1 − 4)2 = √16 + 9 = 5

Note: It does not matter which point is (x1, y1) and which is (x2, y2) .Tip: The Pythagorean theorem is the basis for calculating distance between two points .

The hypotenuse is the distance between the two points . a2 + b2 = c2

Example: Find the distance between the points (-3, 2) and (4, -3) .

- Let: (x1, y1) = (-3, 2) , (x2, y2) = (4, -3)

𝑑𝑑 = �(4 − (𝑡𝑡3))2+(𝑡𝑡3 − 2)2 = √49 + 25 = √𝟕𝟕𝟕𝟕 𝑑𝑑 = �(𝑥𝑥2 − 𝑥𝑥1)2+(𝑦𝑦𝑦𝑦2 − 𝑦𝑦𝑦𝑦1)2

- Let: (x1, y1) = (4, -3) , (x2, y2) = (-3, 2)

𝑑𝑑 = �(𝑡𝑡3 − 4)2+(2 − (𝑡𝑡3))2 = √49 + 25 = √𝟕𝟕𝟕𝟕

• The midpoint formula for a segment determines the midpoint of a line segment .Midpoint Formula Example

�𝑥𝑥1 + 𝑥𝑥2

2 , 𝑦𝑦𝑦𝑦1 + 𝑦𝑦𝑦𝑦2

2�

(x1, y1) = (4, 3) , (x2, y2) = (-4, 1)

midpoint = �4+�-4�2

, 3+12� = (𝟎𝟎,𝟐𝟐𝟐𝟐)

x∙ (x2, y2)

d

0

∙ (x1, y1)

y

A midpoint divides a line segment into two equal parts.

∙ (4, 3) (-4, 1) ∙

0

y

x∙

Midpoint

The intersection of a plane with a double cone.

c b a

UNIT 9 CONICS




9-1 CIRCLES








- Let: (x1, y1) = (-3, 2) , (x2, y2) = (4, -3)


- Let: (x1, y1) = (4, -3) , (x2, y2) = (-3, 2)

𝑑𝑑 = �(𝑡𝑡3 − 4)2+(2 − (𝑡𝑡3))2 = √49 + 25 = √𝟕𝟕𝟕𝟕




2�

(x1, y1) = (4, 3) , (x2, y2) = (-4, 1)


, 3+12� = (𝟎𝟎,𝟐𝟐𝟐𝟐)

x∙ (x2, y2)

d

0

∙ (x1, y1)

y


∙ (4, 3) (-4, 1) ∙

0

y

x∙

Midpoint


c b a

UNIT 9 CONICS




9-1 CIRCLES








- Let: (x1, y1) = (-3, 2) , (x2, y2) = (4, -3)


- Let: (x1, y1) = (4, -3) , (x2, y2) = (-3, 2)

𝑑𝑑 = �(𝑡𝑡3 − 4)2+(2 − (𝑡𝑡3))2 = √49 + 25 = √𝟕𝟕𝟕𝟕




2�

(x1, y1) = (4, 3) , (x2, y2) = (-4, 1)


, 3+12� = (𝟎𝟎,𝟐𝟐𝟐𝟐)

x∙ (x2, y2)

d

0

∙ (x1, y1)

y


∙ (4, 3) (-4, 1) ∙

0

y

x∙

Midpoint


c b a

UNIT 9 CONICS




9-1 CIRCLES








- Let: (x1, y1) = (-3, 2) , (x2, y2) = (4, -3)


- Let: (x1, y1) = (4, -3) , (x2, y2) = (-3, 2)

𝑑𝑑 = �(𝑡𝑡3 − 4)2+(2 − (𝑡𝑡3))2 = √49 + 25 = √𝟕𝟕𝟕𝟕




2�

(x1, y1) = (4, 3) , (x2, y2) = (-4, 1)


, 3+12� = (𝟎𝟎,𝟐𝟐𝟐𝟐)

x∙ (x2, y2)

d

0

∙ (x1, y1)

y


∙ (4, 3) (-4, 1) ∙

0

y

x∙

Midpoint


c b a

(4 – (-3))2 + (-3 – 2)2

(-3 – 4)2 + (2 – (-3))2





The Circle

• Circle: every point on the curve is equally distant from the center .

• Circles in the real worldRing Pizza Clock Smiley face

• Equation of circles

Center of a Circle The Standard Form Equation Examplecenter at origin (0, 0) x2 + y2 = r2 x2 + y2 = 32 r = 3

center at (h, k) (x – h)2 + (y – k)2 = r2 (x –2)2 + (y –1)2 = 22

(h, k) = (2, 1) , r = 2 r – radius

Example: Identify the center and radius of the following circles .

Equation Center Radius Graphx2 + y2 = 1

x2 + y2 = 12 (0, 0) 1

x2 + y2 = 4x2 + y2 = 22 (0, 0) 2

(x –3)2 + (y + 2)2 = 9(x - 3)2 + [y – (-2)]2 = 32

(3, -2) 3

• General form equation for a circle

The General Form Examplex2 + y2 + Cx + Dy + E = 0 x2 + y2 – 2x + 4y – 20 = 0

Example: Identify the center and the radius of the following circle .x2 + y2 – 2x + 4y – 20 = 0 Add 20 to both sides .

x2 + y2 – 2x + 4y = 20 Regroup x and y terms together .

(x2 – 2x ) + (y2 + 4y ) = 20 Complete the square .

�-22�2

= 1 �42�2

= 4 x 2 + bx + c = 0 ; �-𝑏𝑏𝑏𝑏2�2

(x2 – 2x + 1) + (y2 + 4y + 4) = 20 + 1 + 4 Add 1 and 4 to both sides .

(x – 1)2 + (y + 2)2 = 52Factor .

Center: (1, -2) , radius r = 5 (x – h)2 + (y – k)2 = r2 ; center: (h, k) .

(x – 1)2 + [y – (-2)]2 = 52

0 ∙ rC

C – centerr – radius

r = 1

r = 2

∙ r = 3(3, -2)

∙ (1, -2) 5

x

y

0

The Circle

• Circle: every point on the curve is equally distant from the center .

• Circles in the real worldRing Pizza Clock Smiley face

• Equation of circles

Center of a Circle The Standard Form Equation Examplecenter at origin (0, 0) x2 + y2 = r2 x2 + y2 = 32 r = 3


(h, k) = (2, 1) , r = 2 r – radius

Example: Identify the center and radius of the following circles .

Equation Center Radius Graphx2 + y2 = 1

x2 + y2 = 12 (0, 0) 1

x2 + y2 = 4x2 + y2 = 22 (0, 0) 2

(x –3)2 + (y + 2)2 = 9(x - 3)2 + [y – (-2)]2 = 32

(3, -2) 3

• General form equation for a circle

The General Form Examplex2 + y2 + Cx + Dy + E = 0 x2 + y2 – 2x + 4y – 20 = 0

Example: Identify the center and the radius of the following circle .x2 + y2 – 2x + 4y – 20 = 0 Add 20 to both sides .

x2 + y2 – 2x + 4y = 20 Regroup x and y terms together .

(x2 – 2x ) + (y2 + 4y ) = 20 Complete the square .

�-22�2

= 1 �42�2

= 4 x 2 + bx + c = 0 ; �-𝑏𝑏𝑏𝑏2�2

(x2 – 2x + 1) + (y2 + 4y + 4) = 20 + 1 + 4 Add 1 and 4 to both sides .

(x – 1)2 + (y + 2)2 = 52Factor .

Center: (1, -2) , radius r = 5 (x – h)2 + (y – k)2 = r2 ; center: (h, k) .

(x – 1)2 + [y – (-2)]2 = 52

0 ∙ rC

C – centerr – radius

r = 1

r = 2

∙ r = 3(3, -2)

∙ (1, -2) 5

x

y

0





9-2 PARABOLAS

Introduction to Parabolas

• Parabola: a curve (‘U’ shaped curve) where every point is the same distance from a fixed

point (the focus F) and a fixed line (the directrix) .

• Parabolas in the real worldDome Rainbow Suspension Bridge Arch

• Parabola terminology

Term Definition Diagramfocus A fixed point whose relationship with a directrix

defines a parabola .

directrix A fixed straight line perpendicular to the axis of symmetry .

axis of symmetryA line segment that is perpendicular to the directrix and passes through the vertex and focus of a parabola .

vertexThe point where a parabola makes its sharpest turnas it crosses its axis of symmetry . It is where thedistance from the focus and directrix is shortest .

• Recall: equations of parabolasEquations of Parabolas

y = Ax2 + Bx + C x = Ay2 + By + Cy = Ax2 x = Ay2

y = Ax2 + C x = Ay2 + Cy = A(x – h)2 x= A(y – k)2

y = A(x – h)2 + k x= A(y – k)2 + h

Function: f (x) = Ax2 + Bx2 + C f (y) = Ay2 + By2 + C

• The graph of a quadratic function or equation is a parabola .

∙d2

d1 d1= d2

Directrix

∙ F (focus)

Directrix

Axis of symmetry

∙ Vertex

0

(Focus) F ∙

9-2 PARABOLAS

Introduction to Parabolas

• Parabola: a curve (‘U’ shaped curve) where every point is the same distance from a fixed

point (the focus F) and a fixed line (the directrix) .

• Parabolas in the real worldDome Rainbow Suspension Bridge Arch

• Parabola terminology

Term Definition Diagramfocus A fixed point whose relationship with a directrix





• Recall: equations of parabolasEquations of Parabolas

y = Ax2 + Bx + C x = Ay2 + By + Cy = Ax2 x = Ay2

y = Ax2 + C x = Ay2 + Cy = A(x – h)2 x= A(y – k)2

y = A(x – h)2 + k x= A(y – k)2 + h

Function: f (x) = Ax2 + Bx2 + C f (y) = Ay2 + By2 + C


∙d2

d1 d1= d2

Directrix

∙ F (focus)

Directrix

Axis of symmetry

∙ Vertex

0

(Focus) F ∙





Graphing a Parabola

An easy method for graphing parabolas: make a table of x and y values for the equation and

plot the points .

Example: Sketch the graph of y = 3x2 .

- Make a table of x values for the equation .

- Pick some values of x and solve for each corresponding y to get the ordered pairs (points) Tip: Pick points on both sides of the axis .

- Plot the points and connect them with a smooth curve .

x y = 3x2 (x, y)0 y = 3 ∙ 02 = 0 (0, 0)1 y = 3 ∙ 12 = 3 (1, 3)

-1 y = 3 ∙ (-1)2 = 3 (-1, 3)2 y = 3 ∙ 22 = 12 (2, 12)

-2 y = 3 ∙ (-2)2 = 12 (-2, 12)

Choose . Calculate .

∙ (0, 0)

∙ (1, 3) (-1, 3) ∙

∙ (2,12) (-2, 12) ∙

x

y

Graphing a Parabola

An easy method for graphing parabolas: make a table of x and y values for the equation and

plot the points .

Example: Sketch the graph of y = 3x2 .

- Make a table of x values for the equation .

- Pick some values of x and solve for each corresponding y to get the ordered pairs (points) Tip: Pick points on both sides of the axis .

- Plot the points and connect them with a smooth curve .

x y = 3x2 (x, y)0 y = 3 ∙ 02 = 0 (0, 0)1 y = 3 ∙ 12 = 3 (1, 3)

-1 y = 3 ∙ (-1)2 = 3 (-1, 3)2 y = 3 ∙ 22 = 12 (2, 12)

-2 y = 3 ∙ (-2)2 = 12 (-2, 12)

Choose . Calculate .

∙ (0, 0)

∙ (1, 3) (-1, 3) ∙

∙ (2,12) (-2, 12) ∙

x

y





Parabola in the Form f(x) = Ax2 and f(y) = Ay2

• General information for a parabola of the form f(x) = Ax2 and f(y) = Ay2

Equation Vertex Axis of Symmetry Equation for the Axis of Symmetryf(x) = Ax2 (0, 0) symmetry about the y – axis x = 0f(y) = Ay2 (0, 0) symmetry about the x – axis y = 0

Equation General Shape Example Table Graph

f(x)= Ax2

A > 0Opens up

y = 2x2

(A = 2 > 0)

x 0 1 -1 2 -2y 0 2 2 8 8

A < 0Opens down

y = -2x2

(A = -2 < 0)

x 0 1 -1 2 -2y 0 -2 -2 -8 - 8

f(y)= Ay2

A > 0Opens to the right

x = 2y2

(A = 2 > 0)y 0 1 -1 2 -2x 0 2 2 8 8

A < 0Opens to the left

x = -2y2

(A = -2 < 0 )

y 0 1 -1 2 -2x 0 -2 -2 -8 -8

• The line of symmetry divides the parabola into two equal halves .

• The coefficient A in 𝒇𝒇(𝒙𝒙) = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 can shrink or stretch the parabola

The Coefficient A in 𝒚𝒚 = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 & 𝒙𝒙 = 𝑨𝑨𝒚𝒚𝟐𝟐𝟐𝟐 Example

- The larger the |𝐴𝐴|, the narrower the curve .

𝑓𝑓(𝑥𝑥) = Ax2

A > 0

- The smaller the |𝐴𝐴|, the wider the curve .

A < 0

y = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2

y = x2

y = 3x2

y = - 3x2y = -x2

y = - 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2

(-1, 2) ∙ ∙ (1, 2)

(-2, 8) ∙ ∙ (2, 8)

∙ (1,-2)

∙ (2, -8)

(-1, -2) ∙

(-2, -8) ∙

∙ (2,-1) ∙ (8, 2)

∙ (2, 1)

∙ (8,-2)

∙ (-2, 1) ∙ (-2,-1)

∙ (-8,-2)

∙ (-8, 2)

(Pick y values)

(Calculate x values)

Parabola in the Form f(x) = Ax2 and f(y) = Ay2

• General information for a parabola of the form f(x) = Ax2 and f(y) = Ay2

Equation Vertex Axis of Symmetry Equation for the Axis of Symmetryf(x) = Ax2 (0, 0) symmetry about the y – axis x = 0f(y) = Ay2 (0, 0) symmetry about the x – axis y = 0

Equation General Shape Example Table Graph

f(x)= Ax2

A > 0Opens up

y = 2x2

(A = 2 > 0)

x 0 1 -1 2 -2y 0 2 2 8 8

A < 0Opens down

y = -2x2

(A = -2 < 0)

x 0 1 -1 2 -2y 0 -2 -2 -8 - 8

f(y)= Ay2

A > 0Opens to the right

x = 2y2

(A = 2 > 0)y 0 1 -1 2 -2x 0 2 2 8 8

A < 0Opens to the left

x = -2y2

(A = -2 < 0 )

y 0 1 -1 2 -2x 0 -2 -2 -8 -8

• The line of symmetry divides the parabola into two equal halves .

• The coefficient A in 𝒇𝒇(𝒙𝒙) = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 can shrink or stretch the parabola

The Coefficient A in 𝒚𝒚 = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 & 𝒙𝒙 = 𝑨𝑨𝒚𝒚𝟐𝟐𝟐𝟐 Example

- The larger the |𝐴𝐴|, the narrower the curve .


A > 0

- The smaller the |𝐴𝐴|, the wider the curve .

A < 0

y = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2

y = x2

y = 3x2

y = - 3x2y = -x2

y = - 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2

(-1, 2) ∙ ∙ (1, 2)

(-2, 8) ∙ ∙ (2, 8)

∙ (1,-2)

∙ (2, -8)

(-1, -2) ∙

(-2, -8) ∙

∙ (2,-1) ∙ (8, 2)

∙ (2, 1)

∙ (8,-2)

∙ (-2, 1) ∙ (-2,-1)

∙ (-8,-2)

∙ (-8, 2)

(Pick y values)

(Calculate x values)





Parabola in the Form y = Ax2 + C & x = Ay2 + C

• The graph of 𝑦𝑦𝑦𝑦 = 𝐴𝐴𝐴𝐴2 + 𝐶𝐶 or 𝐴𝐴 = 𝐴𝐴𝑦𝑦𝑦𝑦2 + 𝐶𝐶 is a parabola that has the same shape as

𝑦𝑦𝑦𝑦 = 𝐴𝐴𝐴𝐴2 or 𝐴𝐴 = 𝐴𝐴𝑦𝑦𝑦𝑦2 , but is shifted C units vertically or horizontally .

• General information for a parabola of the form y = Ax2 + C and x = Ay2 + C

Parabola Equation Vertex Axis of Symmetry Shifting

y = A x2 + C (0, C) symmetry about the y - axisThe same shape as y = Ax2,but is shifted C units vertically .

x = A y2 + C (C, 0) symmetry about the x - axisThe same shape as y = Ax2,but is shifted C units horizontally .

Equation C Shifting Vertex Graph Example Graph

y = Ax2 + C

A > 0: opens upA < 0: opens down

C >0 y = Ax2 is shifted Cunits up .

(0, C)

y = 2x2 + 1(C = 1 > 0)

C < 0 y = Ax2 is shifted Cunits down .

y = 2x2 – 3(C = -3 < 0)

x = Ay2 + C

A > 0: opens to the rightA < 0: opens to the left

C > 0 x = Ay2 is shifted Cunits to the right .

(C, 0)

x = 2y2 + 4(C = 4 > 0)

C < 0 x = Ay2 is shifted C units to the left .

x = 2y2 – 13

(C = - 13

< 0 )

Tips: - C indicates how far the parabola has been shifted vertically or horizontally .

- If A > 0, the parabola opens up or to the right . If A < 0, the parabola opens down or to the left .

Example: Sketch the graph of y = -3x2 – 2 . (A = -3 < 0, opens down)

- Make a table for y = -3x2

x 0 1 -1y = -3x2 0 -3 -3

- Plot y = -3x2 : A = -3 < 0 , opens down .

- Plot 𝐴𝐴 = -3𝑦𝑦𝑦𝑦2 − 2 : C = -2 < 0 , shift 2 units down from y = -3x2 .

∙ (0, -2) ∙ (0, 0)

y

x

y = -3x2

y = -3x2 − 2

∙ (0, C) ∙ (0, -3)

∙ (0, 1)

∙ (C, 0)

∙(C, 0)

∙ (4, 0)

∙ �-13

, 0)

∙ (0, C)

Parabola in the Form y = Ax2 + C & x = Ay2 + C

• The graph of 𝑦𝑦𝑦𝑦 = 𝐴𝐴𝐴𝐴2 + 𝐶𝐶 or 𝐴𝐴 = 𝐴𝐴𝑦𝑦𝑦𝑦2 + 𝐶𝐶 is a parabola that has the same shape as

𝑦𝑦𝑦𝑦 = 𝐴𝐴𝐴𝐴2 or 𝐴𝐴 = 𝐴𝐴𝑦𝑦𝑦𝑦2 , but is shifted C units vertically or horizontally .

• General information for a parabola of the form y = Ax2 + C and x = Ay2 + C

Parabola Equation Vertex Axis of Symmetry Shifting

y = A x2 + C (0, C) symmetry about the y - axisThe same shape as y = Ax2,but is shifted C units vertically .

x = A y2 + C (C, 0) symmetry about the x - axisThe same shape as y = Ax2,but is shifted C units horizontally .

Equation C Shifting Vertex Graph Example Graph

y = Ax2 + C


C >0 y = Ax2 is shifted Cunits up .

(0, C)

y = 2x2 + 1(C = 1 > 0)

C < 0 y = Ax2 is shifted Cunits down .

y = 2x2 – 3(C = -3 < 0)

x = Ay2 + C


C > 0 x = Ay2 is shifted Cunits to the right .

(C, 0)

x = 2y2 + 4(C = 4 > 0)

C < 0 x = Ay2 is shifted C units to the left .

x = 2y2 – 13

(C = - 13

< 0 )

Tips: - C indicates how far the parabola has been shifted vertically or horizontally .

- If A > 0, the parabola opens up or to the right . If A < 0, the parabola opens down or to the left .

Example: Sketch the graph of y = -3x2 – 2 . (A = -3 < 0, opens down)

- Make a table for y = -3x2

x 0 1 -1y = -3x2 0 -3 -3

- Plot y = -3x2 : A = -3 < 0 , opens down .

- Plot 𝐴𝐴 = -3𝑦𝑦𝑦𝑦2 − 2 : C = -2 < 0 , shift 2 units down from y = -3x2 .

∙ (0, -2) ∙ (0, 0)

y

x

y = -3x2

y = -3x2 − 2

∙ (0, C) ∙ (0, -3)

∙ (0, 1)

∙ (C, 0)

∙(C, 0)

∙ (4, 0)

∙ �-13

, 0)

∙ (0, C)





Parabola in the Form y = A(x – h)2 & x = A(y – h)2

• The graph of y = A(x – h)2 or x = A(y – h)2 is a parabola that has the same shape as

y = Ax2 or x = Ay2 , but is shifted h units horizontally or vertically .

• General information for a parabola of the form y = A(x – h)2 and x = A(y – h)2

Equation Vertex Axis of Symmetry Shifting

y = A(x – h)2 (h, 0) symmetry about the line x = h

The same shape as y = Ax2, but is shifted h units horizontally .

x = A(y – h)2 (0, h) symmetry about the line y = h

The same shape as y = Ax2, but is shifted h units vertically .

Equation h Shifting Vertex Graph Example Graph

y = A(x – h)2


h >0 y = Ax2 is shifted hunits to the right .

(h, 0)

y = 2(x – 3)2

(h = 3 > 0)

h < 0 y = Ax2 is shifted hunits to the left.

y = 2(x + 3)2

(h = -3 < 0)

y = 2[x – (- 3)]2

x = A(y – h)2


h > 0 x = Ay2 is shifted hunits up.

(0, h)

x = 2(y – 3)2

(h = 3 > 0)

h < 0 x = Ay2 is shifted h units down .

x = 2(y + 3)2

(h = - 3 < 0 )

x = 2[y – (- 3)]2

Tip: h shows how far the parabola has been shifted vertically or horizontally .

Example: Sketch the graph of x = -3(y − 4)2 .

- Make a table for x = -3y2

y 0 1 -1x = -3y2 0 -3 -3

(x, y) (0, 0) (-3, 1) (-3, -1)

- Plot x = -3y2 A = -3 > 0 , opens left .

- Plot 𝑥𝑥 = -3(𝑦𝑦𝑦𝑦 − 4)2 : h = 4 > 0 ,

x = -3y2 is shifted 4 units up .- Vertex: (0, 4)

- Axis of symmetry: y = 4

∙ (-3, 1) ∙ (0, 0)

∙ (-3, -1) x

y

∙

x = -3y2

x = -3(y − 4)2

y = 4

∙ (h, 0) ∙ (3, 0)

∙ (-3, 0) ∙(h, 0)

∙ (0, h)

∙ (0, h)

∙ (0, 3)

∙ (0, -3)

(Pick y values.)

(Calculate x values.)

Vertex (0, 4)

Parabola in the Form y = A(x – h)2 & x = A(y – h)2

• The graph of y = A(x – h)2 or x = A(y – h)2 is a parabola that has the same shape as

y = Ax2 or x = Ay2 , but is shifted h units horizontally or vertically .

• General information for a parabola of the form y = A(x – h)2 and x = A(y – h)2

Equation Vertex Axis of Symmetry Shifting

y = A(x – h)2 (h, 0) symmetry about the line x = h

The same shape as y = Ax2, but is shifted h units horizontally .

x = A(y – h)2 (0, h) symmetry about the line y = h

The same shape as y = Ax2, but is shifted h units vertically .

Equation h Shifting Vertex Graph Example Graph

y = A(x – h)2


h >0 y = Ax2 is shifted hunits to the right .

(h, 0)

y = 2(x – 3)2

(h = 3 > 0)

h < 0 y = Ax2 is shifted hunits to the left.

y = 2(x + 3)2

(h = -3 < 0)

y = 2[x – (- 3)]2

x = A(y – h)2


h > 0 x = Ay2 is shifted hunits up.

(0, h)

x = 2(y – 3)2

(h = 3 > 0)

h < 0 x = Ay2 is shifted h units down .

x = 2(y + 3)2

(h = - 3 < 0 )

x = 2[y – (- 3)]2

Tip: h shows how far the parabola has been shifted vertically or horizontally .

Example: Sketch the graph of x = -3(y − 4)2 .

- Make a table for x = -3y2

y 0 1 -1x = -3y2 0 -3 -3

(x, y) (0, 0) (-3, 1) (-3, -1)

- Plot x = -3y2 A = -3 > 0 , opens left .

- Plot 𝑥𝑥 = -3(𝑦𝑦𝑦𝑦 − 4)2 : h = 4 > 0 ,

x = -3y2 is shifted 4 units up .- Vertex: (0, 4)

- Axis of symmetry: y = 4

∙ (-3, 1) ∙ (0, 0)

∙ (-3, -1) x

y

∙

x = -3y2

x = -3(y − 4)2

y = 4

∙ (h, 0) ∙ (3, 0)

∙ (-3, 0) ∙(h, 0)

∙ (0, h)

∙ (0, h)

∙ (0, 3)

∙ (0, -3)

(Pick y values.)

(Calculate x values.)

Vertex (0, 4)





Parabola in the Form y = A(x – h)2 + k & x = A(y – k)2 + h

General information for a parabola of the form y = A(x – h)2 + k and x = A(y – k)2 + h

Equation Vertex Axis of Symmetryy = A(x – h)2 + k (h, k) symmetry about the line x = hx = A(y – k)2 + h symmetry about the line y = h

Equation Graph Example Graph

y = A(x – h)2 + kA > 0: opens upA < 0: opens down

y = 2(x – 3)2 + 4 (h = 3, k = 4)

x = A(y – k)2 + hA > 0: opens to the right A < 0: opens to the left

x = 2(y – 3)2 + 4 (h = 4, k = 3)

Example: Sketch the graph of y = -2x2− 4x − 5 .

- Convert to y = A(x – h)2 + k y = -2x2 – 4x – 5by completing the square . = -2(x2 + 2x ) – 5 Factor out -2 .

�𝑏𝑏𝑏𝑏2�2 = -2(x2 + 2x + 1 – 1) – 5 �𝑏𝑏𝑏𝑏

2�2

= �22�2

= 1

= -2(x2 + 2x + 1) + (-2)(-1) – 5= -2(x2 + 2x + 1) – 3

y = -2(x + 1)2 – 3 y = -2[x – (-1)]2 + (-3)

- Identify the vertex: (h, k) (h, k) = (-1, -3) y = A(x – h)2 + k

- Determine the axis of symmetry: x = h x = h = -1

- Check A �𝐴𝐴 > 0: opens up 𝐴𝐴 < 0: opens down A = -2 < 0: opens down y = -2(x + 1)2 – 3

- Plot the vertex and the axis of symmetry .

- Make a table and find a few points .

x y = -2x2 – 4x – 5 (x, y)0 y = -2 ∙ 02 – 4 ∙ 0 – 5 = -5 (0, -5)

-2 y = -2 (-2)2 – 4 (-2) – 5 = -5 (-2, -5)1 y = -2 ∙ 12 – 4 ∙ 1 – 5 = -11 (1, -11)

-3 y = -2 (-3)2 – 4 (-3) – 5 = -11 (-3, -11)

(-1, -3) ∙

Axis of symmetry

∙ (0, -5)

Vertex

(-2, -5) ∙

∙ (1, -11) (-3, -11) ∙

x

x = -1 y

∙ 0

∙ (h, k) 0

0

∙ (3, 4)

∙ (h, k)

0 ∙ (4, 3)

0

Parabola in the Form y = A(x – h)2 + k & x = A(y – k)2 + h

General information for a parabola of the form y = A(x – h)2 + k and x = A(y – k)2 + h

Equation Vertex Axis of Symmetryy = A(x – h)2 + k (h, k) symmetry about the line x = hx = A(y – k)2 + h symmetry about the line y = h

Equation Graph Example Graph

y = A(x – h)2 + kA > 0: opens upA < 0: opens down

y = 2(x – 3)2 + 4 (h = 3, k = 4)

x = A(y – k)2 + hA > 0: opens to the right A < 0: opens to the left

x = 2(y – 3)2 + 4 (h = 4, k = 3)

Example: Sketch the graph of y = -2x2− 4x − 5 .

- Convert to y = A(x – h)2 + k y = -2x2 – 4x – 5by completing the square . = -2(x2 + 2x ) – 5 Factor out -2 .

�𝑏𝑏𝑏𝑏2�2 = -2(x2 + 2x + 1 – 1) – 5 �𝑏𝑏𝑏𝑏

2�2

= �22�2

= 1

= -2(x2 + 2x + 1) + (-2)(-1) – 5= -2(x2 + 2x + 1) – 3

y = -2(x + 1)2 – 3 y = -2[x – (-1)]2 + (-3)

- Identify the vertex: (h, k) (h, k) = (-1, -3) y = A(x – h)2 + k

- Determine the axis of symmetry: x = h x = h = -1

- Check A �𝐴𝐴 > 0: opens up 𝐴𝐴 < 0: opens down A = -2 < 0: opens down y = -2(x + 1)2 – 3



x y = -2x2 – 4x – 5 (x, y)0 y = -2 ∙ 02 – 4 ∙ 0 – 5 = -5 (0, -5)

-2 y = -2 (-2)2 – 4 (-2) – 5 = -5 (-2, -5)1 y = -2 ∙ 12 – 4 ∙ 1 – 5 = -11 (1, -11)

-3 y = -2 (-3)2 – 4 (-3) – 5 = -11 (-3, -11)

(-1, -3) ∙

Axis of symmetry

∙ (0, -5)

Vertex

(-2, -5) ∙

∙ (1, -11) (-3, -11) ∙

x

x = -1 y

∙ 0

∙ (h, k) 0

0

∙ (3, 4)

∙ (h, k)

0 ∙ (4, 3)

0





Parabola in the Form f(x) = Ax2 + Bx + C & f(y) = Ay2 + By + C

Equation General Shape Axis of Symmetry Vertex (h, k)f (x) = Ax2 + Bx + C

or y = Ax2 + Bx + C

A > 0: opens up x = - 𝐵𝐵2𝐴𝐴 (h, k) = (- 𝐵𝐵

2𝐴𝐴, 𝑓𝑓 � -B

2A�)A < 0: opens down

f (y) = Ay2 + By + C or x = Ay2 + By + C

A > 0: opens to the right y = - 𝐵𝐵2𝐴𝐴 (h, k) = (𝑓𝑓 � -B

2A� , - 𝐵𝐵

2𝐴𝐴)A < 0: opens to the left

Example: Sketch the graph of f (x) = x2− 4x + 6.

Steps Solution

- Determine the axis of symmetry: 𝑥𝑥 = -𝐵𝐵2𝐴𝐴

𝑥𝑥 = -𝐵𝐵2𝐴𝐴

= -(-4)2(1)

= 2 y = Ax2 + Bx+ C

- Identify the vertex: (h, k) = (- 𝐵𝐵2𝐴𝐴

, 𝑓𝑓 � -B2A�) h = -𝐵𝐵

2𝐴𝐴= 𝟐𝟐𝟐𝟐

k = 𝑓𝑓 � -B2A� = 22 − 4(2) + 6 = 𝟐𝟐𝟐𝟐 f(x) = 1∙ x2 – 4x + 6

(h, k) = (2, 2)

- Check A �𝐴𝐴 > 0: opens up 𝐴𝐴 < 0: opens down A = 1 > 0: opens up A = 1


- Find a few more points . x y = x2 − 4x + 6 (x, y)0 y = 02 − 4∙0 + 6 = 6 (0, 6) 1 y = 12 − 4∙1 + 6 = 3 (1, 3) 3 y = 32 − 4∙3 + 6 = 3 (3, 3)

Example: Sketch the graph of f (y) = y2 – 2y – 8 .

- The axis of symmetry: y = -𝐵𝐵2𝐴𝐴

= -(-2)2∙1

= 1 x = Ay2 + By + C

- Vertex: k = -𝐵𝐵2𝐴𝐴

= 1 , h = 𝑓𝑓 � -B2A� = 12 − 2(1) − 8 = −9 (h, k) = �𝑓𝑓 � -𝐵𝐵

2𝐴𝐴� , -𝐵𝐵

2𝐴𝐴� , (h, k) = (-9, 1)

- Check A: A = 1 > 0 , opens to the right .



y x = y2 – 2y – 8 (x, y)0 x = 02− 2 ∙ 0 − 8 = -8 (-8, 0)2 x = 22− 2 ∙ 2 − 8 = -8 (-8, 2)4 x = 42− 2 ∙ 4 − 8 = 0 (0, 4)-2 x = (-2)2− 2 (-2) − 8 = 0 (0, -2)

x

y x = 2 : Axis of symmetry

(1, 3) ∙

(0, 6) ∙

∙ (3, 3)

∙ (0, 4)

∙ (0, -2)

∙ (-9, 1) ∙(-8, 0)

∙ (2, 2) Vertex0

Axis of symmetryVertexx

y

∙ 0

(-8, 2) ∙

Pick y. Calculate x.

Parabola in the Form f(x) = Ax2 + Bx + C & f(y) = Ay2 + By + C

Equation General Shape Axis of Symmetry Vertex (h, k)f (x) = Ax2 + Bx + C

or y = Ax2 + Bx + C

A > 0: opens up x = - 𝐵𝐵2𝐴𝐴 (h, k) = (- 𝐵𝐵

2𝐴𝐴, 𝑓𝑓 � -B

2A�)A < 0: opens down

f (y) = Ay2 + By + C or x = Ay2 + By + C

A > 0: opens to the right y = - 𝐵𝐵2𝐴𝐴 (h, k) = (𝑓𝑓 � -B

2A� , - 𝐵𝐵

2𝐴𝐴)A < 0: opens to the left

Example: Sketch the graph of f (x) = x2− 4x + 6.

Steps Solution

- Determine the axis of symmetry: 𝑥𝑥 = -𝐵𝐵2𝐴𝐴

𝑥𝑥 = -𝐵𝐵2𝐴𝐴

= -(-4)2(1)

= 2 y = Ax2 + Bx+ C

- Identify the vertex: (h, k) = (- 𝐵𝐵2𝐴𝐴

, 𝑓𝑓 � -B2A�) h = -𝐵𝐵

2𝐴𝐴= 𝟐𝟐𝟐𝟐

k = 𝑓𝑓 � -B2A� = 22 − 4(2) + 6 = 𝟐𝟐𝟐𝟐 f(x) = 1∙ x2 – 4x + 6

(h, k) = (2, 2)

- Check A �𝐴𝐴 > 0: opens up 𝐴𝐴 < 0: opens down A = 1 > 0: opens up A = 1


- Find a few more points . x y = x2 − 4x + 6 (x, y)0 y = 02 − 4∙0 + 6 = 6 (0, 6) 1 y = 12 − 4∙1 + 6 = 3 (1, 3) 3 y = 32 − 4∙3 + 6 = 3 (3, 3)

Example: Sketch the graph of f (y) = y2 – 2y – 8 .

- The axis of symmetry: y = -𝐵𝐵2𝐴𝐴

= -(-2)2∙1

= 1 x = Ay2 + By + C

- Vertex: k = -𝐵𝐵2𝐴𝐴

= 1 , h = 𝑓𝑓 � -B2A� = 12 − 2(1) − 8 = −9 (h, k) = �𝑓𝑓 � -𝐵𝐵

2𝐴𝐴� , -𝐵𝐵

2𝐴𝐴� , (h, k) = (-9, 1)

- Check A: A = 1 > 0 , opens to the right .



y x = y2 – 2y – 8 (x, y)0 x = 02− 2 ∙ 0 − 8 = -8 (-8, 0)2 x = 22− 2 ∙ 2 − 8 = -8 (-8, 2)4 x = 42− 2 ∙ 4 − 8 = 0 (0, 4)-2 x = (-2)2− 2 (-2) − 8 = 0 (0, -2)

x

y x = 2 : Axis of symmetry

(1, 3) ∙

(0, 6) ∙

∙ (3, 3)

∙ (0, 4)

∙ (0, -2)

∙ (-9, 1) ∙(-8, 0)

∙ (2, 2) Vertex0

Axis of symmetryVertexx

y

∙ 0

(-8, 2) ∙

Pick y. Calculate x.





Summary of the Parabola

Equation of Parabolas(Standard Form) Axis of Symmetry Vertex Shape Graph

y = Ax2 y - axis

(0, 0)

A > 0: opens up

A < 0: opens down

x = Ay2 x - axisA > 0, opens right

A < 0, opens left

y = Ax2 + C y - axis (0, C)The same shape as y = Ax2

C > 0, C units up

C < 0, C units down

x = Ay2 + C x - axis (C, 0)The same shape as x = Ay2

C > 0: C units to the right

C < 0: C units to the left

y = A(x – h)2 x = h (h, 0)The same shape as y = Ax2

h > 0: h units to the righth < 0: h units to the left

x= A(y – h)2 y = h (0, h)The same shape as x = Ay2

h > 0: h units up

h < 0: h units down

y = A(x – h)2 + k x = h (h, k) Symmetry about the x = h

x = A(y – k)2 + h y = k (h, k) Symmetry about the y = h

y = Ax2 + Bx + C 𝑥𝑥 = -𝐵𝐵2𝐴𝐴

�-𝐵𝐵2𝐴𝐴 , 𝑓𝑓 �

-B2A��

A > 0: opens up

A < 0: opens down

x = Ay2 + By + C 𝑦𝑦𝑦𝑦 =-𝐵𝐵2𝐴𝐴

�𝑓𝑓 �-B2A� ,

-𝐵𝐵2𝐴𝐴�

A > 0: opens right

A < 0: opens left

∙ (0, C)

∙ (0, -C)

∙ (C, 0)

∙ (h, 0)

(h, 0) ∙

∙ (0, h)

∙ (0, h)

(C, 0) ∙

∙ (h, k)

∙ (h, k)

Summary of the Parabola


y = Ax2 y - axis

(0, 0)

A > 0: opens up

A < 0: opens down


A < 0, opens left


C > 0, C units up

C < 0, C units down







h > 0: h units up

h < 0: h units down



y = Ax2 + Bx + C 𝑥𝑥 = -𝐵𝐵2𝐴𝐴

�-𝐵𝐵2𝐴𝐴 , 𝑓𝑓 �

-B2A��

A > 0: opens up

A < 0: opens down

x = Ay2 + By + C 𝑦𝑦𝑦𝑦 =-𝐵𝐵2𝐴𝐴

�𝑓𝑓 �-B2A� ,

-𝐵𝐵2𝐴𝐴�

A > 0: opens right

A < 0: opens left

∙ (0, C)

∙ (0, -C)

∙ (C, 0)

∙ (h, 0)

(h, 0) ∙

∙ (0, h)

∙ (0, h)

(C, 0) ∙

∙ (h, k)

∙ (h, k)





9-3 ELLIPSES

Introduction to Ellipse

• Ellipse: a curve (oval) where every point on it whose sum of the distances from two fixed

points (foci) is a constant .

Graph Example

x1 + x2 = x3 + x4 = constant 2 + 3 = 1 + 4 = 5

• Ellipse in the real world

Ellipse Diagram

football

orbits of the planets (Earth, Mars, etc .)

oval mirror

• Ellipse terminology

Term Definition Diagramfoci Two fixed points (F) inside of an ellipse that define

the curve .major axis The longest diameter of the ellipse .

(The longer axis and passes through both foci)

minor axis The shortest diameter of the ellipse .(The shorter axis)

vertex The point where an ellipse makes its sharpest turn . (On the major axis)

• Equations of ellipse

Standard Form𝑥𝑥2

𝑎𝑎𝑎𝑎2+ 𝑦𝑦2

𝑏𝑏𝑏𝑏2= 1

(𝑥𝑥−ℎ)2

𝑎𝑎𝑎𝑎2+ (𝑦𝑦−𝑘𝑘)2


∙ Focus x2

x1

x3 x4

A ∙

B ∙

- F ∙ F

B ∙

2

1 4

3

∙ F ∙ F

∙-b

-a ∙ ∙ a

∙ b

Major axis

Minor axis

A ∙

∙ Focus

9-3 ELLIPSES

Introduction to Ellipse

• Ellipse: a curve (oval) where every point on it whose sum of the distances from two fixed

points (foci) is a constant .

Graph Example

x1 + x2 = x3 + x4 = constant 2 + 3 = 1 + 4 = 5

• Ellipse in the real world

Ellipse Diagram

football

orbits of the planets (Earth, Mars, etc .)

oval mirror


Term Definition Diagramfoci Two fixed points (F) inside of an ellipse that define

the curve .major axis The longest diameter of the ellipse .

(The longer axis and passes through both foci)


vertex The point where an ellipse makes its sharpest turn . (On the major axis)

• Equations of ellipse




(𝑥𝑥−ℎ)2



∙ Focus x2

x1

x3 x4

A ∙

B ∙

- F ∙ F

B ∙

2

1 4

3

∙ F ∙ F

∙-b

-a ∙ ∙ a

∙ b

Major axis

Minor axis

A ∙

∙ Focus





Ellipse in the Form 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

General information for an ellipse in the form 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ 𝒚𝒚

𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Equation Shape Center Axis of Ellipse Graph Example

𝑥𝑥2



a > b :horizontal ellipse

(0, 0)

major axis: x-axisminor axis: y-axis

𝑥𝑥2

22+ 𝑦𝑦2

12= 1

b > a :vertical ellipse

major axis: y-axisminor axis: x-axis

𝑥𝑥2

22+ 𝑦𝑦2

32= 1

Note: If a = b, then the ellipse is a circle .

Equation Vertex Co-Vertex Focus𝑥𝑥2



a > b(-a, 0) , (a, 0) (0, b) , (0, -b)

(F, 0) , (-F, 0)

F = √𝑎𝑎𝑎𝑎2 − 𝑏𝑏𝑏𝑏2

𝑥𝑥2



b > a(0, b), (0, -b) (-a, 0) , (a, 0)

(0, F) , (0, -F)

F = √𝑏𝑏𝑏𝑏2 − 𝑎𝑎𝑎𝑎2

.

Example: Sketch the graph of 𝒙𝒙𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝟑𝟑𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 . 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1, b > a

- Vertices: (0, 3) , (0, -3) (0, b), (0, -b)

Co-vertex: (-2, 0) , (2, 0) (-a, 0), (a, 0)

- Foci: F = √𝑏𝑏𝑏𝑏2 − 𝑎𝑎𝑎𝑎2 = √32 − 22 = √5

(0, F) = (0,√5) , (0, -F) = (0, -√5) (0, F), (0, -F)

- The major axis: y-axis

- The minor axis: x-axis∙ (0, 3)

∙ (0, -3)

(-2, 0) ∙ ∙ (2, 0)

∙ (0, √5)

∙ (0, -√5)

VertexVertex

∙ b -a ∙

y

∙ a x∙ -b

∙ -b

∙ -a

∙ b

∙ -a ∙ a ∙ b

∙ -b

Co-vertex

Vertex

Vertex

Co-vertex

Co-vertex

∙ a Co-vertex

b

aF

ab ∙F



∙ b

∙ a -a ∙

∙ -b

x

y

Ellipse in the Form 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

General information for an ellipse in the form 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ 𝒚𝒚

𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


𝑥𝑥2




(0, 0)


𝑥𝑥2

22+ 𝑦𝑦2

12= 1



𝑥𝑥2

22+ 𝑦𝑦2

32= 1

Note: If a = b, then the ellipse is a circle .




a > b(-a, 0) , (a, 0) (0, b) , (0, -b)

(F, 0) , (-F, 0)


𝑥𝑥2



b > a(0, b), (0, -b) (-a, 0) , (a, 0)

(0, F) , (0, -F)


.

Example: Sketch the graph of 𝒙𝒙𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝟑𝟑𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 . 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1, b > a

- Vertices: (0, 3) , (0, -3) (0, b), (0, -b)

Co-vertex: (-2, 0) , (2, 0) (-a, 0), (a, 0)

- Foci: F = √𝑏𝑏𝑏𝑏2 − 𝑎𝑎𝑎𝑎2 = √32 − 22 = √5

(0, F) = (0,√5) , (0, -F) = (0, -√5) (0, F), (0, -F)

- The major axis: y-axis

- The minor axis: x-axis∙ (0, 3)

∙ (0, -3)

(-2, 0) ∙ ∙ (2, 0)

∙ (0, √5)

∙ (0, -√5)

VertexVertex

∙ b -a ∙

y

∙ a x∙ -b

∙ -b

∙ -a

∙ b

∙ -a ∙ a ∙ b

∙ -b

Co-vertex

Vertex

Vertex

Co-vertex

Co-vertex

∙ a Co-vertex

b

aF

ab ∙F



∙ b

∙ a -a ∙

∙ -b

x

y





Graph the Ellipse 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Procedure to graph the ellipse: 𝑥𝑥2



Steps Example: Graph 𝒙𝒙𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝟒𝟒= 𝟏𝟏𝟏𝟏.

- Rewrite in standard form: 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 𝑥𝑥2

42+ 𝑦𝑦2

22= 1

and determine a and b. a = 4, b = 2

- Check a and b �𝑎𝑎𝑎𝑎 > 𝑏𝑏𝑏𝑏: horizontal ellipse𝑏𝑏𝑏𝑏 > 𝑎𝑎𝑎𝑎: vertical ellipse a > b: 4 > 2 Horizontal ellipse

- Find the vertices: (-a, 0) , (a, 0) Vertices: (-4, 0) , (4, 0)

Find the co-vertices: (0, b) , (0, -b) Co-vertices: (0, 2) , (0, -2)

- Sketch the ellipse (passes through the vertices and co-vertices) .

Example: Sketch the graph of 9x2 + y2 = 9 .

- Standard form: 𝑥𝑥2

12+ 𝑦𝑦2

32= 1 Divide 9 by each term .

- a = 1, b = 3 𝑥𝑥2

𝑎𝑎𝑎𝑎2+ 𝑦𝑦

2


- 3 > 1 : vertical ellipse b > a

- Vertices: (0, 3) , (0, -3)

(0, 𝑏𝑏𝑏𝑏), (0, -𝑏𝑏𝑏𝑏)

Co-vertices: (-1, 0) , (1, 0) �-𝑎𝑎𝑎𝑎, 0�, (𝑎𝑎𝑎𝑎, 0)

- Sketch .

∙ (0, -2)

∙ (0, 3)

∙ (0, -3)

∙ (1, 0) (-1, 0) ∙

∙ (4, 0)

∙ (0, 2)

∙ (-4, 0)

Graph the Ellipse 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Procedure to graph the ellipse: 𝑥𝑥2



Steps Example: Graph 𝒙𝒙𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏+ 𝒚𝒚𝟐𝟐𝟐𝟐

𝟒𝟒= 𝟏𝟏𝟏𝟏.

- Rewrite in standard form: 𝑥𝑥2



42+ 𝑦𝑦2

22= 1

and determine a and b. a = 4, b = 2

- Check a and b �𝑎𝑎𝑎𝑎 > 𝑏𝑏𝑏𝑏: horizontal ellipse𝑏𝑏𝑏𝑏 > 𝑎𝑎𝑎𝑎: vertical ellipse a > b: 4 > 2 Horizontal ellipse

- Find the vertices: (-a, 0) , (a, 0) Vertices: (-4, 0) , (4, 0)

Find the co-vertices: (0, b) , (0, -b) Co-vertices: (0, 2) , (0, -2)

- Sketch the ellipse (passes through the vertices and co-vertices) .

Example: Sketch the graph of 9x2 + y2 = 9 .

- Standard form: 𝑥𝑥2

12+ 𝑦𝑦2

32= 1 Divide 9 by each term .

- a = 1, b = 3 𝑥𝑥2


2


- 3 > 1 : vertical ellipse b > a

- Vertices: (0, 3) , (0, -3)

(0, 𝑏𝑏𝑏𝑏), (0, -𝑏𝑏𝑏𝑏)

Co-vertices: (-1, 0) , (1, 0) �-𝑎𝑎𝑎𝑎, 0�, (𝑎𝑎𝑎𝑎, 0)

- Sketch .

∙ (0, -2)

∙ (0, 3)

∙ (0, -3)

∙ (1, 0) (-1, 0) ∙

∙ (4, 0)

∙ (0, 2)

∙ (-4, 0)





Ellipse in the Form (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

• General information for an ellipse of the form (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Equation Shape Center Graph Example Graph

(𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


(h, k)

(𝑥𝑥−3)2

42+ (𝑦𝑦−5)2

22= 1

(h, k) = (3, 5) a = 4, b = 2

b > a:vertical ellipse

(𝑥𝑥−3)2

22+ (𝑦𝑦−5)2

42= 1

(h, k) = (3, 5) a = 2, b = 4

• Procedure to graph the ellipse (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Steps Example: Graph (𝒙𝒙+𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏+ (𝒚𝒚−𝟑𝟑)𝟐𝟐𝟐𝟐

𝟒𝟒= 𝟏𝟏𝟏𝟏

- Write in standard form: (𝑥𝑥−ℎ)2

𝑎𝑎𝑎𝑎2 + (𝑦𝑦−𝑘𝑘)2

𝑏𝑏𝑏𝑏2= 1 [𝑥𝑥−(−2)]2

42+ (𝑦𝑦−3)2

22= 1

- Determine the center (h, k) . (h, k) = (-2, 3)

- Determine a and b. a = 4, b = 2

- Check a and b �𝑎𝑎𝑎𝑎 > 𝑏𝑏𝑏𝑏: horizontal ellipse𝑏𝑏𝑏𝑏 > 𝑎𝑎𝑎𝑎: vertical ellipse a > b: 4 > 2 horizontal ellipse

-2 + 4 -2 – 4

- Determine the vertices: (h + a, k), (h – a, k) Vertices : (2, 3), (-6, 3)Determine the co-vertices: (h, k + b), (h, k –b) Co-vertices: (-2, 5), (-2, 1)

- Sketch the ellipse .

a and b Vertex Co-Vertex Focus Graph

a > b (h + a, k), (h −a, k) (h, k+b), (h, k−b)(h + F, k), (h − F, k)


b > a (h, k+b), (h, k−b) (h + a, k), (h − a, k)(h + F, k), (h − F, k)


∙ (h, k)

∙ 0

∙ (h, k)

∙ (3, 5)

-2 ∙

∙ 2 ∙ -4 ∙ 4

∙ -2

∙ 4

∙ 2

∙ -4

∙ (3, 5)

∙ (-2, 5)

∙ (-6, 3) ∙ (2, 3)

∙ (-2, 1)

∙ (-2, 3)

∙ 0

3 + 2 3 – 2

∙ (h, k+b)

∙ (h+a, k) (h-a, k) ∙ ∙ (h, k) ∙ (h, k-b)

∙ 0

∙ (h+a, k)

∙ (h, k+b)

∙ (h, k-b)

(h-a, k) ∙

∙ 0

∙(h, k)

Ellipse in the Form (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

• General information for an ellipse of the form (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


(𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


(h, k)

(𝑥𝑥−3)2

42+ (𝑦𝑦−5)2

22= 1

(h, k) = (3, 5) a = 4, b = 2


(𝑥𝑥−3)2

22+ (𝑦𝑦−5)2

42= 1

(h, k) = (3, 5) a = 2, b = 4

• Procedure to graph the ellipse (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Steps Example: Graph (𝒙𝒙+𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏+ (𝒚𝒚−𝟑𝟑)𝟐𝟐𝟐𝟐

𝟒𝟒= 𝟏𝟏𝟏𝟏


𝑎𝑎𝑎𝑎2 + (𝑦𝑦−𝑘𝑘)2

𝑏𝑏𝑏𝑏2= 1 [𝑥𝑥−(−2)]2

42+ (𝑦𝑦−3)2

22= 1



- Check a and b �𝑎𝑎𝑎𝑎 > 𝑏𝑏𝑏𝑏: horizontal ellipse𝑏𝑏𝑏𝑏 > 𝑎𝑎𝑎𝑎: vertical ellipse a > b: 4 > 2 horizontal ellipse

-2 + 4 -2 – 4

- Determine the vertices: (h + a, k), (h – a, k) Vertices : (2, 3), (-6, 3)Determine the co-vertices: (h, k + b), (h, k –b) Co-vertices: (-2, 5), (-2, 1)

- Sketch the ellipse .




b > a (h, k+b), (h, k−b) (h + a, k), (h − a, k)(h + F, k), (h − F, k)


∙ (h, k)

∙ 0

∙ (h, k)

∙ (3, 5)

-2 ∙

∙ 2 ∙ -4 ∙ 4

∙ -2

∙ 4

∙ 2

∙ -4

∙ (3, 5)

∙ (-2, 5)

∙ (-6, 3) ∙ (2, 3)

∙ (-2, 1)

∙ (-2, 3)

∙ 0

3 + 2 3 – 2

∙ (h, k+b)

∙ (h+a, k) (h-a, k) ∙ ∙ (h, k) ∙ (h, k-b)

∙ 0

∙ (h+a, k)

∙ (h, k+b)

∙ (h, k-b)

(h-a, k) ∙

∙ 0

∙(h, k)





9-4 HYPERBOLAS

Introduction to Hyperbolas

• Hyperbola: a curve (arch) with two branches in which difference of distances of all the

points from two fixed points (foci) is a constant .A hyperbola has two arches — each one is a mirror image of the other .

Example

|𝑥𝑥2 − 𝑥𝑥1| = constant = 2a |6 − 2| = 4 (2a = 2 ∙ 2 = 4)

• Hyperbola in the real world

Ellipse Diagram

hourglass

nuclear cooling tower

• Hyperbola terminology

Note: An asymptote is a line segment whose distance to a given curve approaches zero and shows where the curve would go .

• Equations of hyperbolas


𝑎𝑎𝑎𝑎2− 𝑦𝑦2

𝑏𝑏𝑏𝑏2= 1 𝑦𝑦2

𝑎𝑎𝑎𝑎2− 𝑥𝑥2


(𝑥𝑥−ℎ)2

𝑎𝑎𝑎𝑎2− (𝑦𝑦−𝑘𝑘)2

𝑏𝑏𝑏𝑏2= 1 (𝑦𝑦−𝑘𝑘)2

𝑎𝑎𝑎𝑎2− (𝑥𝑥−ℎ)2


Term Definition Diagramfoci Two fixed points (F) inside each branch of a hyperbola

that define the curve .axis of symmetry The line segment on which a hyperbola is reflected onto itself .

Each hyperbola has two axes of symmetry that intersect at the center .vertex The points (V) where a hyperbola makes its sharpest turns .

asymptote Line segment that is approaching but never touching or crossing the hyperbola . Each hyperbola has two asymptotes .

transverse axis The line segment that passes through the vertices and foci .

∙-F

∙ x2 x1 6 ∙

∙ F -F ∙

Asymptotes

∙ V V ∙

Axis of symmetry

∙-a

∙F

∙a ∙

2 ∙-2

∙F

∙-F

2

9-4 HYPERBOLAS

Introduction to Hyperbolas

• Hyperbola: a curve (arch) with two branches in which difference of distances of all the

points from two fixed points (foci) is a constant .A hyperbola has two arches — each one is a mirror image of the other .

Example

|𝑥𝑥2 − 𝑥𝑥1| = constant = 2a |6 − 2| = 4 (2a = 2 ∙ 2 = 4)

• Hyperbola in the real world

Ellipse Diagram

hourglass

nuclear cooling tower


Note: An asymptote is a line segment whose distance to a given curve approaches zero and shows where the curve would go .

• Equations of hyperbolas



𝑏𝑏𝑏𝑏2= 1 𝑦𝑦2



(𝑥𝑥−ℎ)2


𝑏𝑏𝑏𝑏2= 1 (𝑦𝑦−𝑘𝑘)2

𝑎𝑎𝑎𝑎2− (𝑥𝑥−ℎ)2







∙-F

∙ x2 x1 6 ∙

∙ F -F ∙

Asymptotes

∙ V V ∙

Axis of symmetry

∙-a

∙F

∙a ∙

2 ∙-2

∙F

∙-F

2





Hyperbolas in the Form 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & 𝒚𝒚𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

• General information for hyperbolas of the form 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 & 𝑦𝑦2



Equation Shape Center Axis of Symmetry Graph Example

𝑥𝑥2



Horizontal (x is first)

horizontal transverse axis opens left and right

(0, 0)

y - axis𝑥𝑥2

22− 𝑦𝑦2

32= 1

𝑦𝑦2



Vertical (y is first)

vertical transverse axis opens up and down x - axis

𝑦𝑦2

22− 𝑥𝑥2

32= 1

Recall: The transverse axis is the line that passes through the vertices and foci .

Equation Vertices Foci Asymptotes Graph Example Graph

𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 (-a, 0), (a, 0)

(F, 0) , (-F, 0)

F = √𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏2y = ± 𝑏𝑏𝑏𝑏

𝑎𝑎𝑎𝑎x 𝑥𝑥2

22− 𝑦𝑦2

32= 1

𝑦𝑦2


𝑏𝑏𝑏𝑏2= 1 (0, a), (0, -a)

(0, F) , (0, -F)

F = √𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏2y = ± 𝑎𝑎𝑎𝑎

𝑏𝑏𝑏𝑏x 𝑦𝑦2

22− 𝑥𝑥2

32= 1

Tips: - The equation of the ellipse is 𝑥𝑥2


2

𝑏𝑏𝑏𝑏2= 1 (sum) .

- The equation of the hyperbola is 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 (difference) .

• Procedure to graph

Steps Example: 𝒙𝒙𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏− 𝒚𝒚𝟐𝟐𝟐𝟐

𝟒𝟒= 𝟏𝟏𝟏𝟏

- Write in standard form: 𝑥𝑥2



42− 𝑦𝑦2

22= 1


- Locate the points (-a, 0) , (a, 0) , (0, b), (0, -b) . (-4, 0) , (4, 0) , (0, 2) , (0, -2)

- Sketch a reference rectangle intersecting at abovefour points .

- Sketch the asymptotes by extending the diagonals of the rectangle .

- Determine the vertices: (-a, 0) , (a, 0) . (-4, 0) , (4, 0)

- Sketch the hyperbola . Using the vertices and asymptotes as guides to sketch . 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 ∶ opens left and right (x is first) .

y = 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎

x

y = −𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎

x

y = 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏

x

y = −𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏

x ∙ (3, 0) (-3, 0) ∙

(0, 2) ∙ ∙ (0, -2)

∙ (0, a) ∙ (0, -a)

∙ (a, 0)(-a, 0) ∙

∙ (4, 0) ∙ (-4, 0) ∙ (0, 2)

∙ (0, -2)

∙ (4, 0) (-4, 0) ∙ ∙ (0, 2)

∙ (0, -2)

(-2, 0) ∙ ∙ (0, 3) ∙ (2, 0)

∙ (0, -3)

Hyperbolas in the Form 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & 𝒚𝒚𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏







𝑥𝑥2





(0, 0)

y - axis𝑥𝑥2

22− 𝑦𝑦2

32= 1

𝑦𝑦2





𝑦𝑦2

22− 𝑥𝑥2

32= 1

Recall: The transverse axis is the line that passes through the vertices and foci .


𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 (-a, 0), (a, 0)

(F, 0) , (-F, 0)



22− 𝑦𝑦2

32= 1

𝑦𝑦2


𝑏𝑏𝑏𝑏2= 1 (0, a), (0, -a)

(0, F) , (0, -F)



22− 𝑥𝑥2

32= 1

Tips: - The equation of the ellipse is 𝑥𝑥2


2

𝑏𝑏𝑏𝑏2= 1 (sum) .

- The equation of the hyperbola is 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 (difference) .

• Procedure to graph

Steps Example: 𝒙𝒙𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏− 𝒚𝒚𝟐𝟐𝟐𝟐

𝟒𝟒= 𝟏𝟏𝟏𝟏

- Write in standard form: 𝑥𝑥2



42− 𝑦𝑦2

22= 1


- Locate the points (-a, 0) , (a, 0) , (0, b), (0, -b) . (-4, 0) , (4, 0) , (0, 2) , (0, -2)

- Sketch a reference rectangle intersecting at abovefour points .


- Determine the vertices: (-a, 0) , (a, 0) . (-4, 0) , (4, 0)

- Sketch the hyperbola . Using the vertices and asymptotes as guides to sketch . 𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 ∶ opens left and right (x is first) .


x


x


x


x ∙ (3, 0) (-3, 0) ∙

(0, 2) ∙ ∙ (0, -2)

∙ (0, a) ∙ (0, -a)

∙ (a, 0)(-a, 0) ∙

∙ (4, 0) ∙ (-4, 0) ∙ (0, 2)

∙ (0, -2)

∙ (4, 0) (-4, 0) ∙ ∙ (0, 2)

∙ (0, -2)

(-2, 0) ∙ ∙ (0, 3) ∙ (2, 0)

∙ (0, -3)





Graph the Hyperbola: 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & 𝒚𝒚𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Example: Graph 𝒚𝒚𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝟗𝟗= 𝟏𝟏𝟏𝟏 and identify the vertices, foci and the asymptotic lines .

- Write in standard form: 𝑦𝑦2

52− 𝑥𝑥2

32= 1 Standard form: 𝑦𝑦

2



- a = 5, b = 3

- Locate 4 points: (0, 5), (0, -5), (-3, 0), (3, 0) (0, a), (0, -a), (-b, 0), (b, 0)

- Sketch a reference rectangle intersecting at the four points from above .

- Sketch the asymptotes .

- Determine the vertices: (0, 5), (0, -5) (0, a) , (0, -a)

- Sketch the curve:

It opens up and down . 𝑦𝑦2


𝑏𝑏𝑏𝑏2= 1 (y is first) .

- Calculate the asymptote: y = ± 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏

x = ± 5 3x

- Calculate the foci: F = √𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏2 = √52 + 32 ≈ 5.83

(0, 5 .83), (0, -5 .83) (0, F), (0, -F)

Example: Sketch the graph of y2 – 4x2 = 4 .

- 𝑦𝑦2

4− 4𝑥𝑥2

4= 1 Divide both sides by 4 .

- 𝑦𝑦2

22− 𝑥𝑥2

12= 1 Write in standard form .

- 𝑎𝑎𝑎𝑎 = 2, 𝑏𝑏𝑏𝑏 = 1 𝑦𝑦2



- 4 Points: (0, 2) , (0, -2) , (-1, 0) , (1, 0) (0, a) , (0, -a) , (-b, 0) , (b, 0)

- Vertices: (0, 2) (0, -2) (0, a) , (0, -a)

- Sketch: It opens up and down . 𝑦𝑦2



∙ (3, 0)

∙ (0, 5)

(-3, 0) ∙

∙ (0, -5)

∙ (0, -5)

(-3, 0) ∙ ∙ (3, 0)

∙ (0, 5)

(-1, 0) ∙ ∙ (1, 0)

∙ (0, 2)

∙ (0, -2)

𝑦𝑦𝑦𝑦 = 5 3x 𝑦𝑦𝑦𝑦 = -5

3x

Graph the Hyperbola: 𝒙𝒙𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒚𝒚𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & 𝒚𝒚𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Example: Graph 𝒚𝒚𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝟗𝟗= 𝟏𝟏𝟏𝟏 and identify the vertices, foci and the asymptotic lines .

- Write in standard form: 𝑦𝑦2

52− 𝑥𝑥2

32= 1 Standard form: 𝑦𝑦

2



- a = 5, b = 3

- Locate 4 points: (0, 5), (0, -5), (-3, 0), (3, 0) (0, a), (0, -a), (-b, 0), (b, 0)

- Sketch a reference rectangle intersecting at the four points from above .

- Sketch the asymptotes .

- Determine the vertices: (0, 5), (0, -5) (0, a) , (0, -a)

- Sketch the curve:

It opens up and down . 𝑦𝑦2


𝑏𝑏𝑏𝑏2= 1 (y is first) .

- Calculate the asymptote: y = ± 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏

x = ± 5 3x

- Calculate the foci: F = √𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏2 = √52 + 32 ≈ 5.83

(0, 5 .83), (0, -5 .83) (0, F), (0, -F)

Example: Sketch the graph of y2 – 4x2 = 4 .

- 𝑦𝑦2

4− 4𝑥𝑥2

4= 1 Divide both sides by 4 .

- 𝑦𝑦2

22− 𝑥𝑥2

12= 1 Write in standard form .

- 𝑎𝑎𝑎𝑎 = 2, 𝑏𝑏𝑏𝑏 = 1 𝑦𝑦2



- 4 Points: (0, 2) , (0, -2) , (-1, 0) , (1, 0) (0, a) , (0, -a) , (-b, 0) , (b, 0)

- Vertices: (0, 2) (0, -2) (0, a) , (0, -a)

- Sketch: It opens up and down . 𝑦𝑦2



∙ (3, 0)

∙ (0, 5)

(-3, 0) ∙

∙ (0, -5)

∙ (0, -5)

(-3, 0) ∙ ∙ (3, 0)

∙ (0, 5)

(-1, 0) ∙ ∙ (1, 0)

∙ (0, 2)

∙ (0, -2)

𝑦𝑦𝑦𝑦 = 5 3x 𝑦𝑦𝑦𝑦 = -5

3x





Hyperbolas in the Form (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

• General information for a hyperbola of the form (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


(𝑥𝑥−ℎ)2



horizontal transverse axis opens left and right (x is first)

(h, k)

x = h (𝑥𝑥−5)2

32− (𝑦𝑦−3)2

22= 1

(𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

vertical transverse axis opens left and right (y is first)

y = k (𝑦𝑦−3)2

32− (𝑥𝑥−5)2

22= 1

The transverse axis: the line that passes through the vertices and foci .

Equation Vertices Foci Asymptotes(𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 (h – a, k) (h + a, k)

(h – F, k) (h + F, k)F = √𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏2

y – k = ± 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎

(x – h)

(𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 (h, k – a) (h, k + a)

(h, k – F) (h, k + F)F = √𝑎𝑎𝑎𝑎2 + 𝑏𝑏𝑏𝑏2

y – k = ± 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏

(x – h)

• Procedure to graph hyperbola

Steps Example: (𝒙𝒙+𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐

𝟗𝟗− (𝒚𝒚−𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏



𝑏𝑏𝑏𝑏2= 1 [(𝑥𝑥−(−1)]2

32− (𝑦𝑦−2)2

42= 1


- Identify a and b. a = 3, b = 4

- Determine and plot the vertices: (h – a, k), (h + a, k) (-4, 2), (2, 2)-1-3 -1+3

- Determine the up/down midpoints of a reference Move 4 units up and down from (-1, 2) .

rectangle by moving ‘b’ units up and down from the center (h, k) .

- Sketch the reference rectangle crossing at the 4points from above .


- Sketch the hyperbola . Opens to the left and right: (𝑥𝑥−ℎ)2



Determine if the curve is opening to the left and right (x is first) or up and down (y is first) . x is first

∙ 0

∙ (2, 2) (-1, 2)∙ (-4, 2) ∙

(-1, 6) ∙

x

y

(-1, -2) ∙

∙ h

-a ∙∙ -b

∙ a∙ b

k ∙

∙ 0

∙ a-a ∙∙ -b∙ h

k ∙∙ b

Hyperbolas in the Form (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


(𝑥𝑥−ℎ)2




(h, k)

x = h (𝑥𝑥−5)2

32− (𝑦𝑦−3)2

22= 1

(𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

vertical transverse axis opens left and right (y is first)

y = k (𝑦𝑦−3)2

32− (𝑥𝑥−5)2

22= 1

The transverse axis: the line that passes through the vertices and foci .


𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 (h – a, k) (h + a, k)



(x – h)

(𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 (h, k – a) (h, k + a)



(x – h)

• Procedure to graph hyperbola

Steps Example: (𝒙𝒙+𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐

𝟗𝟗− (𝒚𝒚−𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏



𝑏𝑏𝑏𝑏2= 1 [(𝑥𝑥−(−1)]2

32− (𝑦𝑦−2)2

42= 1


- Identify a and b. a = 3, b = 4

- Determine and plot the vertices: (h – a, k), (h + a, k) (-4, 2), (2, 2)-1-3 -1+3

- Determine the up/down midpoints of a reference Move 4 units up and down from (-1, 2) .

rectangle by moving ‘b’ units up and down from the center (h, k) .

- Sketch the reference rectangle crossing at the 4points from above .


- Sketch the hyperbola . Opens to the left and right: (𝑥𝑥−ℎ)2



Determine if the curve is opening to the left and right (x is first) or up and down (y is first) . x is first

∙ 0

∙ (2, 2) (-1, 2)∙ (-4, 2) ∙

(-1, 6) ∙

x

y

(-1, -2) ∙

∙ h

-a ∙∙ -b

∙ a∙ b

k ∙

∙ 0

∙ a-a ∙∙ -b∙ h

k ∙∙ b





9-5 THE GENERAL CONIC FORM

Function Transformations

• Conic Sections - summary: center at (0, 0)

Conics Standard Form Shape Graphcircle x2 + y2 = r2

parabolay = Ax2 A > 0: opens up

A < 0: opens down

x = Ay2 A > 0, opens rightA < 0, opens left

ellipse 𝑥𝑥2



a > b : horizontal ellipseb > a : vertical ellipse

hyperbola

𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 opens left and right

𝑦𝑦2


𝑏𝑏𝑏𝑏2= 1 opens up and down

• Function transformations: change the position of the graph of the function .

It includes shifting, stretching / shrinking, or reflecting graphs .

• Shifting

Function Shifting Example Diagram

y = f (x) + C

y = f (x) − C

Shift the graph of f (x) C units up .

Shift the graph of f (x) C units down .

y = x2 + 2 Shift f (x) = x2 2 units up .y = x2 − 2Shift f (x) = x2 2 units down .

y = f (x + C)

y = f (x − C)

Shift the graph of f (x) C units to the left .Shift the graph of f (x) C units to the right .

y = (x + 2)2

Shift x2 2 units to the left y = (x − 2)2

Shift x2 2 units to the right

• Reflection

Function Reflection Example Graph

y = -f (x) Reflect the graph of y = f (x) about thex-axis .

y = x2 and y = - (x2)

y = f (-x) Reflect the graph of y = f (x) about the y-axis .

y = 2x+1 andy = 2(-x)+1 = -2x+1

∙ 0 ∙ -2

∙ 2

y = x2 + 2

y = x2 – 2

y = x2

y = (x + 2)2

∙0

∙-2

y = (x −2)2

∙2

y = x2

y = x2

x

y

y = 2x +1

y = -(x2) y

x y = -2x +1

9-5 THE GENERAL CONIC FORM

Function Transformations

• Conic Sections - summary: center at (0, 0)

Conics Standard Form Shape Graphcircle x2 + y2 = r2


A < 0: opens down


ellipse 𝑥𝑥2




hyperbola

𝑥𝑥2



𝑦𝑦2



• Function transformations: change the position of the graph of the function .

It includes shifting, stretching / shrinking, or reflecting graphs .

• Shifting

Function Shifting Example Diagram

y = f (x) + C

y = f (x) − C




y = f (x + C)

y = f (x − C)


y = (x + 2)2



• Reflection


y = -f (x) Reflect the graph of y = f (x) about thex-axis .

y = x2 and y = - (x2)

y = f (-x) Reflect the graph of y = f (x) about the y-axis .

y = 2x+1 andy = 2(-x)+1 = -2x+1

∙ 0 ∙ -2

∙ 2

y = x2 + 2

y = x2 – 2

y = x2

y = (x + 2)2

∙0

∙-2

y = (x −2)2

∙2

y = x2

y = x2

x

y

y = 2x +1

y = -(x2) y

x y = -2x +1





General-Form Conic Equation

• A general-form conic equation (a second-degree equation)

EquationAx2 + Bxy + Cy2 + Dx + Ey + F = 0

• The type of conic sections can be determined from the discriminant B2 – 4AC .

• Identify the type of conic section from the sign of B2 – 4AC

B2 – 4AC The Graph is a:B2 – 4AC = 0 parabolaB2 – 4AC < 0 ellipseB2 – 4AC > 0 hyperbola

Example: Sketch the graph of 9y2 – 4x2 = 36 .

Steps Example

- Write in general conic form . A B C D E F

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 -4x2 + 0xy + 9y2 + 0∙x + 0∙y − 36 = 0

- Calculate B2 – 4AC . 02 – 4 (-4) ∙ 9 = 144 > 0 : Hyperbola

- Convert to standard form . 9y2 – 4x2 = 36 ÷ 36 both sides .

The standard equation of a hyperbola: 𝑦𝑦𝑦𝑦2

𝑎𝑎𝑎𝑎2 −𝑥𝑥2


𝑦𝑦2

4− 𝑥𝑥2

9= 36

36

- Graph . Opens up and down (y is first)𝒚𝒚𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝟑𝟑𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Vertices: (0, a), (0, -a) = (0, 2), (0, -2)

Example: Convert x + 3y2 + 6y + 1 = 0 to standard form and graph it .

Steps Example


Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 0x2 + 0xy + 3y2 + 1 ∙ x + 6y + 1 = 0

- Calculate B2 – 4AC . B2 – 4AC = 02 – 4 ∙ 0 ∙ 3 = 0 Parabola

- Convert to standard form . 3y2 + 6y + x + 1= 0

o Collect y terms on the left-hand side; collect 3y2 + 6y = -x − 1x terms & constants on the right-hand side . 3(y2 + 2y + ) = -x – 1 + 3

Factor out 3 Add 3

∙ (3, 0) (-3, 0) ∙

∙ (0, -2)

∙ (0, 2)

General-Form Conic Equation

• A general-form conic equation (a second-degree equation)

EquationAx2 + Bxy + Cy2 + Dx + Ey + F = 0

• The type of conic sections can be determined from the discriminant B2 – 4AC .

• Identify the type of conic section from the sign of B2 – 4AC

B2 – 4AC The Graph is a:B2 – 4AC = 0 parabolaB2 – 4AC < 0 ellipseB2 – 4AC > 0 hyperbola

Example: Sketch the graph of 9y2 – 4x2 = 36 .

Steps Example


Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 -4x2 + 0xy + 9y2 + 0∙x + 0∙y − 36 = 0

- Calculate B2 – 4AC . 02 – 4 (-4) ∙ 9 = 144 > 0 : Hyperbola

- Convert to standard form . 9y2 – 4x2 = 36 ÷ 36 both sides .

The standard equation of a hyperbola: 𝑦𝑦𝑦𝑦2

𝑎𝑎𝑎𝑎2 −𝑥𝑥2


𝑦𝑦2

4− 𝑥𝑥2

9= 36

36

- Graph . Opens up and down (y is first)𝒚𝒚𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐− 𝒙𝒙𝟐𝟐𝟐𝟐

𝟑𝟑𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

Vertices: (0, a), (0, -a) = (0, 2), (0, -2)

Example: Convert x + 3y2 + 6y + 1 = 0 to standard form and graph it .

Steps Example


Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 0x2 + 0xy + 3y2 + 1 ∙ x + 6y + 1 = 0

- Calculate B2 – 4AC . B2 – 4AC = 02 – 4 ∙ 0 ∙ 3 = 0 Parabola

- Convert to standard form . 3y2 + 6y + x + 1= 0

o Collect y terms on the left-hand side; collect 3y2 + 6y = -x − 1x terms & constants on the right-hand side . 3(y2 + 2y + ) = -x – 1 + 3

Factor out 3 Add 3

∙ (3, 0) (-3, 0) ∙

∙ (0, -2)

∙ (0, 2)





o Completing the square . 3(y2 + 2y + 1) = -x – 1 + 3 ∙ 1 �𝑏𝑏𝑏𝑏2�2

= �22�

2= 1

= �𝑏𝑏𝑏𝑏2�2 1 should be multiplied by 3 on the right-hand side .

3(y + 1)2 = 2 – x Isolate x.

Standard form of a parabola: x = A(y – k)2 + h x = -3(y + 1)2 + 2 = -3[y – (-1)2] + 2

A k h

- Graph . A = -3 < 0: opens leftVertex: (h, k) = (2, -1)Symmetry: y = k = -1

(Find a few more points .) y x = -3(y + 1)2 + 2 (x, y) 0 x = -3(0 + 1)2 + 2 = -1 (-1, 0)

-2 x = -3(-2 + 1)2 + 2 = -1 (-1, -2)

Example: Convert 16x2 + 9y2 – 64x – 18y – 71 = 0 to standard form and graph it .

Steps Example


Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 16x2 + 0xy + 9y2 – 64x – 18y –71 = 0

- Calculate B2 – 4AC. B2 – 4AC = 02 – 4 ∙ 16 ∙ 9= - 576 < 0 Ellipse

- Convert to standard form bycompleting the square . (16x2 – 64x ) + (9y2 – 18y ) –71 = 0 Regroup .

= �𝑏𝑏𝑏𝑏2�2

16(x2 – 4x + ) + 9(y2 – 2y + ) = 71 + 16 + 9 Factor out 16 and 9 . Add 71 ; add 16 , 9

16(x2 – 4x + 4) + 9(y2 – 2y + 1) = 71 + 16 ∙ 4 + 9 ∙ 1 �-4

2�2

= 4 , �-22�2

= 14 and 1 should be multiplied by 16 and 9 on the right-hand side .

16(x – 2)2 + 9(y –1)2 = 144 ÷ 144 both sides .

(𝑥𝑥−2)2

9+ (𝑦𝑦−1)2

16= 144

14416144

= 19

; 9144

= 116

Standard form of an ellipse: (𝑥𝑥−ℎ)2


𝑏𝑏𝑏𝑏2= 1 (𝒙𝒙−𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐

𝟑𝟑𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐

𝟒𝟒𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 b > a

- Graph . - Center: (h, k) = (2, 1)

- Vertices: (h, k ± b) = (2, 5) , (2, -3)

- Co-vertices: (h ± a, k) = (5, 1) , (-1, 1)

∙ (2, -1)

y

x(-1, 0) ∙

∙ (-1, -2)

∙ (2, 1)

x

y

0

2+3 2-3

1+4 1-4

∙ (2, 5)

∙ (2, -3)

(-1, 1) ∙ ∙ (5, 1)





9-6 NONLINEAR SYSTEMS OF EQUATIONS

Nonlinear Systems

• Nonlinear equation: the highest power of the variable is higher than Example

one (an equation whose graph is not a straight line) . 2x2 + y = 3

• Nonlinear system of equations: a system in which the highest power

of the variable is higher than one .

• Solutions of the systems of equations: the particular values of the variables in the system

that make the system true .

Example: Solve the system .

x2 + y2 = 25 (1) x – y + 1 = 0 (2)

Solution: x = y – 1 (3) Solve for x in (2) .

(y – 1)2 + y2 = 25 (3) (1) Substitute y –1 for x in (1) .

y2 – 2y + 1 + y2 = 25 (a – b)2 = a2 – 2ab + b2

2y2 – 2y – 24 = 0

2(y2 – y – 12) = 0 Factor out 2 .

2 (y – 4) (y + 3) = 0 Factor .

(y – 4) = 0 (y + 3) = 0 Zero-product property

y = 4 y = -3 x = y – 1 x = y – 1 Substitute 4 & -3 for y in (3) .

= 4 – 1 = 3 = -3 – 1 = -4Solution sets: (3, 4) (-4, -3)

Check: (x, y) x2 + y2 = 25 x – y + 1 = 0

(3, 4) 32 + 42 = 25 3 – 4 + 1 = 0 9 + 16 = 25 √ 0 = 0 √

? ?

(-4, -3) (-4)2 + (-3)2 = 25 -4 – (-3) + 1 = 0 16 + 9 = 25 0 = 0 Correct!

• Check by graphing: Graph the equations in the system . The point(s) of intersection in the

graph are the solutions to the system .

x

∙ (3, 4)

∙ (-4, -3)

y x – y + 1= 0

x2 + y2 = 25

? ?

√

√

The graphs intersect at the points (3, 4) and (-4, -3), correct!

9-6 NONLINEAR SYSTEMS OF EQUATIONS

Nonlinear Systems

• Nonlinear equation: the highest power of the variable is higher than Example

one (an equation whose graph is not a straight line) . 2x2 + y = 3

• Nonlinear system of equations: a system in which the highest power

of the variable is higher than one .




x2 + y2 = 25 (1) x – y + 1 = 0 (2)

Solution: x = y – 1 (3) Solve for x in (2) .

(y – 1)2 + y2 = 25 (3) (1) Substitute y –1 for x in (1) .

y2 – 2y + 1 + y2 = 25 (a – b)2 = a2 – 2ab + b2

2y2 – 2y – 24 = 0

2(y2 – y – 12) = 0 Factor out 2 .

2 (y – 4) (y + 3) = 0 Factor .

(y – 4) = 0 (y + 3) = 0 Zero-product property

y = 4 y = -3 x = y – 1 x = y – 1 Substitute 4 & -3 for y in (3) .

= 4 – 1 = 3 = -3 – 1 = -4Solution sets: (3, 4) (-4, -3)

Check: (x, y) x2 + y2 = 25 x – y + 1 = 0

(3, 4) 32 + 42 = 25 3 – 4 + 1 = 0 9 + 16 = 25 √ 0 = 0 √

? ?

(-4, -3) (-4)2 + (-3)2 = 25 -4 – (-3) + 1 = 0 16 + 9 = 25 0 = 0 Correct!

• Check by graphing: Graph the equations in the system . The point(s) of intersection in the

graph are the solutions to the system .

x

∙ (3, 4)

∙ (-4, -3)

y x – y + 1= 0

x2 + y2 = 25

? ?

√

√

The graphs intersect at the points (3, 4) and (-4, -3), correct!





Solving Nonlinear Systems


9x2 + 4y2 = 9 (1) y – x = 2 (2)

Solution: y = x + 2 (3) Solve for y in (2) .

9x2 + 4(x + 2)2 = 9 (3) (1) Substitute x + 2 for y in (1) .

9x2 + 4(x2 + 4x + 4) = 9 (a + b)2 = a2 + 2ab + b2

9x2 + 4x2 + 16x + 16 = 9

13x2 + 16x + 7 = 0

x = -16 ± �162−4(13)(7)2 ∙ 13

𝑥𝑥 = -𝑏𝑏𝑏𝑏 ± √𝑏𝑏𝑏𝑏2−4𝑎𝑎𝑎𝑎𝑎𝑎2𝑎𝑎𝑎𝑎

=-16 ± �-108

26

=-16 ± ��-1�27∙ 4

26

= -16 ± 2√27 𝑖𝑖26

x = -𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

± √𝟐𝟐𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏

i i = �-1

x = -813

+ √2713

i x = -813

– √2713

i

y = -813

+ √27 13

𝑖𝑖 + 2 y = -813

– √2713

𝑖𝑖 + 2 Substitute x in (3): y = x + 2 .

= -813

+ √27 13

𝑖𝑖 + 2613

= -813

– √27 13

𝑖𝑖 + 2613

y = 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

+ √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊 y = 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

– √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊

Solution sets: � -𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

+ √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊 , 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

+ √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊�

�−𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

– √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊 , 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

– √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊�These are imaginary solutions . The two graphs do not intersect .

Solving Nonlinear Systems


9x2 + 4y2 = 9 (1) y – x = 2 (2)

Solution: y = x + 2 (3) Solve for y in (2) .

9x2 + 4(x + 2)2 = 9 (3) (1) Substitute x + 2 for y in (1) .

9x2 + 4(x2 + 4x + 4) = 9 (a + b)2 = a2 + 2ab + b2

9x2 + 4x2 + 16x + 16 = 9

13x2 + 16x + 7 = 0

x = -16 ± �162−4(13)(7)2 ∙ 13

𝑥𝑥 = -𝑏𝑏𝑏𝑏 ± √𝑏𝑏𝑏𝑏2−4𝑎𝑎𝑎𝑎𝑎𝑎2𝑎𝑎𝑎𝑎

=-16 ± �-108

26

=-16 ± ��-1�27∙ 4

26

= -16 ± 2√27 𝑖𝑖26

x = -𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

± √𝟐𝟐𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏

i i = �-1

x = -813

+ √2713

i x = -813

– √2713

i

y = -813

+ √27 13

𝑖𝑖 + 2 y = -813

– √2713

𝑖𝑖 + 2 Substitute x in (3): y = x + 2 .

= -813

+ √27 13

𝑖𝑖 + 2613

= -813

– √27 13

𝑖𝑖 + 2613

y = 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

+ √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊 y = 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

– √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊

Solution sets: � -𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

+ √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊 , 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

+ √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊�

�−𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

– √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊 , 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏𝟏𝟏

– √𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏

𝒊𝒊�These are imaginary solutions . The two graphs do not intersect .





Unit 9 Summary

• Distance and midpoint formulasDistance Formula Example

𝑑𝑑 = �(𝑥𝑥2 − 𝑥𝑥1)2+(𝑦𝑦𝑦𝑦2 − 𝑦𝑦𝑦𝑦1)2(x1, y1) = (1, 2) , (x2, y2) = (3, 3)𝑑𝑑 = �(3 − 1)2+(3 − 2)2 = √4 + 1 = √5

Midpoint Formula Example


2 ,

𝑦𝑦𝑦𝑦1 + 𝑦𝑦𝑦𝑦22

�(x1, y1) = (4, 3) , (x2, y2) = (-4, 1)

Midpoint = �4+�-4�2

, 3+12� = (𝟎𝟎,𝟐𝟐𝟐𝟐)

• Equation of circlesCenter of a Circle The Standard Form Equation Example

center at origin (0, 0) x2 + y2 = r2 x2 + y2 = 32 r = 3


(h, k) = (2, 1) , r = 2 r – radius

• The general form equation for a circleThe General Form Example

x2 + y2 + Cx + Dy + E = 0 x2 + y2 – 2x + 4y – 20 = 0

• Parabola terminologyTerm Definition Diagramfocus A fixed point whose relationship with a directrix






• The coefficient A in 𝒇𝒇(𝒙𝒙) = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 can shrink or stretch the parabolaThe Coefficient A in 𝒚𝒚 = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 & 𝒙𝒙 = 𝑨𝑨𝒚𝒚𝟐𝟐𝟐𝟐 Example

The larger the |𝐴𝐴|, the narrower the curve .


A > 0

The smaller the |𝐴𝐴|, the wider the curve .

A < 0

∙ F (focus)

Directrix

Axis of symmetry

∙ Vertex

y = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2

y = x2

y = 3x2

y = - 3x2y = -x2

y = - 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2

Unit 9 Summary

• Distance and midpoint formulasDistance Formula Example

𝑑𝑑 = �(𝑥𝑥2 − 𝑥𝑥1)2+(𝑦𝑦𝑦𝑦2 − 𝑦𝑦𝑦𝑦1)2(x1, y1) = (1, 2) , (x2, y2) = (3, 3)𝑑𝑑 = �(3 − 1)2+(3 − 2)2 = √4 + 1 = √5

Midpoint Formula Example


2 ,

𝑦𝑦𝑦𝑦1 + 𝑦𝑦𝑦𝑦22

�(x1, y1) = (4, 3) , (x2, y2) = (-4, 1)

Midpoint = �4+�-4�2

, 3+12� = (𝟎𝟎,𝟐𝟐𝟐𝟐)

• Equation of circlesCenter of a Circle The Standard Form Equation Example

center at origin (0, 0) x2 + y2 = r2 x2 + y2 = 32 r = 3


(h, k) = (2, 1) , r = 2 r – radius

• The general form equation for a circleThe General Form Example

x2 + y2 + Cx + Dy + E = 0 x2 + y2 – 2x + 4y – 20 = 0

• Parabola terminologyTerm Definition Diagramfocus A fixed point whose relationship with a directrix






• The coefficient A in 𝒇𝒇(𝒙𝒙) = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 can shrink or stretch the parabolaThe Coefficient A in 𝒚𝒚 = 𝑨𝑨𝒙𝒙𝟐𝟐𝟐𝟐 & 𝒙𝒙 = 𝑨𝑨𝒚𝒚𝟐𝟐𝟐𝟐 Example

The larger the |𝐴𝐴|, the narrower the curve .


A > 0

The smaller the |𝐴𝐴|, the wider the curve .

A < 0

∙ F (focus)

Directrix

Axis of symmetry

∙ Vertex

y = 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2

y = x2

y = 3x2

y = - 3x2y = -x2

y = - 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐x2





• Equations of parabolas


y = Ax2 y - axis

(0, 0)

A > 0: opens up

A < 0: opens down


A < 0, opens left


C > 0, C units up

C < 0, C units down







h > 0: h units up

h < 0: h units down



y = Ax2 + Bx + C 𝑥𝑥 = −𝐵𝐵2𝐴𝐴

(-B2A , f �

-B2A� )

A > 0: opens up

A < 0: opens down

x = Ay2 + By + C 𝑦𝑦𝑦𝑦 =−𝐵𝐵2𝐴𝐴

(f �-B2A� ,

-B2A )

A > 0: opens right

A < 0: opens left


Term Definition Diagram

foci Two fixed points (F) inside of an ellipse that define the curve .

major axis The longest diameter of the ellipse .(The longer axis and passes through both foci)


vertex The point where an ellipse makes its sharpest turn .(On the major axis)

∙ (0, C)

∙ (0, -C)

∙ (C, 0)

∙ (h, 0)

(h, 0) ∙

∙ (0, h)

∙ (0, h)

(C, 0) ∙

∙ (h, k)

∙ (h, k)

∙ F ∙ F

∙-b

-a ∙ ∙ a

∙ b

Major axis

Minor axis





• General information for an ellipse in the form 𝑥𝑥2


2



𝑥𝑥2




(0, 0)


𝑥𝑥2

22+ 𝑦𝑦2

12= 1



𝑥𝑥2

22+ 𝑦𝑦2

32= 1




a > b(-a, 0) , (a, 0) (0, b) , (0, -b)

(F, 0) , (-F, 0)

F = √𝑎𝑎𝑎𝑎2 − 𝑏𝑏𝑏𝑏2𝑥𝑥2



b > a(0, b), (0, -b) (-a, 0) , (a, 0)

(0, F) , (0, -F)


• General information for an ellipse of the form (𝑥𝑥−ℎ)2




(𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐+ (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


(h, k)

(𝑥𝑥−3)2

42+ (𝑦𝑦−5)2

22= 1

(h, k) = (3, 5) a = 4, b = 2


(𝑥𝑥−3)2

22+ (𝑦𝑦−5)2

42= 1

(h, k) = (3, 5) a = 2, b = 4





b > a (h, k+b), (h, k−b) (h + a, k), (h − a, k)(h, k + F), (h, k − F)







Axis of symmetry

∙ (h, k+b)

∙ (h+a, k) (h-a, k) ∙ ∙ (h, k) ∙ (h, k-b)

∙ 0

∙ (h+a, k)

∙ (h, k+b)

∙ (h, k-b)

(h-a, k) ∙

∙ 0

∙(h, k)

∙ F -F ∙

Asymptotes

∙ V V ∙

VertexVertex ∙ -a ∙ a ∙ b

∙ -b

Co-vertex

Co-vertex

Vertex∙ -b

∙ -a

∙ b

Co-vertex

Vertex

∙ a Co-vertex

∙ (h, k)

∙ 0

∙ (h, k)

∙ (3, 5)

-2 ∙

∙ 2 ∙ -4 ∙ 4

∙ -2

∙ 4

∙ 2

∙ -4

∙ (3, 5)

∙ b -a ∙

y

∙ a x∙ -b

∙ b

∙ a -a ∙

∙ -b

x

y











𝑥𝑥2





(0, 0)

y - axis𝑥𝑥2

22− 𝑦𝑦2

32= 1

𝑦𝑦2





𝑦𝑦2

22− 𝑥𝑥2

32= 1


𝑥𝑥2


𝑏𝑏𝑏𝑏2= 1 (-a, 0), (a, 0)

(F, 0) , (-F, 0)



22− 𝑦𝑦2

32= 1

𝑦𝑦2


𝑏𝑏𝑏𝑏2= 1 (0, a), (0, -a)

(0, F) , (0, -F)



22− 𝑥𝑥2

32= 1


𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 & (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏


(𝑥𝑥−ℎ)2




(h, k)

x = h (𝑥𝑥−5)2

32− (𝑦𝑦−3)2

22= 1

(𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏

vertical transverse axis opens left and right

(y is first)y = k (𝑦𝑦−3)2

32− (𝑥𝑥−5)2

22= 1


𝒂𝒂𝟐𝟐𝟐𝟐− (𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 (h – a, k) (h + a, k)



(x – h)

(𝒚𝒚−𝒌𝒌)𝟐𝟐𝟐𝟐

𝒂𝒂𝟐𝟐𝟐𝟐− (𝒙𝒙−𝒉𝒉)𝟐𝟐𝟐𝟐

𝒃𝒃𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏 (h, k – a) (h, k + a)



(x – h)

• Summary of conic sections: center at (0, 0) Conics Standard Form Shape Graphcircle x2 + y2 = r2


A < 0: opens down


ellipse 𝑥𝑥2




hyperbola

𝑥𝑥2



𝑦𝑦2




x


x


x


x ∙ (3, 0) (-3, 0) ∙

(0, 2) ∙ ∙ (0, -2)

(-2, 0) ∙ ∙ (0, 3) ∙ (2, 0)

∙ (0, -3)

∙ (0, a) ∙ (0, -a)

∙ (a, 0)(-a, 0) ∙

∙ h

-a ∙∙ -b

∙ a∙ b

k ∙

∙ 0

∙ a-a ∙∙ -b∙ h

k ∙∙ b





• Function transformations: change the position of the graph of the function . It includes shifting, stretching / shrinking, or reflecting graphs .

• ShiftingFunction Shifting Example Diagram

y = f (x) + C

y = f (x) − C




y = f (x + C)

y = f (x − C)


y = (x + 2)2



• ReflectionFunction Reflection Example Graph

y = -f (x) Reflect the graph of y = f (x) about thex –axis .

y = x2 and y = - (x2)

y = f (-x) Reflect the graph of y = f (x) about the y –axis .

y = 2x+1 andy = 2(-x)+1 = -2x+1

• A general-form conic equation (a second - degree equation)Equation

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

• Identify the type of conic section from the sign of B2 – 4ACB2 – 4AC The Graph is a:

B2 – 4AC = 0 parabolaB2 – 4AC < 0 ellipseB2 – 4AC > 0 hyperbola

• Nonlinear system of equations: a system in which the highest power of the variable is

higher than one .



y = x2

x

y

y = 2x +1

y = -(x2) y

x y = -2x +1

∙ 0 ∙ -2

∙ 2 y = x2+2

y = x2-2

y = x2

y = (x + 2)2

∙0

∙-2

y = (x −2)2

∙2

y = x2





PRACTICE QUIZ

Unit 9 Conics

1. Identify the center and radius of the following circle .

x2 + y2 – 4x + 10y + 20 = 0

2. Sketch the graph of f (x) = x2 – 8x + 12 .

3. Sketch the graph of (𝑥𝑥+1)2

25+ (𝑦𝑦−2)2

9= 1 .

4. Sketch the graph of y2 – 9x2 = 9 .

5. Convert 4x2 + 25y2 + 24x – 50y – 39 = 0 to standard form and graph it .

6. Solve the system .

x2 + y2 = 10 2x + y = 1

Page 14

PRACTICE QUIZ

Unit 9 Conics

1. Identify the center and radius of the following circle .

x2 + y2 – 4x + 10y + 20 = 0

2. Sketch the graph of f (x) = x2 – 8x + 12 .

3. Sketch the graph of (𝑥𝑥+1)2

25+ (𝑦𝑦−2)2

9= 1 .

4. Sketch the graph of y2 – 9x2 = 9 .

5. Convert 4x2 + 25y2 + 24x – 50y – 39 = 0 to standard form and graph it .

6. Solve the system .

x2 + y2 = 10 2x + y = 1

Page 14




Algebra I & II Key Concepts, Practice, and Quizzes Unit 10 – Exponential and Logarithmic Functions

UNIT 10 EXPONENTIAL & LOGARITHMIC FUNCTIONS

10-1 EXPONENTIAL FUNCTIONS

Introduction to Exponential Functions

• An exponential function: a function in which the independent variable appears as an exponent .

Exponential Function Example

𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 � 𝑎𝑎 − base 𝑎𝑎 > 0, 𝑎𝑎 ≠ 1 𝑥𝑥𝑥𝑥 − independent variable any real number 𝑓𝑓(𝑥𝑥𝑥𝑥) − function

𝑓𝑓(𝑥𝑥𝑥𝑥) = 5𝑥𝑥𝑥𝑥

Note: - If a = 1: 𝑓𝑓(𝑥𝑥𝑥𝑥) = 1𝑥𝑥𝑥𝑥 = 1

- If a < 0: example: 𝑓𝑓(𝑥𝑥𝑥𝑥) = (-3)𝑥𝑥𝑥𝑥 = (-3)12 = �-3 This is not a real number .

• Graph 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙

Steps Example: Graph y = 3x and y = 3-x

- Make a table . x 0 1 2 -1 -2y = 3x 1 3 9 1

319 3-2 = 1

32= 1

9

y = 3-x 1 13 1

93 9 3-(-1) = 3 ; 3-(-2) = 9

- Plot points .

- Connect points with a smooth curve .Tip: The x-axis is the asymptote .

Note: The graph of f (x) = 3-x is a mirror image or reflection of f (x) = 3x about the y-axis .

• Calculator tip: Use ^ or yx key .

Example: 1. 3π ≈ 31.54 3 yx π = or 3 ^ 2nd π ENTER

2. (�5)√2 ≈ 3.12 √ 5 yx √ 2 = or 2nd √ 5 ^ 2nd √ 2 ENTER

y

∙ (2, 9)

∙ (1, 3)

(-1, 13) ∙ x

∙ (0, 1)

(-2, 9) ∙

(-1, 3) ∙

∙(1, 1

3) ∙ (2, 1

9)

Page 10-1


10-1 EXPONENTIAL FUNCTIONS

Introduction to Exponential Functions

• An exponential function: a function in which the independent variable appears as an exponent .

Exponential Function Example

𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 � 𝑎𝑎 − base 𝑎𝑎 > 0, 𝑎𝑎 ≠ 1 𝑥𝑥𝑥𝑥 − independent variable any real number 𝑓𝑓(𝑥𝑥𝑥𝑥) − function


Note: - If a = 1: 𝑓𝑓(𝑥𝑥𝑥𝑥) = 1𝑥𝑥𝑥𝑥 = 1

- If a < 0: example: 𝑓𝑓(𝑥𝑥𝑥𝑥) = (-3)𝑥𝑥𝑥𝑥 = (-3)12 = �-3 This is not a real number .

• Graph 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙

Steps Example: Graph y = 3x and y = 3-x

- Make a table . x 0 1 2 -1 -2y = 3x 1 3 9 1

319 3-2 = 1

32= 1

9

y = 3-x 1 13 1

93 9 3-(-1) = 3 ; 3-(-2) = 9

- Plot points .

- Connect points with a smooth curve .Tip: The x-axis is the asymptote .

Note: The graph of f (x) = 3-x is a mirror image or reflection of f (x) = 3x about the y-axis .

• Calculator tip: Use ^ or yx key .

Example: 1. 3π ≈ 31.54 3 yx π = or 3 ^ 2nd π ENTER

2. (�5)√2 ≈ 3.12 √ 5 yx √ 2 = or 2nd √ 5 ^ 2nd √ 2 ENTER

y

∙ (2, 9)

∙ (1, 3)

(-1, 13) ∙ x

∙ (0, 1)

(-2, 9) ∙

(-1, 3) ∙

∙(1, 1

3) ∙ (2, 1

9)

Page 10-1






Characteristics of Exponential Functions

• The graph of a typical exponential function

The graph is asymptotic to the x-axis as x approaches ± ∞ .

• Characteristics of exponential functions

Characteristic 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂-𝒙𝒙𝒙𝒙

growth / decay The graph increases (grows) from left to right .

The graph falls (decays) from left to right .

example


Exponential growth

𝑓𝑓(𝑥𝑥𝑥𝑥) = 3-𝑥𝑥𝑥𝑥

Exponential decay

asymptotex-axis (y = 0)

The curve is very close but never touches the x-axis asx approaches -∞.

x-axis (y = 0)The curve is very close but never touches the x-axis as x approaches +∞ .

y - intercept y = 1Curve always passes through (0,1) .

domain x values x = All real numbers or x = (-∞, ∞)

range y values y = (0, ∞) or { y | y > 0 }All positive real numbers (graph is always above the x – axis) .

• Stretching or shifting

Function Stretch or Shrink Example Graph

𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙The larger the a, the narrower the curve .The smaller the a, the wider the curve .

𝑓𝑓(𝑥𝑥𝑥𝑥) = 4𝑥𝑥𝑥𝑥and


• Reflecting (mirror image)


𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂-𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the y – axis .

𝑓𝑓(𝑥𝑥𝑥𝑥) = 2𝑥𝑥𝑥𝑥 and𝑓𝑓(𝑥𝑥𝑥𝑥) = 2-𝑥𝑥𝑥𝑥

𝒇𝒇(𝒙𝒙𝒙𝒙) = -𝒂𝒂𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the x – axis .

𝑓𝑓(𝑥𝑥𝑥𝑥) = 2𝑥𝑥𝑥𝑥 and

𝑓𝑓(𝑥𝑥𝑥𝑥) = -2𝑥𝑥𝑥𝑥

x

y

∙

f (x) = axf (x) = a -x

1

f (x) = 2x

f (x) = - 2-x

1∙

(0, 1) ∙ (0, 1) ∙

f (x) = 4x

f (x) = 2x

1∙

f (x) = 2-x f (x) = 2x

01∙

Page 10-2

Characteristics of Exponential Functions

• The graph of a typical exponential function

The graph is asymptotic to the x-axis as x approaches ± ∞ .





example


Exponential growth


Exponential decay

asymptotex-axis (y = 0)

The curve is very close but never touches the x-axis asx approaches -∞.

x-axis (y = 0)The curve is very close but never touches the x-axis as x approaches +∞ .

y - intercept y = 1Curve always passes through (0,1) .

domain x values x = All real numbers or x = (-∞, ∞)

range y values y = (0, ∞) or { y | y > 0 }All positive real numbers (graph is always above the x – axis) .

• Stretching or shifting

Function Stretch or Shrink Example Graph




• Reflecting (mirror image)


𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂-𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the y – axis .


𝒇𝒇(𝒙𝒙𝒙𝒙) = -𝒂𝒂𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the x – axis .



x

y

∙

f (x) = axf (x) = a -x

1

f (x) = 2x

f (x) = - 2-x

1∙

(0, 1) ∙ (0, 1) ∙

f (x) = 4x

f (x) = 2x

1∙

f (x) = 2-x f (x) = 2x

01∙

Page 10-2





Transformations of Exponential Functions

• Transformations of exponential functions: change the position of the graph of the

exponential function . It includes shifting, stretching / shrinking, or reflecting for exponential

functions .

• Shifting

Exponential Function Shifting Example Graph

𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙 + 𝑪𝑪

𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙 − 𝑪𝑪

Shift the graph of y = 𝑎𝑎𝑥𝑥𝑥𝑥 C units up .

Shift the graph of y = 𝑎𝑎𝑥𝑥𝑥𝑥 C units down .

𝑓𝑓(𝑥𝑥𝑥𝑥) = 2𝑥𝑥𝑥𝑥 + 1

𝑓𝑓(𝑥𝑥𝑥𝑥) = 2𝑥𝑥𝑥𝑥 − 1

𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙+𝑪𝑪

𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂𝒙𝒙𝒙𝒙−𝑪𝑪Shift the graph of y = 𝑎𝑎𝑥𝑥𝑥𝑥 C units to the left .

Shift the graph of y = 𝑎𝑎𝑥𝑥𝑥𝑥 C units to the right .

𝑓𝑓(𝑥𝑥𝑥𝑥) = 2𝑥𝑥𝑥𝑥+1

𝑓𝑓(𝑥𝑥𝑥𝑥) = 2𝑥𝑥𝑥𝑥−1

Example: Sketch the graph of 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐 .

- Make a table .

x 0 1 2 -1 -2

𝒚𝒚𝒚𝒚 = 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐 30 + 2 = 3 31 + 2 = 5 32 + 2 = 11 3-1 + 2 =13 + 2

≈ 2.333−2 + 2 =

132 + 2

≈ 2.11(𝒙𝒙𝒙𝒙,𝒚𝒚𝒚𝒚) (0, 3) (1, 5) (2, 11) (-1, 2 .33) (-2, 2 .11)

- Plot the points and connect them with a smooth curve .Note: The graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 3𝑥𝑥𝑥𝑥 + 2 has the same shape as of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 3𝑥𝑥𝑥𝑥 but is shifted 2 units up . The asymptote is y = 2 .

∙ (0, 3)

∙ (1, 5)

∙ (2, 11)

∙ (-1, 2 .33) ∙ (-2, 2 .11) y = 2

x

y

0

f(x) = 3xf(x) = 3x+ 2

f(x) = 2x

f(x) = 2x+1

f(x) = 2x-1

0

f(x) = 2x + 1

f(x) = 2x – 1

f(x) = 2x

0

Page 10-3

Transformations of Exponential Functions

• Transformations of exponential functions: change the position of the graph of the

exponential function . It includes shifting, stretching / shrinking, or reflecting for exponential

functions .

• Shifting

Exponential Function Shifting Example Graph












Example: Sketch the graph of 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐 .

- Make a table .

x 0 1 2 -1 -2

𝒚𝒚𝒚𝒚 = 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟐𝟐𝟐𝟐 30 + 2 = 3 31 + 2 = 5 32 + 2 = 11 3-1 + 2 =13 + 2

≈ 2.333−2 + 2 =

132 + 2

≈ 2.11(𝒙𝒙𝒙𝒙,𝒚𝒚𝒚𝒚) (0, 3) (1, 5) (2, 11) (-1, 2 .33) (-2, 2 .11)

- Plot the points and connect them with a smooth curve .Note: The graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 3𝑥𝑥𝑥𝑥 + 2 has the same shape as of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 3𝑥𝑥𝑥𝑥 but is shifted 2 units up . The asymptote is y = 2 .

∙ (0, 3)

∙ (1, 5)

∙ (2, 11)

∙ (-1, 2 .33) ∙ (-2, 2 .11) y = 2

x

y

0

f(x) = 3xf(x) = 3x+ 2

f(x) = 2x

f(x) = 2x+1

f(x) = 2x-1

0

f(x) = 2x + 1

f(x) = 2x – 1

f(x) = 2x

0

Page 10-3





Graphing Exponential Functions

Example: Sketch the graph of 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝟒𝟒𝒙𝒙𝒙𝒙 amd 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝟒𝟒-𝒙𝒙𝒙𝒙 .

- Make a table .

x 0 1 2 -1 -2

𝒚𝒚𝒚𝒚 = 𝟒𝟒𝒙𝒙𝒙𝒙 40 = 1 41 = 4 42 = 16 4-1 = 14 4-2 = 1

42= 1

16

(𝒙𝒙𝒙𝒙,𝒚𝒚𝒚𝒚) (0, 1) (1, 4) (2, 16) (-1, 14� (-2, 1

16)

- Plot 𝑓𝑓(𝑥𝑥𝑥𝑥) = 4𝑥𝑥𝑥𝑥 .

- Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 4x about the y – axis for 𝑓𝑓(𝑥𝑥𝑥𝑥) = 4-𝑥𝑥𝑥𝑥 .

X and Y interchanging

Function Shape Example Graph

𝒚𝒚𝒚𝒚 = 𝒂𝒂𝒙𝒙𝒙𝒙 and

𝒙𝒙𝒙𝒙 = 𝒂𝒂𝒚𝒚𝒚𝒚

Reflect the graph of 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥about the line y = x to get

𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦𝑦𝑦𝑦𝑦 = 2𝑥𝑥𝑥𝑥 and𝑥𝑥𝑥𝑥 = 2𝑦𝑦

Example: Sketch the graph of 𝒙𝒙𝒙𝒙 = 𝟒𝟒𝒚𝒚𝒚𝒚 .

- Sketch 𝑦𝑦𝑦𝑦 = 4𝑥𝑥𝑥𝑥 .

- Plot the line y = x .

- Reflect the graph of 𝑦𝑦𝑦𝑦 = 4𝑥𝑥𝑥𝑥 about the

line y = x to get the graph of 𝑥𝑥𝑥𝑥 = 4𝑦𝑦 .

∙ (0, 1)

∙ (1, 4)

∙ (2, 16)

(-1, 14�

∙ ∙ x

y

(-2, 116�

f(x) = 4xf(x) = 4-x

y = x

y = 2x

x = 2y

0

x

y y = 4x

x = 4y

y = x

0

Page 10-4

Graphing Exponential Functions

Example: Sketch the graph of 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝟒𝟒𝒙𝒙𝒙𝒙 amd 𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝟒𝟒-𝒙𝒙𝒙𝒙 .

- Make a table .

x 0 1 2 -1 -2

𝒚𝒚𝒚𝒚 = 𝟒𝟒𝒙𝒙𝒙𝒙 40 = 1 41 = 4 42 = 16 4-1 = 14 4-2 = 1

42= 1

16

(𝒙𝒙𝒙𝒙,𝒚𝒚𝒚𝒚) (0, 1) (1, 4) (2, 16) (-1, 14� (-2, 1

16)

- Plot 𝑓𝑓(𝑥𝑥𝑥𝑥) = 4𝑥𝑥𝑥𝑥 .

- Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 4x about the y – axis for 𝑓𝑓(𝑥𝑥𝑥𝑥) = 4-𝑥𝑥𝑥𝑥 .

X and Y interchanging

Function Shape Example Graph





Example: Sketch the graph of 𝒙𝒙𝒙𝒙 = 𝟒𝟒𝒚𝒚𝒚𝒚 .

- Sketch 𝑦𝑦𝑦𝑦 = 4𝑥𝑥𝑥𝑥 .

- Plot the line y = x .

- Reflect the graph of 𝑦𝑦𝑦𝑦 = 4𝑥𝑥𝑥𝑥 about the

line y = x to get the graph of 𝑥𝑥𝑥𝑥 = 4𝑦𝑦 .

∙ (0, 1)

∙ (1, 4)

∙ (2, 16)

(-1, 14�

∙ ∙ x

y

(-2, 116�

f(x) = 4xf(x) = 4-x

y = x

y = 2x

x = 2y

0

x

y y = 4x

x = 4y

y = x

0

Page 10-4





10-2 INVERSE AND COMPOSITE FUNCTIONS

Inverse Relation

• Recall – relation: a set of ordered pairs (x, y) Domain (x) Range (y) Relation (x, y)

• Inverse relation: a relation formed when the order of the

elements in a given relation is switched . Relation

Example: - Relation: {(2, -1) , (-3, 1) , (-4, 0)}

- Inverse relation: {(-1, 2) , (1, -3) , (0, -4)}

- Graph:

• The graph of inverse relation is a reflection (mirror image) of the relation about the line

y = x .

• Inverse equation: switching x and y in the original equation produces an inverse equation .

Example: - Equation: y = 2x – 3

- Inverse equation: x = 2y – 3 Switch x and y.

- Graph:

x y = 2x – 3 x 𝒚𝒚𝒚𝒚 =𝒙𝒙𝒙𝒙 + 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

0 -3 -3 01 -1 -1 1

y

x∙ (2,-1)

■ (-1, 2)

∙ (-3, 1)

■ (1,-3) (0,-4) ■

∙(-4,0)

y

x∙ (1, -1)

∙ (0, -3)

(-1, 1) ∙ ∙ (-3, 0)

x = 2y – 3

y = 2x – 3

∙ 0

2 -1 (2,-1)

-3 1 (-3, 1)

x y Inverse relation

x = y

x = y

Equation Inverse equation (solve for y from x = 2y -3) .

Page 10-5

10-2 INVERSE AND COMPOSITE FUNCTIONS

Inverse Relation

• Recall – relation: a set of ordered pairs (x, y) Domain (x) Range (y) Relation (x, y)

• Inverse relation: a relation formed when the order of the

elements in a given relation is switched . Relation

Example: - Relation: {(2, -1) , (-3, 1) , (-4, 0)}

- Inverse relation: {(-1, 2) , (1, -3) , (0, -4)}

- Graph:

• The graph of inverse relation is a reflection (mirror image) of the relation about the line

y = x .

• Inverse equation: switching x and y in the original equation produces an inverse equation .

Example: - Equation: y = 2x – 3

- Inverse equation: x = 2y – 3 Switch x and y.

- Graph:

x y = 2x – 3 x 𝒚𝒚𝒚𝒚 =𝒙𝒙𝒙𝒙 + 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

0 -3 -3 01 -1 -1 1

y

x∙ (2,-1)

■ (-1, 2)

∙ (-3, 1)

■ (1,-3) (0,-4) ■

∙(-4,0)

y

x∙ (1, -1)

∙ (0, -3)

(-1, 1) ∙ ∙ (-3, 0)

x = 2y – 3

y = 2x – 3

∙ 0

2 -1 (2,-1)

-3 1 (-3, 1)

x y Inverse relation

x = y

x = y

Equation Inverse equation (solve for y from x = 2y -3) .

Page 10-5





One-to-One Function & Its Inverse

• One-to-one function: a function for which every element of the range (y-value)

corresponds to a unique domain (x-value) .

Example: 1. {(1, 2), (3, -4), (5, 3)} One-to-one function

2. {(1, 2), (3, -4), (5, 2)} Not one-to-one ∵ y = 2 is assigned with two x (1 and 5) .

• The horizontal-line test: If a horizontal line cuts the graph of a function only once, then

the function is one-to-one and its inverse is a function .

Not one-to-one One-to-one (A horizontal line cuts the graph more than once .) (There is no horizontal line that cut the graph more than once .)

Example: Determine whether each function is one-to-one .

1. f (x) = 3x2 + 2

x f (x) = 3x2 + 20 21 5-1 5

Not one-to-one

2. f (x) = 2x

x f (x) = 2x

0 11 22 4

One-to-one

x

f(x)

x

f(x)

∙ (0, 2)

(-1, 5) ∙ ∙ (1, 5)

∙ (2, 4)

∙ (1, 2) ∙ (0, 1)

f(x)f(x)

x

x

∙

∙ ∙

0

0

Page 10-6

One-to-One Function & Its Inverse

• One-to-one function: a function for which every element of the range (y-value)

corresponds to a unique domain (x-value) .

Example: 1. {(1, 2), (3, -4), (5, 3)} One-to-one function

2. {(1, 2), (3, -4), (5, 2)} Not one-to-one ∵ y = 2 is assigned with two x (1 and 5) .

• The horizontal-line test: If a horizontal line cuts the graph of a function only once, then

the function is one-to-one and its inverse is a function .

Not one-to-one One-to-one (A horizontal line cuts the graph more than once .) (There is no horizontal line that cut the graph more than once .)

Example: Determine whether each function is one-to-one .

1. f (x) = 3x2 + 2

x f (x) = 3x2 + 20 21 5-1 5

Not one-to-one

2. f (x) = 2x

x f (x) = 2x

0 11 22 4

One-to-one

x

f(x)

x

f(x)

∙ (0, 2)

(-1, 5) ∙ ∙ (1, 5)

∙ (2, 4)

∙ (1, 2) ∙ (0, 1)

f(x)f(x)

x

x

∙

∙ ∙

0

0

Page 10-6





Inverse Function

• Inverse function f -1(x): the function formed when the order of the elements in a given

function is switched .

• The graph of inverse function f -1 (x) is a reflection the original function f (x) about the

line y = x .

• If a function f (x) is one-to-one, its inverse function f -1(x) can be found as follows:

Steps Example: f (x) = 2x + 3

- Confirm that the function is 1-to-1 . Graph f (x):

x f (x) = 2x + 30 31 5

- Rewrite f (x) as y . y = 2x + 3

- Switch x and y . x = 2y + 3

- Solve for y. y = 𝑥𝑥𝑥𝑥−32 Subtract 3; divide by 2 .

- Replace y with f -1(x) . f -1 (x) = 𝒙𝒙𝒙𝒙−𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

- Graph f -1(x): reflect the graph of f (x) across the line y = x .

Note: A function has an inverse function f -1(x) only if the function is a one-to-one function .

∙

∙

0

Function f(x) x y Inverse function f -1(x)

Yes, 1-to-1

y = xf (x)

f -1(x)

Page 10-7

Inverse Function

• Inverse function f -1(x): the function formed when the order of the elements in a given

function is switched .

• The graph of inverse function f -1 (x) is a reflection the original function f (x) about the

line y = x .


Steps Example: f (x) = 2x + 3

- Confirm that the function is 1-to-1 . Graph f (x):

x f (x) = 2x + 30 31 5

- Rewrite f (x) as y . y = 2x + 3

- Switch x and y . x = 2y + 3

- Solve for y. y = 𝑥𝑥𝑥𝑥−32 Subtract 3; divide by 2 .

- Replace y with f -1(x) . f -1 (x) = 𝒙𝒙𝒙𝒙−𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

- Graph f -1(x): reflect the graph of f (x) across the line y = x .

Note: A function has an inverse function f -1(x) only if the function is a one-to-one function .

∙

∙

0

Function f(x) x y Inverse function f -1(x)

Yes, 1-to-1

y = xf (x)

f -1(x)

Page 10-7





Page 10-8

Graph the Function and Its Inverse

Example: Determine whether the function is one-to-one. If it is, find its inverse function.

f (x) = 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙

x y = 4/x 1 4 2 2 4 1 -1 -4 -2 -2 -4 -1

- Let y = f (x). y = 𝟒𝟒𝟒𝟒

𝒙𝒙𝒙𝒙

- Switch x and y. x = 𝟒𝟒𝟒𝟒𝒚𝒚𝒚𝒚

- Solve with y. y = 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙 Divide by x; multiply by y.

- Replace y with f -1 (x). f -1(x) = 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙

Example: Sketch the graph of the function h(x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏 and its inverse.

- Graph h(x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏.

x h(x) = 23𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏

0 1 3 3

Yes, 1-to-1

- Replace h(x) with y. y = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏 Rewrite h(x) as y.

- Switch x and y. x = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚 + 𝟏𝟏𝟏𝟏

- Solve for y. x – 1 = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚 , 3(x – 1) = 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚

y = = 𝟑𝟑𝟑𝟑(𝒙𝒙𝒙𝒙−𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐

- Replace y with h -1(x). h -1(x) = 𝟑𝟑𝟑𝟑(𝒙𝒙𝒙𝒙−𝟏𝟏𝟏𝟏)

𝟐𝟐𝟐𝟐

- Graph h -1(x): reflect the graph of h(x) across the line y = x .

h (x)

x

∙

∙

y = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝑥𝑥𝑥𝑥 + 1

h -1 (x) = 𝟑𝟑𝟑𝟑(𝒙𝒙𝒙𝒙−𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐

y = x

- Graph f (x) = 4𝑥𝑥𝑥𝑥 .

0

Yes, 1-to-1.

x

f (x)

∙

∙ ∙

0 f (x) = 4

𝑥𝑥𝑥𝑥

∙ ∙

∙

f (x) = 4𝑥𝑥𝑥𝑥

Page 10-8

Graph the Function and Its Inverse

Example: Determine whether the function is one-to-one. If it is, find its inverse function.

f (x) = 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙

x y = 4/x 1 4 2 2 4 1 -1 -4 -2 -2 -4 -1

- Let y = f (x). y = 𝟒𝟒𝟒𝟒

𝒙𝒙𝒙𝒙

- Switch x and y. x = 𝟒𝟒𝟒𝟒𝒚𝒚𝒚𝒚

- Solve with y. y = 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙 Divide by x; multiply by y.

- Replace y with f -1 (x). f -1(x) = 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙

Example: Sketch the graph of the function h(x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏 and its inverse.

- Graph h(x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏.

x h(x) = 23𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏

0 1 3 3

Yes, 1-to-1

- Replace h(x) with y. y = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏 Rewrite h(x) as y.

- Switch x and y. x = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚 + 𝟏𝟏𝟏𝟏

- Solve for y. x – 1 = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚 , 3(x – 1) = 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚

y = = 𝟑𝟑𝟑𝟑(𝒙𝒙𝒙𝒙−𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐

- Replace y with h -1(x). h -1(x) = 𝟑𝟑𝟑𝟑(𝒙𝒙𝒙𝒙−𝟏𝟏𝟏𝟏)

𝟐𝟐𝟐𝟐

- Graph h -1(x): reflect the graph of h(x) across the line y = x .

h (x)

x

∙

∙

y = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝑥𝑥𝑥𝑥 + 1

h -1 (x) = 𝟑𝟑𝟑𝟑(𝒙𝒙𝒙𝒙−𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐

y = x

- Graph f (x) = 4𝑥𝑥𝑥𝑥 .

0

Yes, 1-to-1.

x

f (x)

∙

∙ ∙

0 f (x) = 4

𝑥𝑥𝑥𝑥

∙ ∙

∙

f (x) = 4𝑥𝑥𝑥𝑥





Composition of Functions

Composite function f ∘ g(x): a combination of two or more functions in which the result of

one function is applied to another function (substitute a function into another function) .

Composite Function Formula Comments

f ∘ g (x) f ∘ g(x) = f [g(x)]g (x) (inner function)

The result of g ( ) is applying to f ( ) .

f ( ) (outer function)

g ∘ f (x) g ∘ f (x) = g [f(x)]f (x) (inner function)

The result of f ( ) is applying to g ( ) .

g ( ) (outer function)

Tips - Composite function: a function within another function .- Read: f ∘ g(x): “f of g of x” ; g ∘ f (x): “g of f of x”

Example: Find f ∘ g(x) and g ∘ f (x) .

If f (x) = 3 – 2x,

and g (x) = x – 4

- f ∘ g(x) = f [g(x)] g (x) = x – 4

= f [x – 4] f ( ) Replace g(x) with (x – 4) .

= 3 – 2 (x – 4) 3 – 2( ) Replace (x – 4) with x in f (x) .

= 3 – 2x + 8

f ∘ g(x) = 11 – 2x

- g ∘ f (x) = g [f(x)] f (x) = 3 – 2x

= g (3 – 2x) g ( ) Replace f (x) with (3–2x) .

= (3 – 2x) – 4 ( ) – 4 Replace (3 – 2x) with x in g(x) .

= 3 – 2x – 4

g ∘ f (x) = - 2x – 1

f (x) = 3 – 2 x

g (x) = x – 4

Page 10-9

Composition of Functions

Composite function f ∘ g(x): a combination of two or more functions in which the result of









Tips - Composite function: a function within another function .- Read: f ∘ g(x): “f of g of x” ; g ∘ f (x): “g of f of x”


If f (x) = 3 – 2x,

and g (x) = x – 4

- f ∘ g(x) = f [g(x)] g (x) = x – 4

= f [x – 4] f ( ) Replace g(x) with (x – 4) .

= 3 – 2 (x – 4) 3 – 2( ) Replace (x – 4) with x in f (x) .

= 3 – 2x + 8

f ∘ g(x) = 11 – 2x

- g ∘ f (x) = g [f(x)] f (x) = 3 – 2x

= g (3 – 2x) g ( ) Replace f (x) with (3–2x) .

= (3 – 2x) – 4 ( ) – 4 Replace (3 – 2x) with x in g(x) .

= 3 – 2x – 4

g ∘ f (x) = - 2x – 1

f (x) = 3 – 2 x

g (x) = x – 4

Page 10-9






If f (x) = 𝟏𝟏𝟏𝟏𝒙𝒙𝒙𝒙

,

and g (x) = 2 – 3x2

- f ∘ g(x) = f [g (x)] g (x) = 2 – 3x2

= f [2 – 3x2] f ( ) Replace g(x) with (2 – 3x2) .

= 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐−𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏( )

Replace (2 – 3x2) with x in f (x) .

f (x) = 𝟏𝟏𝟏𝟏𝒙𝒙𝒙𝒙

- g ∘ f (x) = g [f (x)] f (x) =𝟏𝟏𝟏𝟏𝒙𝒙𝒙𝒙

= g �𝟏𝟏𝟏𝟏𝒙𝒙𝒙𝒙� g ( ) Replace f (x) with 𝟏𝟏𝟏𝟏

𝒙𝒙𝒙𝒙 .

= 2 – 𝟑𝟑𝟑𝟑 �𝟏𝟏𝟏𝟏𝒙𝒙𝒙𝒙�

𝟐𝟐𝟐𝟐2 – 3( )2 Replace 𝟏𝟏𝟏𝟏

𝒙𝒙𝒙𝒙with x in g (x) .

g (x) = 2 – 3x2

= 𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐

Example: Determine f (x) and g (x) such that h (x) = f ∘ g (x), h (x) = (3 − 2x)2 .

Solution: ∵ h (x) = f ∘ g(x) = f [g(x)] h (x) = (3 − 2x)2

g (x) = 3 − 2x

∴ f (x) = x2 f ( ) = ( )2 h (x) = (3 − 2x)2

g (x) = 3 − 2x g (x)

Check: h (x) = f ∘ g(x) = f [g(x)] g (x) = 3 − 2x

= f (3 − 2x) f ( ) Replace g(x) with (3 − 2x) .

= (3 − 2x)2 ( )2 = h (x) f (x) = x2

√ ∴ h (x) = (3 − 2x)2

Correct!

Page 10-10





Page 10-11

Inverse Functions and Composition

Inverse Function If a function is one-to-one, then f -1 ∘ f (x) = x and f ∘ f -1 (x) = x.

Example: Use composition to show that the inverse is correct.

f (x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 ,

f -1 (x) = 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙

- f -1 ∘ f (x) = f -1 [f (x)] f (x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

𝒙𝒙𝒙𝒙 ∵ f ∘ g (x) = f [g (x)]

= f -1 [𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙] f -1( )

= 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐

(𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 𝒙𝒙𝒙𝒙) 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐 ( ) f -1 (x) = 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙

= 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔

𝒙𝒙𝒙𝒙

= x

- f ∘ f -1 (x) = f [f -1 (x)] f -1(x) = 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 ∵ f ∘ g (x) = f [g (x)]

= f [ 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

𝒙𝒙𝒙𝒙 ] f ( )

= 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

(𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

𝒙𝒙𝒙𝒙) 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑

( ) f (x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙

= 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔

𝒙𝒙𝒙𝒙

= x

f -1 ∘ f (x) = f ∘ f -1(x) = x Correct!

∴ f -1 ∘ f (x) = f -1 [f (x)]

∴ f ∘ f -1 (x) = f [f -1 (x)]

Page 10-11

Inverse Functions and Composition

Inverse Function If a function is one-to-one, then f -1 ∘ f (x) = x and f ∘ f -1 (x) = x.

Example: Use composition to show that the inverse is correct.

f (x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 ,

f -1 (x) = 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙

- f -1 ∘ f (x) = f -1 [f (x)] f (x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

𝒙𝒙𝒙𝒙 ∵ f ∘ g (x) = f [g (x)]

= f -1 [𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙] f -1( )

= 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐

(𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 𝒙𝒙𝒙𝒙) 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐 ( ) f -1 (x) = 𝟑𝟑𝟑𝟑

𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙

= 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔

𝒙𝒙𝒙𝒙

= x

- f ∘ f -1 (x) = f [f -1 (x)] f -1(x) = 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 ∵ f ∘ g (x) = f [g (x)]

= f [ 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

𝒙𝒙𝒙𝒙 ] f ( )

= 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑

(𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐

𝒙𝒙𝒙𝒙) 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑

( ) f (x) = 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙

= 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔

𝒙𝒙𝒙𝒙

= x

f -1 ∘ f (x) = f ∘ f -1(x) = x Correct!

∴ f -1 ∘ f (x) = f -1 [f (x)]

∴ f ∘ f -1 (x) = f [f -1 (x)]





Example: Determine the inverse of the given function . Then use composition to show

whether the inverse is correct .

f (x) = - 2x

- Find the inverse function f -1 (x) .

- Rewrite f (x) with y . y = -2x f (x) = -2x

- Switch x and y . x = -2y

- Solve for y . y = - 𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐

- Replace y with f -1 (x) . f -1 (x) = - 𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐

- Determine f -1 ∘ f (x) and f ∘ f -1 (x)

f -1 (x) ∘ f (x) = f -1 [f (x)] f (x) = -𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 f ∘ f -1 (x) = f [f -1 (x)] f -1(x) = - 𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐

= f -1 [-2x] f -1( ) = 𝒇𝒇[- 𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐] f ( )

= - (-𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙)𝟐𝟐𝟐𝟐

- ( )𝟐𝟐𝟐𝟐

= - 𝟐𝟐𝟐𝟐(- 𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐

) -𝟐𝟐𝟐𝟐( )

= x = x

?

- Check: f -1 ∘ f (x) = f ∘ f -1(x) = x Definition

√ x = x Correct!

Page 10-12





10-3 LOGARITHMIC FUNCTIONS

Introduction to Logarithms

• The logarithmic function f(x) = loga x: a function that is the inverse of an exponential

function (y = ax) .

Exponent question: Logarithmic question:

32 = ? 3? = 9 3 to what power gives 9?

32 = 9 Two multiples of 3s are required to get 9 .

log39 = 2

• Definition of logarithm

Logarithmic Function Definition of Logarithm Example

f(x) = log𝑎𝑎 x

(x > 0 , a > 0, a ≠ 1)

if 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥, then log𝑎𝑎 𝑦𝑦𝑦𝑦 = 𝑥𝑥𝑥𝑥 .

Or if 𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦, then log𝑎𝑎 x = y.

If 9 = 32,

then log3 9 = 2 .Read: “the log, base 3 of 9, is 2” or “log of 9, base 3, equals 2 .”

• Logarithm of zero log a (0): the logarithm of 0 is undefined . Example∵ if a x = 0 , x does not exist . 3 x = 0 is undefined .

x does not exist for 3x = 0 .

• Logarithm of negative number log a (-y): the logarithm of

negative numbers is undefined . Log 3 (-9) is undefined .∵ base a > 0, y = a x > 0 , y must be positive for any real x . ∵ 3 > 0, 32 > 0

• Converting between exponential and logarithmic forms Example

Exponential to log form: 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦 32 = 9

log𝑎𝑎 y = x log3 9 = 2

Log to exponential form: log𝑎𝑎 y = x log3 9 = 2

𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦 32 = 9

Page 10-13

10-3 LOGARITHMIC FUNCTIONS

Introduction to Logarithms

• The logarithmic function f(x) = loga x: a function that is the inverse of an exponential

function (y = ax) .

Exponent question: Logarithmic question:

32 = ? 3? = 9 3 to what power gives 9?

32 = 9 Two multiples of 3s are required to get 9 .

log39 = 2



f(x) = log𝑎𝑎 x

(x > 0 , a > 0, a ≠ 1)

if 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥, then log𝑎𝑎 𝑦𝑦𝑦𝑦 = 𝑥𝑥𝑥𝑥 .

Or if 𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦, then log𝑎𝑎 x = y.

If 9 = 32,

then log3 9 = 2 .Read: “the log, base 3 of 9, is 2” or “log of 9, base 3, equals 2 .”

• Logarithm of zero log a (0): the logarithm of 0 is undefined . Example∵ if a x = 0 , x does not exist . 3 x = 0 is undefined .

x does not exist for 3x = 0 .

• Logarithm of negative number log a (-y): the logarithm of

negative numbers is undefined . Log 3 (-9) is undefined .∵ base a > 0, y = a x > 0 , y must be positive for any real x . ∵ 3 > 0, 32 > 0

• Converting between exponential and logarithmic forms Example

Exponential to log form: 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦 32 = 9

log𝑎𝑎 y = x log3 9 = 2

Log to exponential form: log𝑎𝑎 y = x log3 9 = 2

𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦 32 = 9

Page 10-13





Evaluating Logarithms

Example: Write in logarithmic form .

1. 23 = 8

log2 ( 8 ) = ( 3 ) 𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐8 = 3

2. 105 = 100,000

log10 (100,000) = ( 5 ) log10100,000 = 5

3. 𝑎𝑎𝑡𝑡 = b

log𝑎𝑎 (b) = (t) 𝐥𝐥𝐥𝐥𝐥𝐥𝒂𝒂 b = t

Example: Write in exponential form .

1. log3 81 = 4

3(4) = ( 81 ) 𝟑𝟑𝟑𝟑𝟒𝟒 = 81

2. log 10 0 .0001 = -4

10(-4) = ( 0 .0001) 𝟏𝟏𝟏𝟏𝟏𝟏-𝟒𝟒 = 0.0001Note: The exponent can be negative, but the base must be positive (𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦 , a > 0) .

3. log5125

= -2

5(-2) = � 125� 𝟓𝟓-𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟓𝟓


Steps Example: 𝐥𝐥𝐥𝐥𝐥𝐥𝟒𝟒 16 = ?

- Let log𝑎𝑎 y = x. Let log416 = 𝑥𝑥𝑥𝑥

- Convert to exponential form . 4(x) = (16) 4𝑥𝑥𝑥𝑥 = 16

- Write x in an exponent . 4𝑥𝑥𝑥𝑥 = 42

x = 2 If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦 , then x = y . The exponents are the same .

Example: Find the value of 𝐥𝐥𝐥𝐥𝐥𝐥𝟓𝟓125 .

Let log5 125 = x Let log𝑎𝑎 y = x

5 (x) = (125) 5𝑥𝑥𝑥𝑥 = 125 Convert to exponential form .

5𝑥𝑥𝑥𝑥 = 53 125 = 53

x = 3 If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦, then x = y .

Page 10-14


Example: Write in logarithmic form .

1. 23 = 8

log2 ( 8 ) = ( 3 ) 𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐8 = 3

2. 105 = 100,000

log10 (100,000) = ( 5 ) log10100,000 = 5

3. 𝑎𝑎𝑡𝑡 = b

log𝑎𝑎 (b) = (t) 𝐥𝐥𝐥𝐥𝐥𝐥𝒂𝒂 b = t

Example: Write in exponential form .

1. log3 81 = 4

3(4) = ( 81 ) 𝟑𝟑𝟑𝟑𝟒𝟒 = 81

2. log 10 0 .0001 = -4

10(-4) = ( 0 .0001) 𝟏𝟏𝟏𝟏𝟏𝟏-𝟒𝟒 = 0.0001Note: The exponent can be negative, but the base must be positive (𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦 , a > 0) .

3. log5125

= -2

5(-2) = � 125� 𝟓𝟓-𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟓𝟓


Steps Example: 𝐥𝐥𝐥𝐥𝐥𝐥𝟒𝟒 16 = ?

- Let log𝑎𝑎 y = x. Let log416 = 𝑥𝑥𝑥𝑥

- Convert to exponential form . 4(x) = (16) 4𝑥𝑥𝑥𝑥 = 16

- Write x in an exponent . 4𝑥𝑥𝑥𝑥 = 42

x = 2 If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦 , then x = y . The exponents are the same .

Example: Find the value of 𝐥𝐥𝐥𝐥𝐥𝐥𝟓𝟓125 .

Let log5 125 = x Let log𝑎𝑎 y = x

5 (x) = (125) 5𝑥𝑥𝑥𝑥 = 125 Convert to exponential form .

5𝑥𝑥𝑥𝑥 = 53 125 = 53

x = 3 If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦, then x = y .

Page 10-14





Solving Logarithmic Equations

• Logarithmic equation: an equation that contains a logarithmic expression .

• The key to solve a logarithmic equation is to convert log into exponential form .

Steps Example: Solve log x 25 = 2 .log x 25 = 2

- Convert to an exponential equation . x (2) = (25) , 𝑥𝑥𝑥𝑥2 = 25- Take the square root of both sides and solve for x . √𝑥𝑥𝑥𝑥2 = ±√25

x = ±5- Check . x = 5 x = -5

? ?

log5 25 = 2 log(-5)25 = 2

52 = 25 Correct! It is not defined

x = 5 is a solution y = log a x , a > 0Example: Solve each of the following equations .

1. 𝐥𝐥𝐥𝐥𝐥𝐥𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 = -4 : log3𝑥𝑥𝑥𝑥 = -4

3-4 = x Log exponent .134

= 𝑥𝑥𝑥𝑥 𝑎𝑎-𝑥𝑥𝑥𝑥 = 1𝑎𝑎𝑥𝑥

𝒙𝒙𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏

34 = 81

?

Check: log3181

= -4 √

3-4 = 181

, 134

= 181

Correct!

2. 𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐16 = x : log216 = x

2𝑥𝑥𝑥𝑥 = 16 Log exponent .

2𝒙𝒙𝒙𝒙 = 2𝟒𝟒 Write 16 in an exponent: 16 = 2𝟒𝟒

x = 4 If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦 , then x = y.

3. 𝐥𝐥𝐥𝐥𝐥𝐥𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒

= x : log1614

= x Log exponent .

16𝑥𝑥𝑥𝑥 = 14

(42)𝑥𝑥𝑥𝑥 = 4-1 , 42𝑥𝑥𝑥𝑥= 4-1 16 = 42

2𝑥𝑥𝑥𝑥 = -1 , x = -12

If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦, then x = y.

4. 𝐥𝐥𝐥𝐥𝐥𝐥𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐𝟐𝟐 = 3 : log𝑥𝑥𝑥𝑥27 = 3 Log exponent .

𝑥𝑥𝑥𝑥3 = 27√𝑥𝑥𝑥𝑥33 = √273

Take the cube root of both sides .

√𝑥𝑥𝑥𝑥33 = √333 , x = 3 27 = 33

Page 10-15

Solving Logarithmic Equations


• The key to solve a logarithmic equation is to convert log into exponential form .

Steps Example: Solve log x 25 = 2 .log x 25 = 2

- Convert to an exponential equation . x (2) = (25) , 𝑥𝑥𝑥𝑥2 = 25- Take the square root of both sides and solve for x . √𝑥𝑥𝑥𝑥2 = ±√25

x = ±5- Check . x = 5 x = -5

? ?

log5 25 = 2 log(-5)25 = 2

52 = 25 Correct! It is not defined

x = 5 is a solution y = log a x , a > 0Example: Solve each of the following equations .

1. 𝐥𝐥𝐥𝐥𝐥𝐥𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 = -4 : log3𝑥𝑥𝑥𝑥 = -4

3-4 = x Log exponent .134

= 𝑥𝑥𝑥𝑥 𝑎𝑎-𝑥𝑥𝑥𝑥 = 1𝑎𝑎𝑥𝑥

𝒙𝒙𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟏𝟏

34 = 81

?

Check: log3181

= -4 √

3-4 = 181

, 134

= 181

Correct!

2. 𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐16 = x : log216 = x

2𝑥𝑥𝑥𝑥 = 16 Log exponent .

2𝒙𝒙𝒙𝒙 = 2𝟒𝟒 Write 16 in an exponent: 16 = 2𝟒𝟒

x = 4 If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦 , then x = y.

3. 𝐥𝐥𝐥𝐥𝐥𝐥𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒

= x : log1614

= x Log exponent .

16𝑥𝑥𝑥𝑥 = 14

(42)𝑥𝑥𝑥𝑥 = 4-1 , 42𝑥𝑥𝑥𝑥= 4-1 16 = 42

2𝑥𝑥𝑥𝑥 = -1 , x = -12

If 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦, then x = y.

4. 𝐥𝐥𝐥𝐥𝐥𝐥𝒙𝒙𝒙𝒙𝟐𝟐𝟐𝟐𝟐𝟐 = 3 : log𝑥𝑥𝑥𝑥27 = 3 Log exponent .

𝑥𝑥𝑥𝑥3 = 27√𝑥𝑥𝑥𝑥33 = √273

Take the cube root of both sides .

√𝑥𝑥𝑥𝑥33 = √333 , x = 3 27 = 33

Page 10-15





Graphing Logarithmic Functions

• The graph of a function is the reflection of its inverse about the y = x line .

• Logarithmic functions and exponential functions are inverse functions.

• The graph of the logarithmic function is the reflection of the graph of the exponential

function about the y = x line .

• Graphing a logarithmic function using its inverse

Steps Example: Graph f(x) = 𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 .

- Convert to exponential form . log2𝑥𝑥𝑥𝑥 = y

2𝑦𝑦 = x- Switch x and y . 2𝑥𝑥𝑥𝑥 = y or y = 2𝑥𝑥𝑥𝑥

- Make a table for 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥 .x 0 1 2 -1 -2

𝒚𝒚𝒚𝒚 = 𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 20 = 1 21 = 2 22 = 4 2-1 = 12

2-2 = 14

(x, y) (0, 1) (1, 2) (2, 4) (-1, 12� (-2, 1

4�

- Graph 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥 . Graph y = 2𝑥𝑥𝑥𝑥

- Graph f(x) = log𝑎𝑎𝑥𝑥𝑥𝑥 by reflecting the curve Graph f(x) = log2𝑥𝑥𝑥𝑥 by reflecting 𝑦𝑦𝑦𝑦 = 2𝑥𝑥𝑥𝑥

of 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥 about the y = x line . about the y = x line .

- Or switch the x and y values in the table, Switch the x and y values in the table,then graph y = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥 . then graph y = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑥𝑥𝑥𝑥 .

𝑦𝑦𝑦𝑦 = 2𝑥𝑥𝑥𝑥 (0, 1) (1, 2) (2, 4) (-1, 12� (-2, 1

4�

log 2 x = y (1, 0) (2, 1) (4, 2) �12, -1) �1

4, -2)

Tip: Switch the x and y values to get log2 x = y

∙

∙

∙ ∙

y= 2 x

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 = y

y = x

y

x0

Choose x .

Calculate y .

∙

Page 10-16

Graphing Logarithmic Functions

• The graph of a function is the reflection of its inverse about the y = x line .

• Logarithmic functions and exponential functions are inverse functions.



• Graphing a logarithmic function using its inverse

Steps Example: Graph f(x) = 𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 .

- Convert to exponential form . log2𝑥𝑥𝑥𝑥 = y

2𝑦𝑦 = x- Switch x and y . 2𝑥𝑥𝑥𝑥 = y or y = 2𝑥𝑥𝑥𝑥

- Make a table for 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥 .x 0 1 2 -1 -2

𝒚𝒚𝒚𝒚 = 𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 20 = 1 21 = 2 22 = 4 2-1 = 12

2-2 = 14

(x, y) (0, 1) (1, 2) (2, 4) (-1, 12� (-2, 1

4�

- Graph 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥 . Graph y = 2𝑥𝑥𝑥𝑥

- Graph f(x) = log𝑎𝑎𝑥𝑥𝑥𝑥 by reflecting the curve Graph f(x) = log2𝑥𝑥𝑥𝑥 by reflecting 𝑦𝑦𝑦𝑦 = 2𝑥𝑥𝑥𝑥

of 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥 about the y = x line . about the y = x line .

- Or switch the x and y values in the table, Switch the x and y values in the table,then graph y = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥 . then graph y = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑥𝑥𝑥𝑥 .

𝑦𝑦𝑦𝑦 = 2𝑥𝑥𝑥𝑥 (0, 1) (1, 2) (2, 4) (-1, 12� (-2, 1

4�

log 2 x = y (1, 0) (2, 1) (4, 2) �12, -1) �1

4, -2)

Tip: Switch the x and y values to get log2 x = y

∙

∙

∙ ∙

y= 2 x

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟐𝟐𝟐𝟐𝒙𝒙𝒙𝒙 = y

y = x

y

x0

Choose x .

Calculate y .

∙

Page 10-16





Properties of Logarithms

• Comparing properties of logarithmic and exponential functions

Exponential Functiony = f (x) = ax

Logarithmic Functionlog𝑎𝑎 x

Exampley = 2x log2 x

Domain (x-values) (-∞, ∞)All real numbers .

(0, ∞) or x > 0 (-∞, ∞) (0, ∞)

Range (y-values) (0, ∞) or y > 0 (-∞, ∞)All real numbers .

y > 0 (-∞, ∞)

Intercept y-intercept = 1 x-intercept = 1

Asymptote x - axis y - axis

• Basic properties of logarithms

Property Example Proof

log𝑎𝑎𝟏𝟏𝟏𝟏 = 0 log41 = 0log41 = 0

40 = 1

log𝒂𝒂𝒂𝒂 = 1 log77 = 1log77 = 1

71 = 7

log𝑎𝑎 𝑎𝑎𝒙𝒙𝒙𝒙 = x log2 23 = 3log2 23 = 3

23 = 23

𝑎𝑎log𝑎𝑎𝑥𝑥𝑥𝑥 = x 3log34 = 4 3log34 = 4log34 = log34

Example: Evaluate each of the following .

1. log91 log91 = 0 loga1 = 0

2. log2121 log2121 = 1 loga𝑎𝑎 = 1

3. log5 57 log5 5𝟕𝟕 = 7 log𝑎𝑎 𝑎𝑎𝑥𝑥𝑥𝑥 = x

4. 6log65 6log6𝟓𝟓 = 5 𝑎𝑎log𝑎𝑎𝑥𝑥𝑥𝑥 = x

5. log749 log749 = log7 72 = 2 log𝑎𝑎 𝑎𝑎𝑥𝑥𝑥𝑥 = x

6. log100.001 log10𝟎𝟎.𝟎𝟎𝟎𝟎𝟏𝟏𝟏𝟏 = log10𝟏𝟏𝟏𝟏𝟎𝟎−𝟑𝟑𝟑𝟑 = -3 log𝑎𝑎 𝑎𝑎𝑥𝑥𝑥𝑥 = x

∙ 1 ∙ 1

Page 10-17

Properties of Logarithms






(0, ∞) or x > 0 (-∞, ∞) (0, ∞)


y > 0 (-∞, ∞)




Property Example Proof

log𝑎𝑎𝟏𝟏𝟏𝟏 = 0 log41 = 0log41 = 0

40 = 1

log𝒂𝒂𝒂𝒂 = 1 log77 = 1log77 = 1

71 = 7

log𝑎𝑎 𝑎𝑎𝒙𝒙𝒙𝒙 = x log2 23 = 3log2 23 = 3

23 = 23

𝑎𝑎log𝑎𝑎𝑥𝑥𝑥𝑥 = x 3log34 = 4 3log34 = 4log34 = log34


1. log91 log91 = 0 loga1 = 0

2. log2121 log2121 = 1 loga𝑎𝑎 = 1

3. log5 57 log5 5𝟕𝟕 = 7 log𝑎𝑎 𝑎𝑎𝑥𝑥𝑥𝑥 = x

4. 6log65 6log6𝟓𝟓 = 5 𝑎𝑎log𝑎𝑎𝑥𝑥𝑥𝑥 = x

5. log749 log749 = log7 72 = 2 log𝑎𝑎 𝑎𝑎𝑥𝑥𝑥𝑥 = x

6. log100.001 log10𝟎𝟎.𝟎𝟎𝟎𝟎𝟏𝟏𝟏𝟏 = log10𝟏𝟏𝟏𝟏𝟎𝟎−𝟑𝟑𝟑𝟑 = -3 log𝑎𝑎 𝑎𝑎𝑥𝑥𝑥𝑥 = x

∙ 1 ∙ 1

Page 10-17





10-4 RULES OF LOGARITHMS

Rules

• Recall rules of exponents

Name Rule Exampleproduct rule am an = am + n 23 · 24 = 23 + 4

quotient rule 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛 35

32= 35−2

power of a power (am)n = am n (43)2 = 43∙2

• The rules of logarithms are very similar to the rules of exponents, because the log

function is the inverse of the exponential function .

• Rules of logarithms

Name Rule Example

product rule log𝑎𝑎(𝑨𝑨 ∙ 𝑩𝑩) = log𝑎𝑎𝑨𝑨 + log𝑎𝑎𝑩𝑩The log of a product is the sum of the logs .

log5(3 ∙ 4) = log53 + log54

quotient rule log𝑎𝑎 �𝑨𝑨𝑩𝑩� = log𝑎𝑎𝑨𝑨 − log𝑎𝑎𝑩𝑩

The log of a quotient is the difference of the logs .log3 �

72� = log37 − log32

power rule log𝑎𝑎 𝐴𝐴𝒏𝒏 = n log𝑎𝑎𝐴𝐴The log of a power is equal to the product of the exponentof the power and the log of its base .

log2 3𝟒𝟒 = 4 log23

Where a > 0, A > 0, B > 0, log a ≠ 1

Example: Write each of the following as simpler logarithms .

1. 𝐥𝐥𝐥𝐥𝐥𝐥𝟒𝟒(𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝟑𝟑𝟑𝟑) = log43 + log4𝑦𝑦𝑦𝑦𝟑𝟑𝟑𝟑 log𝑎𝑎(𝐴𝐴𝐴𝐴) = log𝑎𝑎𝐴𝐴 + log𝑎𝑎𝐴𝐴

= log43 + 𝟑𝟑𝟑𝟑𝑙𝑙𝑙𝑙og4𝑦𝑦𝑦𝑦 log𝑎𝑎 𝐴𝐴𝑛𝑛 = n log𝑎𝑎𝐴𝐴

2. 𝐥𝐥𝐥𝐥𝐥𝐥𝟓𝟓 �𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐� = log53 ∙ 𝑥𝑥𝑥𝑥 − log5𝑦𝑦𝑦𝑦2 log𝑎𝑎 �

𝐴𝐴𝐵𝐵� = loga𝐴𝐴 − log𝑎𝑎𝐴𝐴

= log53 + log5𝑥𝑥𝑥𝑥 − log5𝑦𝑦𝑦𝑦2 log𝑎𝑎(𝐴𝐴𝐴𝐴) = log𝑎𝑎𝐴𝐴 + log𝑎𝑎𝐴𝐴

= log53 + log5𝑥𝑥𝑥𝑥 − 2log5𝑦𝑦𝑦𝑦 log𝑎𝑎 𝐴𝐴𝑛𝑛 = n log𝑎𝑎𝐴𝐴

3. 𝟒𝟒𝒍𝒍𝒍𝒍𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 + 𝟐𝟐𝟐𝟐𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 − 𝟑𝟑𝟑𝟑𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝒛𝒛

= log2𝑥𝑥𝑥𝑥4+ log2𝑦𝑦𝑦𝑦2 − log2𝑧𝑧3 𝑙𝑙𝑙𝑙 log𝑎𝑎𝐴𝐴 = log𝑎𝑎 𝐴𝐴𝑛𝑛

= log2(𝑥𝑥𝑥𝑥4 ∙ 𝑦𝑦𝑦𝑦2) − log2𝑧𝑧3 log𝑎𝑎𝐴𝐴 + log𝑎𝑎𝐴𝐴 = log𝑎𝑎(𝐴𝐴𝐴𝐴)

= log2 �𝑥𝑥𝑥𝑥4𝑦𝑦𝑦𝑦2

𝑧𝑧3 � log𝑎𝑎 �𝐴𝐴𝐵𝐵� = log𝑎𝑎𝐴𝐴 − log𝑎𝑎𝐴𝐴

Page 10-18

10-4 RULES OF LOGARITHMS

Rules

• Recall rules of exponents

Name Rule Exampleproduct rule am an = am + n 23 · 24 = 23 + 4

quotient rule 𝑎𝑎𝑚𝑚

𝑎𝑎𝑛𝑛= 𝑎𝑎𝑚𝑚−𝑛𝑛 35

32= 35−2

power of a power (am)n = am n (43)2 = 43∙2

• The rules of logarithms are very similar to the rules of exponents, because the log

function is the inverse of the exponential function .


Name Rule Example

product rule log𝑎𝑎(𝑨𝑨 ∙ 𝑩𝑩) = log𝑎𝑎𝑨𝑨 + log𝑎𝑎𝑩𝑩The log of a product is the sum of the logs .

log5(3 ∙ 4) = log53 + log54

quotient rule log𝑎𝑎 �𝑨𝑨𝑩𝑩� = log𝑎𝑎𝑨𝑨 − log𝑎𝑎𝑩𝑩

The log of a quotient is the difference of the logs .log3 �

72� = log37 − log32

power rule log𝑎𝑎 𝐴𝐴𝒏𝒏 = n log𝑎𝑎𝐴𝐴The log of a power is equal to the product of the exponentof the power and the log of its base .

log2 3𝟒𝟒 = 4 log23

Where a > 0, A > 0, B > 0, log a ≠ 1

Example: Write each of the following as simpler logarithms .

1. 𝐥𝐥𝐥𝐥𝐥𝐥𝟒𝟒(𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝟑𝟑𝟑𝟑) = log43 + log4𝑦𝑦𝑦𝑦𝟑𝟑𝟑𝟑 log𝑎𝑎(𝐴𝐴𝐴𝐴) = log𝑎𝑎𝐴𝐴 + log𝑎𝑎𝐴𝐴

= log43 + 𝟑𝟑𝟑𝟑𝑙𝑙𝑙𝑙og4𝑦𝑦𝑦𝑦 log𝑎𝑎 𝐴𝐴𝑛𝑛 = n log𝑎𝑎𝐴𝐴

2. 𝐥𝐥𝐥𝐥𝐥𝐥𝟓𝟓 �𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝟐𝟐𝟐𝟐� = log53 ∙ 𝑥𝑥𝑥𝑥 − log5𝑦𝑦𝑦𝑦2 log𝑎𝑎 �

𝐴𝐴𝐵𝐵� = loga𝐴𝐴 − log𝑎𝑎𝐴𝐴

= log53 + log5𝑥𝑥𝑥𝑥 − log5𝑦𝑦𝑦𝑦2 log𝑎𝑎(𝐴𝐴𝐴𝐴) = log𝑎𝑎𝐴𝐴 + log𝑎𝑎𝐴𝐴

= log53 + log5𝑥𝑥𝑥𝑥 − 2log5𝑦𝑦𝑦𝑦 log𝑎𝑎 𝐴𝐴𝑛𝑛 = n log𝑎𝑎𝐴𝐴

3. 𝟒𝟒𝒍𝒍𝒍𝒍𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑 + 𝟐𝟐𝟐𝟐𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 − 𝟑𝟑𝟑𝟑𝐥𝐥𝐥𝐥𝐥𝐥𝟐𝟐𝟐𝟐𝒛𝒛

= log2𝑥𝑥𝑥𝑥4+ log2𝑦𝑦𝑦𝑦2 − log2𝑧𝑧3 𝑙𝑙𝑙𝑙 log𝑎𝑎𝐴𝐴 = log𝑎𝑎 𝐴𝐴𝑛𝑛

= log2(𝑥𝑥𝑥𝑥4 ∙ 𝑦𝑦𝑦𝑦2) − log2𝑧𝑧3 log𝑎𝑎𝐴𝐴 + log𝑎𝑎𝐴𝐴 = log𝑎𝑎(𝐴𝐴𝐴𝐴)

= log2 �𝑥𝑥𝑥𝑥4𝑦𝑦𝑦𝑦2

𝑧𝑧3 � log𝑎𝑎 �𝐴𝐴𝐵𝐵� = log𝑎𝑎𝐴𝐴 − log𝑎𝑎𝐴𝐴

Page 10-18





Proof of the Logarithms Rules

• Proof: 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂(𝑨𝑨𝑨𝑨) = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 + 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

- Let 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 = 𝑥𝑥𝑥𝑥 and 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵 = 𝑦𝑦𝑦𝑦 .

- Convert to exponential form . 𝑎𝑎𝑥𝑥𝑥𝑥 = A 𝑎𝑎𝑦𝑦 = B

- Multiply A and B . A · B = 𝑎𝑎𝑥𝑥𝑥𝑥 ∙ 𝑎𝑎𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥+𝑦𝑦

- Convert to logarithmic form . 𝑎𝑎𝑥𝑥𝑥𝑥+𝑦𝑦 = AB

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐵𝐵) = x + y

- Replace x and y by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 and 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵 . 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂(𝑨𝑨𝑨𝑨) = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 + 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

• Proof: 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 �𝑨𝑨𝑨𝑨� = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 − 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨



- Divide A and B. 𝐴𝐴𝐵𝐵

= 𝑎𝑎𝑥𝑥

𝑎𝑎𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥−𝑦𝑦

- Convert to logarithmic form . 𝑎𝑎𝑥𝑥𝑥𝑥−𝑦𝑦 = 𝐴𝐴𝐵𝐵

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �𝐴𝐴𝐵𝐵� = x − y

- Replace x and y by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 and 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵 . 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 �𝑨𝑨𝑨𝑨� = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 − 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

• Proof: 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 𝑨𝑨𝒏𝒏 = n 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

- Let 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 = 𝑥𝑥𝑥𝑥 .

- Convert to exponential form . 𝑎𝑎𝑥𝑥𝑥𝑥 = A

- Raise both sides to nth power . (𝑎𝑎𝑥𝑥𝑥𝑥)𝑛𝑛 = 𝐴𝐴𝑛𝑛 or 𝑎𝑎𝑥𝑥𝑥𝑥𝑛𝑛 = 𝐴𝐴𝑛𝑛

- Convert to logarithmic form . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥

- Replace x by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴)𝑥𝑥𝑥𝑥

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 𝑨𝑨𝒏𝒏 = n 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

Page 10-19

Proof of the Logarithms Rules

• Proof: 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂(𝑨𝑨𝑨𝑨) = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 + 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨



- Multiply A and B . A · B = 𝑎𝑎𝑥𝑥𝑥𝑥 ∙ 𝑎𝑎𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥+𝑦𝑦

- Convert to logarithmic form . 𝑎𝑎𝑥𝑥𝑥𝑥+𝑦𝑦 = AB

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐵𝐵) = x + y

- Replace x and y by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 and 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵 . 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂(𝑨𝑨𝑨𝑨) = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 + 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

• Proof: 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 �𝑨𝑨𝑨𝑨� = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 − 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨



- Divide A and B. 𝐴𝐴𝐵𝐵

= 𝑎𝑎𝑥𝑥

𝑎𝑎𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥−𝑦𝑦

- Convert to logarithmic form . 𝑎𝑎𝑥𝑥𝑥𝑥−𝑦𝑦 = 𝐴𝐴𝐵𝐵

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �𝐴𝐴𝐵𝐵� = x − y

- Replace x and y by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 and 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵 . 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 �𝑨𝑨𝑨𝑨� = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨 − 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

• Proof: 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 𝑨𝑨𝒏𝒏 = n 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

- Let 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 = 𝑥𝑥𝑥𝑥 .

- Convert to exponential form . 𝑎𝑎𝑥𝑥𝑥𝑥 = A

- Raise both sides to nth power . (𝑎𝑎𝑥𝑥𝑥𝑥)𝑛𝑛 = 𝐴𝐴𝑛𝑛 or 𝑎𝑎𝑥𝑥𝑥𝑥𝑛𝑛 = 𝐴𝐴𝑛𝑛

- Convert to logarithmic form . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥

- Replace x by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴)𝑥𝑥𝑥𝑥

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂 𝑨𝑨𝒏𝒏 = n 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑨𝑨

Page 10-19





Applying Rules of Logarithm

Example: Expand 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒�𝒙𝒙𝒙𝒙𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝒛𝒛

𝟑𝟑𝟑𝟑.

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4�𝑥𝑥𝑥𝑥3𝑦𝑦𝑧𝑧

3= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4 �

𝑥𝑥𝑥𝑥3𝑦𝑦𝑧𝑧�13

√𝑎𝑎𝑛𝑛 = 𝑎𝑎1𝑛𝑛

= 13𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4

𝑥𝑥𝑥𝑥3𝑦𝑦𝑧𝑧

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 13

[𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4(𝑥𝑥𝑥𝑥3 ∙ 𝑦𝑦𝑦𝑦) − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑧𝑧] 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �𝐴𝐴𝐵𝐵� = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

= 13

(𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑥𝑥𝑥𝑥3 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑦𝑦𝑦𝑦 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑧𝑧) 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐵𝐵) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

= 13

(3𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑥𝑥𝑥𝑥 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑦𝑦𝑦𝑦 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑧𝑧) 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒𝒚𝒚𝒚𝒚 −

𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒𝒛𝒛

Example: Given 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟒𝟒 = 𝟎𝟎.𝟔𝟔 and 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟓𝟓 = 𝟎𝟎.𝟕𝟕, find the following .

1. 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟐𝟐𝟐𝟐𝟎𝟎 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(4 ∙ 5)

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐵𝐵) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

≈ 0 .6 + 0 .7 = 0.13 Given 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 = 0.6 , 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 = 0.7

2. 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟐𝟐𝟐𝟐𝟓𝟓𝟒𝟒

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎25 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �𝐴𝐴𝐵𝐵� = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎52 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4

= 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 2(0.7) − 0.6 = 0.8 Given 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 = 0.6 , 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 = 0.7

3. 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂√𝟓𝟓 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎512 √𝑎𝑎𝑛𝑛 = 𝑎𝑎

1𝑛𝑛

= 12 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 12

(0.7) = 𝟎𝟎.𝟑𝟑𝟑𝟑𝟓𝟓 Given 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 = 0.7

Page 10-20

Applying Rules of Logarithm

Example: Expand 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒�𝒙𝒙𝒙𝒙𝟑𝟑𝟑𝟑𝒚𝒚𝒚𝒚𝒛𝒛

𝟑𝟑𝟑𝟑.

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4�𝑥𝑥𝑥𝑥3𝑦𝑦𝑧𝑧

3= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4 �

𝑥𝑥𝑥𝑥3𝑦𝑦𝑧𝑧�13

√𝑎𝑎𝑛𝑛 = 𝑎𝑎1𝑛𝑛

= 13𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4

𝑥𝑥𝑥𝑥3𝑦𝑦𝑧𝑧

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 13

[𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4(𝑥𝑥𝑥𝑥3 ∙ 𝑦𝑦𝑦𝑦) − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑧𝑧] 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �𝐴𝐴𝐵𝐵� = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

= 13

(𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑥𝑥𝑥𝑥3 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑦𝑦𝑦𝑦 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑧𝑧) 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐵𝐵) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

= 13

(3𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑥𝑥𝑥𝑥 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑦𝑦𝑦𝑦 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4𝑧𝑧) 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒𝒙𝒙𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒𝒚𝒚𝒚𝒚 −

𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟒𝟒𝒛𝒛

Example: Given 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟒𝟒 = 𝟎𝟎.𝟔𝟔 and 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟓𝟓 = 𝟎𝟎.𝟕𝟕, find the following .

1. 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟐𝟐𝟐𝟐𝟎𝟎 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(4 ∙ 5)

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐵𝐵) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

≈ 0 .6 + 0 .7 = 0.13 Given 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 = 0.6 , 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 = 0.7

2. 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝟐𝟐𝟐𝟐𝟓𝟓𝟒𝟒

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎25 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �𝐴𝐴𝐵𝐵� = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎52 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4

= 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 2(0.7) − 0.6 = 0.8 Given 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎4 = 0.6 , 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 = 0.7

3. 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂√𝟓𝟓 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎512 √𝑎𝑎𝑛𝑛 = 𝑎𝑎

1𝑛𝑛

= 12 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

= 12

(0.7) = 𝟎𝟎.𝟑𝟑𝟑𝟑𝟓𝟓 Given 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎5 = 0.7

Page 10-20





10-5 COMMON AND NATURAL LOGARITHMS

Common Logarithm

• The common logarithm is a logarithm with base 10, i .e . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10𝑥𝑥𝑥𝑥 . It is usually denoted as

log 𝑥𝑥𝑥𝑥 (without writing the base 10) .

Notation for Common Log Example𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥

Base 10 No base means base 10 .𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙103 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3

Example: 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10000 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙1010000

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10104 = 𝟒𝟒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 0.0001 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙100.0001

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙1010-4 = -𝟒𝟒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥

• The log of a negative number has no real number answers.

Example: 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10�-3� = 𝑥𝑥𝑥𝑥

10𝑥𝑥𝑥𝑥 = -3

No matter what the value of x may be, 10𝑥𝑥𝑥𝑥 will never be negative .

∴ log (-3) is undefined .

• Calculator tip: for the common logarithm, use the LOG key .

Example: 1. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 312 ≈ 2.494 LOG 312 ENTER ENTER or = x

2. log 0.146 ≈ -0.836 LOG 0 .146 = = or ENTER

3. 103.1532 ≈ 1423 2nd 10x 3 .1532 ENTER 2nd or INV

Page 10-21

10-5 COMMON AND NATURAL LOGARITHMS

Common Logarithm

• The common logarithm is a logarithm with base 10, i .e . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10𝑥𝑥𝑥𝑥 . It is usually denoted as

log 𝑥𝑥𝑥𝑥 (without writing the base 10) .


Base 10 No base means base 10 .𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙103 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3

Example: 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10000 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙1010000

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10104 = 𝟒𝟒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 0.0001 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙100.0001

= 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙1010-4 = -𝟒𝟒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥

• The log of a negative number has no real number answers.

Example: 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10�-3� = 𝑥𝑥𝑥𝑥

10𝑥𝑥𝑥𝑥 = -3

No matter what the value of x may be, 10𝑥𝑥𝑥𝑥 will never be negative .

∴ log (-3) is undefined .

• Calculator tip: for the common logarithm, use the LOG key .

Example: 1. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 312 ≈ 2.494 LOG 312 ENTER ENTER or = x

2. log 0.146 ≈ -0.836 LOG 0 .146 = = or ENTER

3. 103.1532 ≈ 1423 2nd 10x 3 .1532 ENTER 2nd or INV

Page 10-21





Natural Logarithm

• The natural logarithm: the logarithm with base e, i .e . log𝑒𝑒𝑥𝑥𝑥𝑥 . It is usually denoted as ln x .

• The number e: e is not a whole number (an irrational number) .

e ≈ 2 .718281828 … The decimal expansion of e never ends nor repeats .

Notation for natural log Example𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑒𝑒𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥

Base e 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑒𝑒3 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3

• Calculator tip: for the natural logarithm, use the LN key .

Example

1. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3 ≈ 1 .099 LN 3 = or ENTER

2. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 0.03 ≈ - 3 .51 LN 0 .03 ENTER or = x

3. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 (-7) is undefined

4. 𝑒𝑒2.153 ≈ 8.61 2nd ex 2 .153 = 2nd or INV

• Graphing of f (x) = ln x , ex and e-x

The graph of the exponential function (ex) is a reflection of the graph of (e-x ) about

the y-axis .Recall: The graph of 𝑎𝑎-𝑥𝑥𝑥𝑥 is the reflection of graph of 𝑎𝑎𝑥𝑥𝑥𝑥 about the y-axis .

The graph of the natural logarithmic function ln x is the reflection of the graph of the

exponential function ex about the y = x line .

Example: Sketch the graphs of y = e x , y = e -x and y = ln x.

- Make a table for y = e x Calculator tip: 2nd e x …

x 0 1 -1 -2y = ex 𝑒𝑒0 = 1 𝑒𝑒1 ≈ 2.72 𝑒𝑒-1 ≈ 0.37 𝑒𝑒-2 ≈ 0.14

𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏 (x, y) (0, 1) (1, 2 .72) (-1, 0 .37) (-2, 0 .14)

- Graph y = ex.

- Graph y = e-x: reflect the curve of y = ex about the y-axis .

- Graph y = ln x: reflect the curve of y = ex about the y = x line .𝑦𝑦𝑦𝑦 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑥𝑥𝑥𝑥

𝒚𝒚𝒚𝒚 = 𝒆𝒆𝒆𝒆-𝒙𝒙𝒙𝒙

𝑦𝑦𝑦𝑦 = 𝑥𝑥𝑥𝑥

𝒚𝒚𝒚𝒚 = 𝒆𝒆𝒆𝒆𝒙𝒙𝒙𝒙

Page 10-22

Natural Logarithm

• The natural logarithm: the logarithm with base e, i .e . log𝑒𝑒𝑥𝑥𝑥𝑥 . It is usually denoted as ln x .

• The number e: e is not a whole number (an irrational number) .

e ≈ 2 .718281828 … The decimal expansion of e never ends nor repeats .

Notation for natural log Example𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑒𝑒𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥


• Calculator tip: for the natural logarithm, use the LN key .

Example

1. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3 ≈ 1 .099 LN 3 = or ENTER

2. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 0.03 ≈ - 3 .51 LN 0 .03 ENTER or = x

3. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 (-7) is undefined

4. 𝑒𝑒2.153 ≈ 8.61 2nd ex 2 .153 = 2nd or INV



the y-axis .Recall: The graph of 𝑎𝑎-𝑥𝑥𝑥𝑥 is the reflection of graph of 𝑎𝑎𝑥𝑥𝑥𝑥 about the y-axis .

The graph of the natural logarithmic function ln x is the reflection of the graph of the

exponential function ex about the y = x line .

Example: Sketch the graphs of y = e x , y = e -x and y = ln x.

- Make a table for y = e x Calculator tip: 2nd e x …

x 0 1 -1 -2y = ex 𝑒𝑒0 = 1 𝑒𝑒1 ≈ 2.72 𝑒𝑒-1 ≈ 0.37 𝑒𝑒-2 ≈ 0.14

𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏 (x, y) (0, 1) (1, 2 .72) (-1, 0 .37) (-2, 0 .14)

- Graph y = ex.

- Graph y = e-x: reflect the curve of y = ex about the y-axis .

- Graph y = ln x: reflect the curve of y = ex about the y = x line .𝑦𝑦𝑦𝑦 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑥𝑥𝑥𝑥

𝒚𝒚𝒚𝒚 = 𝒆𝒆𝒆𝒆-𝒙𝒙𝒙𝒙

𝑦𝑦𝑦𝑦 = 𝑥𝑥𝑥𝑥

𝒚𝒚𝒚𝒚 = 𝒆𝒆𝒆𝒆𝒙𝒙𝒙𝒙

Page 10-22





Changing the Base of a Logarithm

• Most scientific calculators have keys for only base 10 (the common log LOG) and base e

(the natural log LN) .

• Change of base formula evaluates logarithms with different bases other than 10 or e . Such as 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙23128 = ?

Change of Base Formula 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒃𝒃𝒙𝒙𝒙𝒙 =𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒙𝒙𝒙𝒙𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒃𝒃

Example

change to base 10 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝒃𝒃𝒙𝒙𝒙𝒙 =𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝒙𝒙𝒙𝒙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝒃𝒃 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙35 = 𝑙𝑙𝑙𝑙𝑙𝑙 5

𝑙𝑙𝑙𝑙𝑙𝑙3≈ 1.465

change to base e 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝒃𝒃𝒙𝒙𝒙𝒙 =𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝒙𝒙𝒙𝒙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝒃𝒃

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙35 = 𝑙𝑙𝑙𝑙 5𝑙𝑙𝑙𝑙 3

≈ 1.465

Note: x, a, and b are positive, a ≠ 1, b ≠ 1.

Tip: The same answer will result regardless of the logarithm (log or ln) that we use in the change of base .

Example: Derivative of 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒃𝒃𝒙𝒙𝒙𝒙 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒙𝒙𝒙𝒙𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒃𝒃

.

Let 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑢𝑢

𝑏𝑏𝑢𝑢 = x Convert log to exponential form .

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑏𝑏𝑢𝑢 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝑎𝑎𝑥𝑥𝑥𝑥 Take log both sides .

u 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑏𝑏 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑙𝑙 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

𝑢𝑢 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑏𝑏

Divide both sides by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑏𝑏.

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒃𝒃𝒙𝒙𝒙𝒙 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒙𝒙𝒙𝒙𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒃𝒃

Replace u with 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 .


1. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙617 = 𝑙𝑙𝑙𝑙𝑙𝑙17𝑙𝑙𝑙𝑙𝑙𝑙6

≈ 1.581 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏

2. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙0.49 = 𝑙𝑙𝑙𝑙9𝑙𝑙𝑙𝑙0.4

≈ -2.398 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙 𝑏𝑏

3. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝜋𝜋3.2 = 𝑙𝑙𝑙𝑙𝑙𝑙3.2𝑙𝑙𝑙𝑙𝑙𝑙𝜋𝜋

≈ 1.016 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙 𝑏𝑏

LOG 5 ) ÷ LOG 3

LN 5 ) ÷ LN 3

Page 10-23

Changing the Base of a Logarithm

• Most scientific calculators have keys for only base 10 (the common log LOG) and base e

(the natural log LN) .

• Change of base formula evaluates logarithms with different bases other than 10 or e . Such as 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙23128 = ?


Example





≈ 1.465

Note: x, a, and b are positive, a ≠ 1, b ≠ 1.

Tip: The same answer will result regardless of the logarithm (log or ln) that we use in the change of base .

Example: Derivative of 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒃𝒃𝒙𝒙𝒙𝒙 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒙𝒙𝒙𝒙𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒃𝒃

.

Let 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑢𝑢

𝑏𝑏𝑢𝑢 = x Convert log to exponential form .

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝑏𝑏𝑢𝑢 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝑎𝑎𝑥𝑥𝑥𝑥 Take log both sides .

u 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑏𝑏 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑙𝑙 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

𝑢𝑢 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑏𝑏

Divide both sides by 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑏𝑏.

𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒃𝒃𝒙𝒙𝒙𝒙 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒙𝒙𝒙𝒙𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂𝒃𝒃

Replace u with 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 .


1. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙617 = 𝑙𝑙𝑙𝑙𝑙𝑙17𝑙𝑙𝑙𝑙𝑙𝑙6

≈ 1.581 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏

2. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙0.49 = 𝑙𝑙𝑙𝑙9𝑙𝑙𝑙𝑙0.4

≈ -2.398 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙 𝑏𝑏

3. 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝜋𝜋3.2 = 𝑙𝑙𝑙𝑙𝑙𝑙3.2𝑙𝑙𝑙𝑙𝑙𝑙𝜋𝜋

≈ 1.016 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙 𝑏𝑏

LOG 5 ) ÷ LOG 3

LN 5 ) ÷ LN 3

Page 10-23





10-6 EXPONENTIAL AND LOGARITHMIC EQUATIONS

Exponential Equations

• Exponential equation: an equation that contains Example

variable(s) in the exponent . 3𝑥𝑥𝑥𝑥 = 5 , 8𝑦𝑦 = 54𝑥𝑥𝑥𝑥 , 4𝑥𝑥𝑥𝑥+5 = 2

• The key to solve an exponential equation: take the logarithm of both sides .

• Procedure to solve an exponential equation

Steps Example: Solve 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 = 𝟗𝟗.

Common Log Method Natural Log Method

- Take the log or ln of both sides . log3𝑥𝑥𝑥𝑥 = log9 ln3𝑥𝑥𝑥𝑥 = ln 9

- Use the power rule . (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴𝑛𝑛 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴) 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙9 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙9

- Isolate the variable. 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥3𝑥𝑥𝑥𝑥𝑥𝑥3

= 𝑥𝑥𝑥𝑥𝑥𝑥9𝑥𝑥𝑥𝑥𝑥𝑥3

𝑥𝑥𝑥𝑥𝑥𝑥𝑛𝑛3𝑥𝑥𝑛𝑛3

= 𝑥𝑥𝑛𝑛9𝑥𝑥𝑛𝑛3

Divide both sides by log 3 . Divide both sides by ln 3 .

- Use a calculator to evaluate the log . 𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥𝑥𝑥9𝑥𝑥𝑥𝑥𝑥𝑥3

= 2 𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑛𝑛9𝑥𝑥𝑛𝑛3

= 2

- Check . ? √

32 = 9 , 9 = 9 Correct!

Example: Solve the given equation .

𝟓𝟓𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 = 𝟒𝟒

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙53𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 4 Take the log of both sides .

3x 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4 log𝑎𝑎 𝐴𝐴𝑛𝑛 = n log𝑎𝑎𝐴𝐴

𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥𝑥𝑥43𝑥𝑥𝑥𝑥𝑥𝑥5

Divide both sides by 3log 5 .

𝒙𝒙𝒙𝒙 ≈ 𝟎𝟎.𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 LOG 4 ) ÷ ( 3 × LOG 5 ) x

? √

Check: 53(0.287) = 4, 50.861 ≈ 4 Correct! 5 yx 0 .861 = yx or ^ x

Page 10-24

10-6 EXPONENTIAL AND LOGARITHMIC EQUATIONS

Exponential Equations

• Exponential equation: an equation that contains Example

variable(s) in the exponent . 3𝑥𝑥𝑥𝑥 = 5 , 8𝑦𝑦 = 54𝑥𝑥𝑥𝑥 , 4𝑥𝑥𝑥𝑥+5 = 2


• Procedure to solve an exponential equation

Steps Example: Solve 𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 = 𝟗𝟗.

Common Log Method Natural Log Method

- Take the log or ln of both sides . log3𝑥𝑥𝑥𝑥 = log9 ln3𝑥𝑥𝑥𝑥 = ln 9

- Use the power rule . (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴𝑛𝑛 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴) 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙9 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙9

- Isolate the variable. 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥3𝑥𝑥𝑥𝑥𝑥𝑥3

= 𝑥𝑥𝑥𝑥𝑥𝑥9𝑥𝑥𝑥𝑥𝑥𝑥3

𝑥𝑥𝑥𝑥𝑥𝑥𝑛𝑛3𝑥𝑥𝑛𝑛3

= 𝑥𝑥𝑛𝑛9𝑥𝑥𝑛𝑛3

Divide both sides by log 3 . Divide both sides by ln 3 .

- Use a calculator to evaluate the log . 𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥𝑥𝑥9𝑥𝑥𝑥𝑥𝑥𝑥3

= 2 𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑛𝑛9𝑥𝑥𝑛𝑛3

= 2

- Check . ? √

32 = 9 , 9 = 9 Correct!


𝟓𝟓𝟑𝟑𝟑𝟑𝒙𝒙𝒙𝒙 = 𝟒𝟒

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙53𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 4 Take the log of both sides .

3x 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙4 log𝑎𝑎 𝐴𝐴𝑛𝑛 = n log𝑎𝑎𝐴𝐴

𝑥𝑥𝑥𝑥 = 𝑥𝑥𝑥𝑥𝑥𝑥43𝑥𝑥𝑥𝑥𝑥𝑥5

Divide both sides by 3log 5 .

𝒙𝒙𝒙𝒙 ≈ 𝟎𝟎.𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 LOG 4 ) ÷ ( 3 × LOG 5 ) x

? √

Check: 53(0.287) = 4, 50.861 ≈ 4 Correct! 5 yx 0 .861 = yx or ^ x

Page 10-24





Solving Exponential Equations

Example: Solve each of the following equations .

1. 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐+𝟓𝟓 = 𝟕𝟕

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙32𝑥𝑥𝑥𝑥+5 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙7 Take the ln of both sides .

(2x + 5) 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙7 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

2𝑥𝑥𝑥𝑥 + 5 = 𝑙𝑙𝑛𝑛7𝑙𝑙𝑛𝑛3

Divide both sides by ln3 .

2𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑛𝑛7𝑙𝑙𝑛𝑛3

– 5 Subtract 5 .

𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑛𝑛22𝑙𝑙𝑛𝑛3

– 2 .5 Divide by 2 .

x ≈ -2.185

2. 𝟑𝟑𝟑𝟑(𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐+𝟏𝟏𝟏𝟏) = 𝟓𝟓𝟐𝟐𝟐𝟐

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3(2𝑥𝑥𝑥𝑥+1) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5𝑥𝑥𝑥𝑥 Take the log of both sides .

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑥𝑥𝑥𝑥+1 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5𝑥𝑥𝑥𝑥 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐴𝐴) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 + (𝑥𝑥𝑥𝑥 + 1)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 = 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 + 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 = 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 Distribute .

x 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 = -𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 Isolate x terms .

x (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5) = -𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 Factor out x .

𝑥𝑥𝑥𝑥 = -𝑙𝑙𝑙𝑙𝑙𝑙2−𝑙𝑙𝑙𝑙𝑙𝑙3𝑙𝑙𝑙𝑙𝑙𝑙2−𝑙𝑙𝑙𝑙𝑙𝑙5

Divide by (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5)

x ≈ -0.778-0.3979

𝟐𝟐𝟐𝟐 ≈ 1.96

Page 10-25

Solving Exponential Equations

Example: Solve each of the following equations .

1. 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐+𝟓𝟓 = 𝟕𝟕

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙32𝑥𝑥𝑥𝑥+5 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙7 Take the ln of both sides .

(2x + 5) 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙7 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

2𝑥𝑥𝑥𝑥 + 5 = 𝑙𝑙𝑛𝑛7𝑙𝑙𝑛𝑛3

Divide both sides by ln3 .

2𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑛𝑛7𝑙𝑙𝑛𝑛3

– 5 Subtract 5 .

𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑛𝑛22𝑙𝑙𝑛𝑛3

– 2 .5 Divide by 2 .

x ≈ -2.185

2. 𝟑𝟑𝟑𝟑(𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐+𝟏𝟏𝟏𝟏) = 𝟓𝟓𝟐𝟐𝟐𝟐

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3(2𝑥𝑥𝑥𝑥+1) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5𝑥𝑥𝑥𝑥 Take the log of both sides .

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑥𝑥𝑥𝑥+1 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5𝑥𝑥𝑥𝑥 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐴𝐴) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 + (𝑥𝑥𝑥𝑥 + 1)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 = 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 𝐴𝐴𝑛𝑛 = n 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 + 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 = 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 Distribute .

x 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑥𝑥𝑥𝑥𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5 = -𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 Isolate x terms .

x (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5) = -𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3 Factor out x .

𝑥𝑥𝑥𝑥 = -𝑙𝑙𝑙𝑙𝑙𝑙2−𝑙𝑙𝑙𝑙𝑙𝑙3𝑙𝑙𝑙𝑙𝑙𝑙2−𝑙𝑙𝑙𝑙𝑙𝑙5

Divide by (𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙5)

x ≈ -0.778-0.3979

𝟐𝟐𝟐𝟐 ≈ 1.96

Page 10-25





Logarithmic Equations

• Logarithmic equation: an equation that contains Examplelogarithms of the expression (s) . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3(𝑥𝑥𝑥𝑥 + 5) = 7

• The key to solve a log equation: convert the logarithmic 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦

form into exponential form . 𝑎𝑎𝑦𝑦 = 𝑥𝑥𝑥𝑥

• Procedure to solve a logarithmic equation

Steps Example: Solve 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟑𝟑𝟑𝟑(𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒𝟒𝟒) = 𝟐𝟐𝟐𝟐

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3(3 + 4𝑥𝑥𝑥𝑥) = 2

- Convert the log equation to an exponential equation . 32 = 3 + 4𝑥𝑥𝑥𝑥

- Isolate x term . 32 − 3 = 4𝑥𝑥𝑥𝑥

- Solve for x . 6 = 4x , x = 1.5 ?

- Check . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3[3 + 4(1.5)] = 2 ?𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙39 = 2

? 𝑙𝑙𝑙𝑙𝑙𝑙9𝑙𝑙𝑙𝑙𝑙𝑙3

= 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏

√2 = 2 Correct!


𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟐𝟐𝟐𝟐𝟒𝟒𝟒𝟒 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟐𝟐𝟐𝟐(𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 − 𝟐𝟐𝟐𝟐) + 𝟑𝟑𝟑𝟑

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑥𝑥𝑥𝑥 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2(3𝑥𝑥𝑥𝑥 − 2) = 3 Collect log terms on one side .

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑙𝑙𝑙𝑙

3𝑙𝑙𝑙𝑙−2= 3 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �

𝐴𝐴𝐵𝐵� = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵

23 = 𝑙𝑙𝑙𝑙3𝑙𝑙𝑙𝑙−2

log equation exponential equation


8(3𝑥𝑥𝑥𝑥 − 2) = 𝑥𝑥𝑥𝑥 Solve for x.

24x – 16 = x

x ≈ 0.7

Page 10-26

Logarithmic Equations

• Logarithmic equation: an equation that contains Examplelogarithms of the expression (s) . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3(𝑥𝑥𝑥𝑥 + 5) = 7

• The key to solve a log equation: convert the logarithmic 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦

form into exponential form . 𝑎𝑎𝑦𝑦 = 𝑥𝑥𝑥𝑥

• Procedure to solve a logarithmic equation

Steps Example: Solve 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟑𝟑𝟑𝟑(𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒𝟒𝟒) = 𝟐𝟐𝟐𝟐

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3(3 + 4𝑥𝑥𝑥𝑥) = 2

- Convert the log equation to an exponential equation . 32 = 3 + 4𝑥𝑥𝑥𝑥

- Isolate x term . 32 − 3 = 4𝑥𝑥𝑥𝑥

- Solve for x . 6 = 4x , x = 1.5 ?

- Check . 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙3[3 + 4(1.5)] = 2 ?𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙39 = 2

? 𝑙𝑙𝑙𝑙𝑙𝑙9𝑙𝑙𝑙𝑙𝑙𝑙3

= 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏

√2 = 2 Correct!


𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟐𝟐𝟐𝟐𝟒𝟒𝟒𝟒 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝟐𝟐𝟐𝟐(𝟑𝟑𝟑𝟑𝟒𝟒𝟒𝟒 − 𝟐𝟐𝟐𝟐) + 𝟑𝟑𝟑𝟑

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑥𝑥𝑥𝑥 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2(3𝑥𝑥𝑥𝑥 − 2) = 3 Collect log terms on one side .

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙2𝑙𝑙𝑙𝑙

3𝑙𝑙𝑙𝑙−2= 3 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 �

𝐴𝐴𝐵𝐵� = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵


log equation exponential equation


8(3𝑥𝑥𝑥𝑥 − 2) = 𝑥𝑥𝑥𝑥 Solve for x.

24x – 16 = x

x ≈ 0.7

Page 10-26






𝐥𝐥𝐥𝐥𝐥𝐥 (𝒙𝒙𝒙𝒙 + 𝟗𝟗) = 𝟏𝟏𝟏𝟏 − 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒙𝒙𝒙𝒙 Collect log terms on one side .

log (𝑥𝑥𝑥𝑥 + 9) + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑥𝑥𝑥𝑥 = 1 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐴𝐴 + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎𝐵𝐵 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎(𝐴𝐴𝐵𝐵)

log [(𝑥𝑥𝑥𝑥 + 9) ∙ 𝑥𝑥𝑥𝑥] = 1 log equation exponential equation

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10𝑥𝑥𝑥𝑥 (𝑥𝑥𝑥𝑥 + 9) = 1 log (𝑥𝑥𝑥𝑥 + 9)𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙10𝑥𝑥𝑥𝑥 (𝑥𝑥𝑥𝑥 + 9)

101 = 𝑥𝑥𝑥𝑥(𝑥𝑥𝑥𝑥 + 9) Distributive property .

𝑥𝑥𝑥𝑥2 + 9𝑥𝑥𝑥𝑥 − 10 = 0 Solve for x.

(x − 1)(x + 10) = 0 Factor .

x − 1 = 0 x + 10 = 0 Zero-product property

x = 1 x = -10

Check: log (𝑥𝑥𝑥𝑥 + 9) = 1 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑥𝑥𝑥𝑥

x = 1 x = -10 ? ?

log (1 + 9) = 1 − 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙1 log �-10 + 9� = 1 − log (-10) ? ?

log 10 = 1 − 0 log ( -1) = 1 − log (-10)

√ Undefined

1 = 1 Correct! x = -10 is not a solution .

Solution: x = 1

Page 10-27





Unit 10 Summary





example


Exponential growth


Exponential decay

asymptote x-axis (y = 0) x-axis (y = 0)

y - intercept y = 1domain x values x = all real numbers or x = (-∞, ∞)range y values y = (0, ∞) or { y | y > 0 }

• Stretching or shiftingFunction Stretch or Shrink Example Graph




• Reflecting (mirror image)Function Reflection Example Graph

𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂-𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the y-axis .


𝒇𝒇(𝒙𝒙𝒙𝒙) = -𝒂𝒂𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the x-axis .



• ShiftingExponential

Function Shifting Example Graph












(0, 1) ∙ (0, 1) ∙

f (x) = 4x

f (x) = 2x

1∙

f(x) = 2x

f(x) = 2x+1

f(x) = 2x-1

0

f(x) = 2x + 1

f(x) = 2x - 1

f(x) = 2x

0

f (x) = 2x

f (x) = - 2-x

1∙

f (x) = 2-x f (x) = 2x

01∙

Page 10-28

Unit 10 Summary





example


Exponential growth


Exponential decay

asymptote x-axis (y = 0) x-axis (y = 0)

y - intercept y = 1domain x values x = all real numbers or x = (-∞, ∞)range y values y = (0, ∞) or { y | y > 0 }

• Stretching or shiftingFunction Stretch or Shrink Example Graph




• Reflecting (mirror image)Function Reflection Example Graph

𝒇𝒇(𝒙𝒙𝒙𝒙) = 𝒂𝒂-𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the y-axis .


𝒇𝒇(𝒙𝒙𝒙𝒙) = -𝒂𝒂𝒙𝒙𝒙𝒙 Reflect the graph of 𝑓𝑓(𝑥𝑥𝑥𝑥) = 𝑎𝑎𝑥𝑥𝑥𝑥 about the x-axis .



• ShiftingExponential

Function Shifting Example Graph












(0, 1) ∙ (0, 1) ∙

f (x) = 4x

f (x) = 2x

1∙

f(x) = 2x

f(x) = 2x+1

f(x) = 2x-1

0

f(x) = 2x + 1

f(x) = 2x - 1

f(x) = 2x

0

f (x) = 2x

f (x) = - 2-x

1∙

f (x) = 2-x f (x) = 2x

01∙

Page 10-28





• X and Y interchanging Function Shape Example Graph





• One-to-one function: a function for which every element of the range (y-value) corresponds to a unique domain (x-value) .

• The horizontal-line test: if a horizontal line cuts the graph of a function only once, then the function is one-to-one and its inverse is a function .

• Inverse function f -1(x): the function formed when the order of the elements in a given function is switched .

• The graph of inverse function f -1(x) is a reflection the original function f (x) about the line y = x .


- Confirm that the function is 1-to-1 .- Rewrite f (x) as y .- Switch x and y .- Solve for y.- Replace y with f -1 (x) . - Graph f -1 (x): reflect the graph of f (x) across the line y = x.

• Composite function f ∘ g(x): a combination of two or more functions in which the result of









Inverse FunctionIf a function is one-to-one, then f -1 ∘ f (x) = x and f ∘ f -1 (x) = x.

• The logarithmic function f (x) = loga x: a function that is the inverse of an exponential function (y = ax) .

y = x

y = 2x

x = 2y

0

Page 10-29







f(x) = log𝑎𝑎 x(x > 0 , a > 0, a ≠ 1)

if 𝑦𝑦𝑦𝑦 = 𝑎𝑎𝑥𝑥𝑥𝑥, then log𝑎𝑎 𝑦𝑦𝑦𝑦 = 𝑥𝑥𝑥𝑥 .Or if 𝑥𝑥𝑥𝑥 = 𝑎𝑎𝑦𝑦, then log𝑎𝑎 x = y.

If 9 = 32, then 2 = log3 9 .Read: “the log base 3 of 9 is 2” or “log of 9, base 3, equals 2” .

• Logarithm of zero log a (0): the logarithm of 0 is undefined .

• Logarithm of negative number log a (-x): the logarithm of negative numbers is undefined .

• Converting between exponential and logarithmic forms

Exponential to log form: 𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦log𝑎𝑎 y = x

Log to exponential form: log𝑎𝑎 y = x𝑎𝑎𝑥𝑥𝑥𝑥 = 𝑦𝑦𝑦𝑦


• The key to solve a logarithmic equation: convert log into exponential form .

• Logarithmic functions and exponential functions are inverse functions .








(0, ∞) or x > 0 (-∞, ∞) (0, ∞)


y > 0 (-∞, ∞)




Property Examplelog𝑎𝑎𝟏𝟏𝟏𝟏 = 0 log41 = 0log𝒂𝒂𝒂𝒂 = 1 log77 = 1log𝑎𝑎 𝑎𝑎𝒙𝒙𝒙𝒙 = x log2 23 = 3𝑎𝑎log𝑎𝑎𝑥𝑥𝑥𝑥 = x 3log34 = 4

∙ 1 ∙ 1

Page 10-30






Name Rule Exampleproduct rule log𝑎𝑎(𝑨𝑨 ∙ 𝑩𝑩) = log𝑎𝑎𝑨𝑨 + log𝑎𝑎𝑩𝑩 log5(3 ∙ 4) = log53 + log54

quotient rule log𝑎𝑎 �𝑨𝑨𝑩𝑩� = log𝑎𝑎𝑨𝑨 − log𝑎𝑎𝑩𝑩 log3 �

72� = log37 − log32

power rule log𝑎𝑎 𝐴𝐴𝒏𝒏 = n log𝑎𝑎𝐴𝐴 log2 3𝟒𝟒 = 4 log23

• The common logarithm


Base 10 No base means base 10𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙103 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3

• The natural logarithm

Notation for Natural Log Example𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑒𝑒𝑥𝑥𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥𝑥𝑥




the y-axis .

The graph of the natural logarithmic function ln x is the reflection of the graph of

the exponential function ex about the y = x line .

• Change of base formula


Example





≈ 1.465

• Exponential equation: an equation that contains variable(s) in the exponent .


LOG 5 ) ÷ LOG 3

LN 5 ) ÷ LN 3

Page 10-31





PRACTICE QUIZ

Unit 10 Exponential & Logarithmic Functions

1. Sketch the graph of 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 amd 𝑓𝑓(𝑥𝑥) = 2-𝑥𝑥 .

2. Determine f (x) and g (x) such that h (x) = f ○ g (x), h (x) = (4 – 3x)3

3. Determine the inverse of the given function f (x) = -5x .

4. Solve each of the following equations .

a . log381 = x

b . log2713

= x

5. Find the value of 8log83 .

6. Write each of the following as simpler logarithms .

a . log4 �6𝑥𝑥𝑦𝑦5�

b . log5�𝑎𝑎2𝑏𝑏𝑐𝑐

4

7. Evaluate each of the following .

a . log0.312

b . log𝜋𝜋4.7


a . 4(2𝑥𝑥+1) = 6𝑥𝑥

b . 𝑙𝑙𝑙𝑙𝑙𝑙3𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙3(4𝑥𝑥 − 1) + 2

Page 15

PRACTICE QUIZ

Unit 10 Exponential & Logarithmic Functions

1. Sketch the graph of 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 amd 𝑓𝑓(𝑥𝑥) = 2-𝑥𝑥 .

2. Determine f (x) and g (x) such that h (x) = f ○ g (x), h (x) = (4 – 3x)3

3. Determine the inverse of the given function f (x) = -5x .


a . log381 = x

b . log2713

= x

5. Find the value of 8log83 .

6. Write each of the following as simpler logarithms .

a . log4 �6𝑥𝑥𝑦𝑦5�

b . log5�𝑎𝑎2𝑏𝑏𝑐𝑐

4

7. Evaluate each of the following .

a . log0.312

b . log𝜋𝜋4.7


a . 4(2𝑥𝑥+1) = 6𝑥𝑥

b . 𝑙𝑙𝑙𝑙𝑙𝑙3𝑥𝑥 = 𝑙𝑙𝑙𝑙𝑙𝑙3(4𝑥𝑥 − 1) + 2

Page 15




Algebra I & II Key Concepts, Practice, and Quizzes Unit 11 – Determinants and Matrices

UNIT 11 DETERMINANTS AND MATRICES

11-1 DETERMINANTS

Second-Order Determinants

• Determinant : a square set of numbers (called elements) enclosed in two lines that

represents the sum of the products of numbers, and is useful for solving systems of linear

equations .

• Dimensions of a determinant: a determinant has m rows and n columns (m × n) Rows Columns

• A second-order determinant (2 × 2): �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2

�

2 rows 2 columns Column 1 Column 2

• Evaluate a 2 × 2 determinant

Determinant Evaluation Example

�𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2

� �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2

�+

= 𝑎𝑎1𝑏𝑏2 − 𝑏𝑏1𝑎𝑎2 �2 13 4�+

= 2 ∙ 4 − 1 ∙ 3 = 5

- Draw a diagonal from the first element of the top row downward to the right .

- Draw a diagonal from the second element of the top row downward to the left .

- Multiply the elements on the diagonals, and subtract the products .

�descending from left to right is positive

descending from right to left is negative

Example: Evaluate the following determinants .

1. �2 4-2 -3� = 2 �-3� − 4 �-2� = -6 + 8 = 𝟐𝟐

2. �2 𝑥𝑥-3 5� = 2 ∙ 5 − 𝑥𝑥 �-3� = 𝟏𝟏𝟏𝟏 + 𝟑𝟑𝟑𝟑

Row 1

Row 2

is +

is −

Page 11-1


11-1 DETERMINANTS




equations .



�






�+

= 𝑎𝑎1𝑏𝑏2 − 𝑏𝑏1𝑎𝑎2 �2 13 4�+

= 2 ∙ 4 − 1 ∙ 3 = 5







1. �2 4-2 -3� = 2 �-3� − 4 �-2� = -6 + 8 = 𝟐𝟐

2. �2 𝑥𝑥-3 5� = 2 ∙ 5 − 𝑥𝑥 �-3� = 𝟏𝟏𝟏𝟏 + 𝟑𝟑𝟑𝟑

Row 1

Row 2

is +

is −

Page 11-1


11-1 DETERMINANTS




equations .



�






�+

= 𝑎𝑎1𝑏𝑏2 − 𝑏𝑏1𝑎𝑎2 �2 13 4�+

= 2 ∙ 4 − 1 ∙ 3 = 5







1. �2 4-2 -3� = 2 �-3� − 4 �-2� = -6 + 8 = 𝟐𝟐

2. �2 𝑥𝑥-3 5� = 2 ∙ 5 − 𝑥𝑥 �-3� = 𝟏𝟏𝟏𝟏 + 𝟑𝟑𝟑𝟑

Row 1

Row 2

is +

is −

Page 11-1


11-1 DETERMINANTS




equations .



�






�+

= 𝑎𝑎1𝑏𝑏2 − 𝑏𝑏1𝑎𝑎2 �2 13 4�+

= 2 ∙ 4 − 1 ∙ 3 = 5







1. �2 4-2 -3� = 2 �-3� − 4 �-2� = -6 + 8 = 𝟐𝟐

2. �2 𝑥𝑥-3 5� = 2 ∙ 5 − 𝑥𝑥 �-3� = 𝟏𝟏𝟏𝟏 + 𝟑𝟑𝟑𝟑

Row 1

Row 2

is +

is −

Page 11-1





Third-Order DeterminantsExpansion by Diagonals

• A third-order determinant (3 × 3): �𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

�

Column 1 Column 2 Column 3

• Evaluate a 3 × 3 Determinant – Method I: Using Diagonals

Steps

- Copy the first two columns of the determinant to 𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝒂𝒂𝟏𝟏 𝒃𝒃𝟏𝟏𝒂𝒂𝟐𝟐 𝒃𝒃𝟐𝟐𝒂𝒂𝟑𝟑 𝒃𝒃𝟑𝟑

its right .

- Draw three diagonals from each element of the top 𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

row downward to the right .

- Draw three diagonals from each element of the top𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


row downward to the left .

- Multiply the elements on the diagonals, and sum the products .

�descending from left to right is positive. descending from right to left is negative. 𝑎𝑎1𝑏𝑏2𝑐𝑐3+𝑏𝑏1𝑐𝑐2𝑎𝑎3 + 𝑐𝑐1𝑎𝑎2𝑏𝑏3

-𝑐𝑐1𝑏𝑏2𝑎𝑎3 − 𝑎𝑎1𝑐𝑐2𝑏𝑏3 − 𝑏𝑏1𝑎𝑎2𝑐𝑐3

3 × 3 Determinant Expansion by Diagonals Example

𝒂𝒂𝟏𝟏 𝒃𝒃𝟏𝟏 𝒄𝒄𝟏𝟏 𝒂𝒂𝟐𝟐 𝒃𝒃𝟐𝟐 𝒄𝒄𝟐𝟐 𝒂𝒂𝟑𝟑 𝒃𝒃𝟑𝟑 𝒄𝒄𝟑𝟑



= 𝑎𝑎1𝑏𝑏2𝑐𝑐3+𝑏𝑏1𝑐𝑐2𝑎𝑎3 + 𝑐𝑐1𝑎𝑎2𝑏𝑏3 −𝑐𝑐1𝑏𝑏2𝑎𝑎3 − 𝑎𝑎1𝑐𝑐2𝑏𝑏3 − 𝑏𝑏1𝑎𝑎2𝑐𝑐3

1 1 3 4 3 2 3 1 2

1 14 33 1

= 1∙3∙2 + 1∙2∙3 + 3∙4∙1−3∙3∙3 − 1∙2∙1 − 1∙4∙2

= 6 + 6 + 12 – 27 – 2 – 8 = -13

Note: ‘Expansion by diagonals’ does not work with 4 × 4 or higher-order determinants .

Example: Evaluate the determinant . 𝟑𝟑 𝟐𝟐 𝟎𝟎 -𝟏𝟏 𝟎𝟎 𝟏𝟏 𝟒𝟒 𝟏𝟏 𝟓𝟓

3 2 0 -1 0 1 4 1 5

3 2-1 04 1

= 3∙0∙5 + 2∙1∙4 + 0∙(-1)∙1 – 0∙0∙4 –3∙1∙1 – 2∙(-1)∙5

= 0 + 8 + 0 – 0 – 3 + 10 = 15

Row 1

Row 3 Row 2

is +

is −

Page 11-2

Third-Order DeterminantsExpansion by Diagonals

• A third-order determinant (3 × 3): �𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

�


• Evaluate a 3 × 3 Determinant – Method I: Using Diagonals

Steps

- Copy the first two columns of the determinant to 𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


𝒂𝒂𝟏𝟏 𝒃𝒃𝟏𝟏𝒂𝒂𝟐𝟐 𝒃𝒃𝟐𝟐𝒂𝒂𝟑𝟑 𝒃𝒃𝟑𝟑

its right .

- Draw three diagonals from each element of the top 𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


row downward to the right .

- Draw three diagonals from each element of the top𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


row downward to the left .

- Multiply the elements on the diagonals, and sum the products .

�descending from left to right is positive. descending from right to left is negative. 𝑎𝑎1𝑏𝑏2𝑐𝑐3+𝑏𝑏1𝑐𝑐2𝑎𝑎3 + 𝑐𝑐1𝑎𝑎2𝑏𝑏3

-𝑐𝑐1𝑏𝑏2𝑎𝑎3 − 𝑎𝑎1𝑐𝑐2𝑏𝑏3 − 𝑏𝑏1𝑎𝑎2𝑐𝑐3

3 × 3 Determinant Expansion by Diagonals Example





1 1 3 4 3 2 3 1 2

1 14 33 1

= 1∙3∙2 + 1∙2∙3 + 3∙4∙1−3∙3∙3 − 1∙2∙1 − 1∙4∙2

= 6 + 6 + 12 – 27 – 2 – 8 = -13


Example: Evaluate the determinant . 𝟑𝟑 𝟐𝟐 𝟎𝟎 -𝟏𝟏 𝟎𝟎 𝟏𝟏 𝟒𝟒 𝟏𝟏 𝟓𝟓

3 2 0 -1 0 1 4 1 5

3 2-1 04 1

= 3∙0∙5 + 2∙1∙4 + 0∙(-1)∙1 – 0∙0∙4 –3∙1∙1 – 2∙(-1)∙5

= 0 + 8 + 0 – 0 – 3 + 10 = 15

Row 1

Row 3 Row 2

is +

is −

Page 11-2





Third-Order DeterminantsExpansion by Minors

• The minor of an element in a determinant is a determinant of the next lower order . It is

the determinant that results from crossing out the row and the column that contain

that element.

Example: Find the minor of element a1 and b2 .Cross out this column .

𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

A1 = �𝑏𝑏2 𝑐𝑐2𝑏𝑏3 𝑐𝑐3

� ,𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝒃𝒃𝟐𝟐 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

B2 = �𝑎𝑎1 𝑐𝑐1𝑎𝑎3 𝑐𝑐3�

Cross out this column The minor in a 3×3 determinant is a 2×2 determinant .

• Placing signs for minor: the sign of a minor is determined by its position in the determinant .

- A 3×3 determinant: + − +− + −+ − +

- A 4×4 determinant: + − + −− + − ++−

−+

+ −− +

• Cofactors: minors + place signs = cofactors

𝐴𝐴1 𝐵𝐵1 𝐶𝐶1𝐴𝐴2 𝐵𝐵2 𝐶𝐶2𝐴𝐴3 𝐵𝐵3 𝐶𝐶3

+ + − +− + −+ − +

𝐴𝐴1 -𝐵𝐵1 𝐶𝐶1 -𝐴𝐴2 𝐵𝐵2 -𝐶𝐶2 𝐴𝐴3 -𝐵𝐵3 𝐶𝐶3

minors place signs cofactors

• Evaluate a 3 × 3 determinant – Method II: expansion by minorsSteps

- Choose any row or column in the determinant . Choose this column .

𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1𝒂𝒂𝟐𝟐 𝑏𝑏2 𝑐𝑐2𝒂𝒂𝟑𝟑 𝑏𝑏3 𝑐𝑐3

- Multiply each element in the chosen row/column by its cofactor .𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1 𝒂𝒂𝟐𝟐 𝑏𝑏2 𝑐𝑐2 =𝒂𝒂𝟑𝟑 𝑏𝑏3 𝑐𝑐3

𝒂𝒂𝟏𝟏 �𝑏𝑏2 𝑐𝑐2𝑏𝑏3 𝑐𝑐3

� − 𝒂𝒂𝟐𝟐 �𝑏𝑏1 𝑐𝑐1𝑏𝑏3 𝑐𝑐3

� + 𝒂𝒂𝟑𝟑 �𝑏𝑏1 𝑐𝑐1𝑏𝑏2 𝑐𝑐2

�

+ − + − + − + − +

a1 a2 a3

𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝒂𝒂𝟐𝟐 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝒂𝒂𝟑𝟑 𝑏𝑏3 𝑐𝑐3

Tip: Multiply a1, a2, and a3 by a minor that is not in its row or column .

Cross out this row .


Page 11-3

Third-Order DeterminantsExpansion by Minors

• The minor of an element in a determinant is a determinant of the next lower order . It is

the determinant that results from crossing out the row and the column that contain

that element.

Example: Find the minor of element a1 and b2 .Cross out this column .

𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

A1 = �𝑏𝑏2 𝑐𝑐2𝑏𝑏3 𝑐𝑐3

� ,𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝒃𝒃𝟐𝟐 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

B2 = �𝑎𝑎1 𝑐𝑐1𝑎𝑎3 𝑐𝑐3�

Cross out this column The minor in a 3×3 determinant is a 2×2 determinant .

• Placing signs for minor: the sign of a minor is determined by its position in the determinant .

- A 3×3 determinant: + − +− + −+ − +

- A 4×4 determinant: + − + −− + − ++−

−+

+ −− +



+ + − +− + −+ − +



• Evaluate a 3 × 3 determinant – Method II: expansion by minorsSteps

- Choose any row or column in the determinant . Choose this column .

𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1𝒂𝒂𝟐𝟐 𝑏𝑏2 𝑐𝑐2𝒂𝒂𝟑𝟑 𝑏𝑏3 𝑐𝑐3

- Multiply each element in the chosen row/column by its cofactor .𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1 𝒂𝒂𝟐𝟐 𝑏𝑏2 𝑐𝑐2 =𝒂𝒂𝟑𝟑 𝑏𝑏3 𝑐𝑐3




�

+ − + − + − + − +

a1 a2 a3




Tip: Multiply a1, a2, and a3 by a minor that is not in its row or column .



Page 11-3





Expansion by MinorsExpansion by any Row / Column

Evaluate a determinant that can be expanded by any row or column• Expanding along column 1

Expansion by Minors (choose column 1) Example

𝒂𝒂𝟏𝟏 𝑏𝑏1 𝑐𝑐1 𝒂𝒂𝟐𝟐 𝑏𝑏2 𝑐𝑐2 =𝒂𝒂𝟑𝟑 𝑏𝑏3 𝑐𝑐3




�

𝟐𝟐 0 3 𝟎𝟎 -2 1 =𝟐𝟐 1 4

𝟐𝟐 �-2 11 4

� − 𝟎𝟎 �0 31 4� + 𝟐𝟐 �

0 3-2 1�

= 2(-2∙4−1∙1) − 0(0∙4−3∙1) + 2[0∙1− 3(-2)] = 2(-9) − 0 + 2∙ 6 = -6

+ − + − + − + − +

a1 a2 a3




• Expanding along row 1Expansion by Minors (choose row 1) Example

𝒂𝒂𝟏𝟏 𝒃𝒃𝟏𝟏 𝒄𝒄𝟏𝟏 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 =𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


� − 𝒃𝒃𝟏𝟏 �𝑎𝑎2 𝑐𝑐2𝑎𝑎3 𝑐𝑐3� + 𝒄𝒄𝟏𝟏 �

𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

� 𝟐𝟐 0 𝟑𝟑 0 -2 1 =2 1 4

𝟐𝟐 �-2 11 4

� − 𝟎𝟎 �0 12 4� + 𝟑𝟑 �0 -2

2 1�

= 2[(-2)∙4 - 1∙1] – 0(0∙4 –1∙2)+3[0∙1– (-2) ∙2] = 2(-9) + 0 + 3∙4 = -6

(The same result as choosing column 1 .)

+ − + − + − + − +

a1 b2 c1


𝑎𝑎1 𝒃𝒃𝟏𝟏 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑎𝑎1 𝑏𝑏1 𝒄𝒄𝟏𝟏 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

• Expanding along column 2Expansion by Minors (choose column 2) Example

𝑎𝑎1 𝒃𝒃𝟏𝟏 𝑐𝑐1 𝑎𝑎2 𝒃𝒃𝟐𝟐 𝑐𝑐2 =𝑎𝑎3 𝒃𝒃𝟑𝟑 𝑐𝑐3

-𝒃𝒃𝟏𝟏 �𝑎𝑎2 𝑐𝑐2𝑎𝑎3 𝑐𝑐3� + 𝒃𝒃𝟐𝟐 �

𝑎𝑎1 𝑐𝑐1𝑎𝑎3 𝑐𝑐3� − 𝒃𝒃𝟑𝟑 �

𝑎𝑎1 𝑐𝑐1𝑎𝑎2 𝑐𝑐2�

2 0 3 0 -𝟐𝟐 1 =2 𝟏𝟏 4

-𝟎𝟎 �0 12 4� + �-𝟐𝟐� �2 3

2 4� − 𝟏𝟏 �2 30 1�

= -0(0∙4 – 1∙2) – 2(2∙4 – 2∙3) – 1(2∙1– 3∙0) = 0 – 4 – 2 = -6

(The same result as choosing column 1 or row 1 .)

+ − + − + − + − +

b1 b2 b3


𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝒃𝒃𝟐𝟐 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝒃𝒃𝟑𝟑 𝑐𝑐3

Page 11-4

Expansion by MinorsExpansion by any Row / Column

Evaluate a determinant that can be expanded by any row or column• Expanding along column 1

Expansion by Minors (choose column 1) Example





�

𝟐𝟐 0 3 𝟎𝟎 -2 1 =𝟐𝟐 1 4

𝟐𝟐 �-2 11 4

� − 𝟎𝟎 �0 31 4� + 𝟐𝟐 �

0 3-2 1�

= 2(-2∙4−1∙1) − 0(0∙4−3∙1) + 2[0∙1− 3(-2)] = 2(-9) − 0 + 2∙ 6 = -6

+ − + − + − + − +

a1 a2 a3




• Expanding along row 1Expansion by Minors (choose row 1) Example

𝒂𝒂𝟏𝟏 𝒃𝒃𝟏𝟏 𝒄𝒄𝟏𝟏 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 =𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


� − 𝒃𝒃𝟏𝟏 �𝑎𝑎2 𝑐𝑐2𝑎𝑎3 𝑐𝑐3� + 𝒄𝒄𝟏𝟏 �


� 𝟐𝟐 0 𝟑𝟑 0 -2 1 =2 1 4

𝟐𝟐 �-2 11 4

� − 𝟎𝟎 �0 12 4� + 𝟑𝟑 �0 -2

2 1�

= 2[(-2)∙4 - 1∙1] – 0(0∙4 –1∙2)+3[0∙1– (-2) ∙2] = 2(-9) + 0 + 3∙4 = -6

(The same result as choosing column 1 .)

+ − + − + − + − +

a1 b2 c1



𝑎𝑎1 𝑏𝑏1 𝒄𝒄𝟏𝟏 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

• Expanding along column 2Expansion by Minors (choose column 2) Example

𝑎𝑎1 𝒃𝒃𝟏𝟏 𝑐𝑐1 𝑎𝑎2 𝒃𝒃𝟐𝟐 𝑐𝑐2 =𝑎𝑎3 𝒃𝒃𝟑𝟑 𝑐𝑐3

-𝒃𝒃𝟏𝟏 �𝑎𝑎2 𝑐𝑐2𝑎𝑎3 𝑐𝑐3� + 𝒃𝒃𝟐𝟐 �

𝑎𝑎1 𝑐𝑐1𝑎𝑎3 𝑐𝑐3� − 𝒃𝒃𝟑𝟑 �

𝑎𝑎1 𝑐𝑐1𝑎𝑎2 𝑐𝑐2�

2 0 3 0 -𝟐𝟐 1 =2 𝟏𝟏 4

-𝟎𝟎 �0 12 4� + �-𝟐𝟐� �2 3

2 4� − 𝟏𝟏 �2 30 1�

= -0(0∙4 – 1∙2) – 2(2∙4 – 2∙3) – 1(2∙1– 3∙0) = 0 – 4 – 2 = -6

(The same result as choosing column 1 or row 1 .)

+ − + − + − + − +

b1 b2 b3


𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝒃𝒃𝟐𝟐 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝒃𝒃𝟑𝟑 𝑐𝑐3

Page 11-4





11-2 CRAMER’S RULE

Cramer’s Rule to Solve a 2 × 2 System

• Cramer’s rule: Use the determinant method to solve the system of linear equations .

• Using Cramer’s rule to solve a 2×2 systemA 2×2 System Cramer’s Rule

�𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 = 𝑘𝑘1𝑎𝑎2𝑥𝑥 + 𝑏𝑏2 𝑦𝑦 = 𝑘𝑘2

The solution of the system: x = 𝐷𝐷𝑥𝑥𝐷𝐷

, y = 𝐷𝐷𝑦𝑦𝐷𝐷

(D ≠ 0)

𝐷𝐷 = �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2

� , 𝐷𝐷𝑥𝑥 = �𝑘𝑘1 𝑏𝑏1𝑘𝑘2 𝑏𝑏2

� , 𝐷𝐷𝑦𝑦 = �𝑎𝑎1 𝑘𝑘1𝑎𝑎2 𝑘𝑘2

�

coefficients of x Replace the column a in D with k.

coefficients of y Replace the column b in D with k.

constant

Example Using Cramer’s Rule to Solve a 2×2 System

�3𝑥𝑥 + 𝑦𝑦 = 1 2𝑥𝑥 − 3𝑦𝑦 = 4

𝑫𝑫 = �3 12 -3� = 3(-3) – 1 ∙ 2 = -11 , 𝑫𝑫𝒙𝒙 = �

1 14 -3� = 1 ∙ �-3� − 1 ∙ 4 = -𝟕𝟕, 𝑫𝑫𝒚𝒚 = �3 1

2 4� = 3 ∙ 4 − 1 ∙ 2 =10

x = 𝐷𝐷𝑥𝑥𝐷𝐷

= -7-11

= 711


= 10-11

= - 1011

, Solution: � 𝟕𝟕𝟏𝟏𝟏𝟏

, - 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏�

Two equations in two unknowns (x & y) .

Steps Example: Solve . �2𝑥𝑥 = 2 − 3𝑦𝑦 𝑦𝑦 = -3 − 4𝑥𝑥

- Write equations in standard form .


�2𝑥𝑥 + 3𝑦𝑦 = 2 4𝑥𝑥 + 𝑦𝑦 = -3

- Determine the determinants D, Dx and Dy . (the coefficients of the system)


� = a1 b2 – b1a2 𝐷𝐷 = �2 34 1� = 2 ∙ 1 – 3 ∙ 4 = -10

𝐷𝐷𝑥𝑥 = �𝑘𝑘1 𝑏𝑏1𝑘𝑘2 𝑏𝑏2

� = k1 b2 – b1k2 𝐷𝐷𝑥𝑥 = �2 3-3 1� = 2 ∙ 1 − 3(-3) = 11

Replace the column a in D with the column k.

𝐷𝐷𝑦𝑦 = �𝑎𝑎1 𝑘𝑘1𝑎𝑎2 𝑘𝑘2

�= a1 k2 – k1a2 𝐷𝐷𝑦𝑦 = �2 24 -3� = 2�-3� − 2 ∙ 4 = -14

Replace the column b in D with the column k.

- Solve for x and y. x = 𝐷𝐷𝑥𝑥𝐷𝐷


= 11-10

= - 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

y = 𝐷𝐷𝑦𝑦𝐷𝐷


= -14-10

= 𝟕𝟕𝟓𝟓

Page 11-5

11-2 CRAMER’S RULE

Cramer’s Rule to Solve a 2 × 2 System

• Cramer’s rule: Use the determinant method to solve the system of linear equations .





(D ≠ 0)




�



constant

Example Using Cramer’s Rule to Solve a 2×2 System

�3𝑥𝑥 + 𝑦𝑦 = 1 2𝑥𝑥 − 3𝑦𝑦 = 4

𝑫𝑫 = �3 12 -3� = 3(-3) – 1 ∙ 2 = -11 , 𝑫𝑫𝒙𝒙 = �

1 14 -3� = 1 ∙ �-3� − 1 ∙ 4 = -𝟕𝟕, 𝑫𝑫𝒚𝒚 = �3 1

2 4� = 3 ∙ 4 − 1 ∙ 2 =10


= -7-11

= 711


= 10-11

= - 1011

, Solution: � 𝟕𝟕𝟏𝟏𝟏𝟏

, - 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏�

Two equations in two unknowns (x & y) .

Steps Example: Solve . �2𝑥𝑥 = 2 − 3𝑦𝑦 𝑦𝑦 = -3 − 4𝑥𝑥

- Write equations in standard form .


�2𝑥𝑥 + 3𝑦𝑦 = 2 4𝑥𝑥 + 𝑦𝑦 = -3

- Determine the determinants D, Dx and Dy . (the coefficients of the system)


� = a1 b2 – b1a2 𝐷𝐷 = �2 34 1� = 2 ∙ 1 – 3 ∙ 4 = -10

𝐷𝐷𝑥𝑥 = �𝑘𝑘1 𝑏𝑏1𝑘𝑘2 𝑏𝑏2

� = k1 b2 – b1k2 𝐷𝐷𝑥𝑥 = �2 3-3 1� = 2 ∙ 1 − 3(-3) = 11

Replace the column a in D with the column k.

𝐷𝐷𝑦𝑦 = �𝑎𝑎1 𝑘𝑘1𝑎𝑎2 𝑘𝑘2

�= a1 k2 – k1a2 𝐷𝐷𝑦𝑦 = �2 24 -3� = 2�-3� − 2 ∙ 4 = -14

Replace the column b in D with the column k.

- Solve for x and y. x = 𝐷𝐷𝑥𝑥𝐷𝐷


= 11-10

= - 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏



= -14-10

= 𝟕𝟕𝟓𝟓

Page 11-5





Cramer’s Rule to Solve a 3 × 3 SystemExpansion by Diagonals

• Using Cramer’s rule to solve a 3×3 systemA 3×3 System Cramer’s Rule D Dx , Dy and Dz

�𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1𝑧𝑧 = 𝑘𝑘1𝑎𝑎2𝑥𝑥 + 𝑏𝑏2 𝑦𝑦+ 𝑐𝑐2𝑦𝑦 = 𝑘𝑘2𝑎𝑎3𝑥𝑥 + 𝑏𝑏3 𝑦𝑦+𝑐𝑐3𝑦𝑦 = 𝑘𝑘3



z = 𝐷𝐷𝑧𝑧𝐷𝐷

D = 𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

D ≠ 0

𝐷𝐷𝑥𝑥 = 𝒌𝒌𝟏𝟏 𝑏𝑏1 𝑐𝑐1 𝒌𝒌𝟏𝟏 𝑏𝑏2 𝑐𝑐2 𝒌𝒌𝟏𝟏 𝑏𝑏3 𝑐𝑐3

𝐷𝐷𝑦𝑦 = 𝑎𝑎1 𝒌𝒌𝟏𝟏 𝑐𝑐1 𝑎𝑎2 𝒌𝒌𝟐𝟐 𝑐𝑐2 𝑎𝑎3 𝒌𝒌𝟑𝟑 𝑐𝑐3

𝐷𝐷𝑧𝑧 = 𝑎𝑎1 𝑏𝑏1 𝒌𝒌𝟏𝟏 𝑎𝑎2 𝑏𝑏2 𝒌𝒌𝟐𝟐 𝑎𝑎3 𝑏𝑏3 𝒌𝒌𝟑𝟑

Tip: Use either the ‘expansion by minors’ or ‘expansion by diagonals’ to solve the determinants D, Dx, Dy and Dz .

D The determinant of the coefficient of variables x , y and z .Dx Replacing the coefficients of x in D with the column of constants k .Dy Replacing the coefficients of y in D with the column of constants k .Dz Replacing the coefficients of z in D with the column of constants k .

• Using Cramer’s rule to solve a 3×3 system (use expansion by diagonals)A 3×3 System Determinants & Solutions Example


Example:

�2𝑥𝑥 − 𝑦𝑦 + 3𝑧𝑧 = 1 3𝑥𝑥 + 2𝑦𝑦 + 0𝑧𝑧 = −4 0𝑥𝑥 + 2𝑦𝑦 − 3𝑧𝑧 = 2

𝐷𝐷 =𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


𝐷𝐷𝑥𝑥 = 𝑘𝑘1 𝑏𝑏1 𝑐𝑐1𝑘𝑘2 𝑏𝑏2 𝑐𝑐2𝑘𝑘3 𝑏𝑏3 𝑐𝑐3

𝑘𝑘1 𝑏𝑏1𝑘𝑘2 𝑏𝑏2𝑘𝑘3 𝑏𝑏3

𝐷𝐷𝑦𝑦 = 𝑎𝑎1 𝑘𝑘1 𝑐𝑐1𝑎𝑎2 𝑘𝑘2 𝑐𝑐2𝑎𝑎3 𝑘𝑘3 𝑐𝑐3

𝑎𝑎1 𝑘𝑘1𝑎𝑎2 𝑘𝑘2𝑎𝑎3 𝑘𝑘3

𝐷𝐷𝑧𝑧 =𝑎𝑎1 𝑏𝑏1 𝑘𝑘1𝑎𝑎2 𝑏𝑏2 𝑘𝑘2𝑎𝑎3 𝑏𝑏3 𝑘𝑘3


Solutions: x = 𝐷𝐷𝑥𝑥

𝐷𝐷, y = 𝐷𝐷𝑦𝑦

𝐷𝐷, z = 𝐷𝐷𝑧𝑧

𝐷𝐷

𝐷𝐷 = 2 -1 3 3 2 0 0 2 -3

2 -13 20 2

= 2∙2(-3) + (-1)∙0∙0 + 3∙3∙2 – 3∙2∙0 – 2∙0∙2 – (-1)∙3(-3) = -12 + 0 + 18 – 0 – 0 – 9 = -3

𝐷𝐷𝑥𝑥 = 1 -1 3 -4 2 0 2 2 -3

1 -1-4 22 2

= 1∙2(-3) + (-1)∙0∙2 +3(-4)∙2 – 3∙2∙2 –1∙0∙2 – (-1)(-4)(-3) = -6 + 0 – 24 – 12 – 0 + 12 = -30

𝐷𝐷𝑦𝑦 = 2 1 3 3 -4 0 0 2 -3

2 13 -40 2

= 2(-4)(-3) + 1∙0∙0 + 3∙3∙2 – 3(-4)∙0 – 2∙0∙2 –1∙3(-3) = 24 + 0 +18 + 0 – 0 + 9 = 51

𝐷𝐷𝑧𝑧 = 2 -1 1 3 2 -4 0 2 2

2 -13 20 2

= 2∙2∙2 + (-1)(-4)∙0 + 1∙3∙2 – 1∙2∙0 – 2(-4)2 – (-1)∙3∙2 = 8 + 0 + 6 – 0 + 16 + 6 = 36

Solutions: x = 𝐷𝐷𝑥𝑥𝐷𝐷

= -30-3

= 𝟏𝟏𝟏𝟏 , y = 𝐷𝐷𝑦𝑦𝐷𝐷

= 51-3

= -𝟏𝟏𝟏𝟏


= 36-3

= -𝟏𝟏𝟐𝟐

Page 11-6

Cramer’s Rule to Solve a 3 × 3 SystemExpansion by Diagonals







D ≠ 0




Tip: Use either the ‘expansion by minors’ or ‘expansion by diagonals’ to solve the determinants D, Dx, Dy and Dz .

D The determinant of the coefficient of variables x , y and z .Dx Replacing the coefficients of x in D with the column of constants k .Dy Replacing the coefficients of y in D with the column of constants k .Dz Replacing the coefficients of z in D with the column of constants k .

• Using Cramer’s rule to solve a 3×3 system (use expansion by diagonals)A 3×3 System Determinants & Solutions Example


Example:

�2𝑥𝑥 − 𝑦𝑦 + 3𝑧𝑧 = 1 3𝑥𝑥 + 2𝑦𝑦 + 0𝑧𝑧 = −4 0𝑥𝑥 + 2𝑦𝑦 − 3𝑧𝑧 = 2

𝐷𝐷 =𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3


𝐷𝐷𝑥𝑥 = 𝑘𝑘1 𝑏𝑏1 𝑐𝑐1𝑘𝑘2 𝑏𝑏2 𝑐𝑐2𝑘𝑘3 𝑏𝑏3 𝑐𝑐3

𝑘𝑘1 𝑏𝑏1𝑘𝑘2 𝑏𝑏2𝑘𝑘3 𝑏𝑏3

𝐷𝐷𝑦𝑦 = 𝑎𝑎1 𝑘𝑘1 𝑐𝑐1𝑎𝑎2 𝑘𝑘2 𝑐𝑐2𝑎𝑎3 𝑘𝑘3 𝑐𝑐3

𝑎𝑎1 𝑘𝑘1𝑎𝑎2 𝑘𝑘2𝑎𝑎3 𝑘𝑘3

𝐷𝐷𝑧𝑧 =𝑎𝑎1 𝑏𝑏1 𝑘𝑘1𝑎𝑎2 𝑏𝑏2 𝑘𝑘2𝑎𝑎3 𝑏𝑏3 𝑘𝑘3


Solutions: x = 𝐷𝐷𝑥𝑥

𝐷𝐷, y = 𝐷𝐷𝑦𝑦

𝐷𝐷, z = 𝐷𝐷𝑧𝑧

𝐷𝐷

𝐷𝐷 = 2 -1 3 3 2 0 0 2 -3

2 -13 20 2

= 2∙2(-3) + (-1)∙0∙0 + 3∙3∙2 – 3∙2∙0 – 2∙0∙2 – (-1)∙3(-3) = -12 + 0 + 18 – 0 – 0 – 9 = -3

𝐷𝐷𝑥𝑥 = 1 -1 3 -4 2 0 2 2 -3

1 -1-4 22 2

= 1∙2(-3) + (-1)∙0∙2 +3(-4)∙2 – 3∙2∙2 –1∙0∙2 – (-1)(-4)(-3) = -6 + 0 – 24 – 12 – 0 + 12 = -30

𝐷𝐷𝑦𝑦 = 2 1 3 3 -4 0 0 2 -3

2 13 -40 2

= 2(-4)(-3) + 1∙0∙0 + 3∙3∙2 – 3(-4)∙0 – 2∙0∙2 –1∙3(-3) = 24 + 0 +18 + 0 – 0 + 9 = 51

𝐷𝐷𝑧𝑧 = 2 -1 1 3 2 -4 0 2 2

2 -13 20 2

= 2∙2∙2 + (-1)(-4)∙0 + 1∙3∙2 – 1∙2∙0 – 2(-4)2 – (-1)∙3∙2 = 8 + 0 + 6 – 0 + 16 + 6 = 36

Solutions: x = 𝐷𝐷𝑥𝑥𝐷𝐷

= -30-3

= 𝟏𝟏𝟏𝟏 , y = 𝐷𝐷𝑦𝑦𝐷𝐷

= 51-3

= -𝟏𝟏𝟏𝟏


= 36-3

= -𝟏𝟏𝟐𝟐

Page 11-6





Cramer’s Rule to Solve a 3 × 3 SystemExpansion by Minors

Steps Example: Solve . �𝟐𝟐𝟐𝟐 = 𝒚𝒚 − 𝟑𝟑𝟑𝟑 + 𝟏𝟏𝟑𝟑𝟐𝟐 + 𝟒𝟒 + 𝟐𝟐𝒚𝒚 = 𝟎𝟎 𝟐𝟐𝒚𝒚 − 𝟑𝟑𝟑𝟑 = 𝟐𝟐

- Write equations in standardform .

�𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1𝑧𝑧 = 𝑘𝑘1 𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦 + 𝑐𝑐2𝑧𝑧 = 𝑘𝑘2 𝑎𝑎3𝑥𝑥 + 𝑏𝑏3𝑦𝑦 + 𝑐𝑐3𝑧𝑧 = 𝑘𝑘3

�2𝑥𝑥 − 𝑦𝑦 + 3𝑧𝑧 = 1 3𝑥𝑥 + 2𝑦𝑦 + 0𝑧𝑧 = -4 0𝑥𝑥 + 2𝑦𝑦 − 3𝑧𝑧 = 2

- Determine D, Dx, Dy and Dz . Use ‘expansion by minors .’


D = 𝟐𝟐 -1 3 𝟑𝟑 2 0 =𝟎𝟎 2 -3

𝟐𝟐 �2 02 -3� − 𝟑𝟑 �-1 3

2 -3� + 𝟎𝟎 �-1 3

2 0�

= 2[2(-3) – 0(2)] –3[(-1)(-3) – 2∙3] + 0[(-1)∙0–3∙2]

Choose column 1 . = 2(-6) – 3(3 – 6) + 0 = -12 + 9 = -3

Dx = 𝑘𝑘1 𝑏𝑏1 𝑐𝑐1 𝑘𝑘2 𝑏𝑏2 𝑐𝑐2𝑘𝑘3 𝑏𝑏3 𝑐𝑐3

Dx =𝟏𝟏 -1 3 -𝟒𝟒 2 0 =𝟐𝟐 2 -3

𝟏𝟏 �2 02 -3� − (-𝟒𝟒) �-1 3

2 -3� + 𝟐𝟐 �-1 3

2 0�

= 1[2(-3) – 0∙2] – (-4)[(-1)(-3) – 3∙2] + 2[(-1)∙0 – 3∙2]

= -6 + 4(3–6) + 2(-6) = -6 – 12 – 12 = -30

Dy = 𝑎𝑎1 𝑘𝑘1 𝑐𝑐1 𝑎𝑎2 𝑘𝑘2 𝑐𝑐2 𝑎𝑎3 𝑘𝑘3 𝑐𝑐3

Dy = 2 1 𝟑𝟑 3 -4 𝟎𝟎 =0 2 -𝟑𝟑

𝟑𝟑 �3 -40 2

� − 𝟎𝟎 �2 10 2� + (-𝟑𝟑) �

2 13 -4�

Choose column 3 . = 3[3∙2 – (-4)∙0] – 0 + (-3)[2(-4) – 1∙3]

= 3∙6 – 3(-8 – 3) = 18 + 33 = 51

DZ = 𝑎𝑎1 𝑏𝑏1 𝑘𝑘1 𝑎𝑎2 𝑏𝑏2 𝑘𝑘2 𝑎𝑎3 𝑏𝑏3 𝑘𝑘3

DZ =2 -1 1 3 2 -4 = 𝟎𝟎 𝟐𝟐 𝟐𝟐

𝟎𝟎 �-1 12 -4

� − 𝟐𝟐 �2 13 -4� + 𝟐𝟐 �2 -1

3 2�

Choose row 3 . = 0 – 2[2(- 4) –1∙3] + 2[2∙2 – (-1)3] = -2(-8 – 3) + 2(4 + 3) = 22 + 14 = 36

- Solve for x , y and z .



= -30-3

= 10


𝒚𝒚 = 𝐷𝐷𝑦𝑦𝐷𝐷

= 51-3

= -17


z = 36 -3

= -12

Note: It gives the same answers as expansion by diagonals .

+ − + − + − + − +

Tip: Add a coefficient for the missing variable .

+ − + − + − + − +

+ − + − + − + − +

Page 11-7

Cramer’s Rule to Solve a 3 × 3 SystemExpansion by Minors

Steps Example: Solve . �𝟐𝟐𝟐𝟐 = 𝒚𝒚 − 𝟑𝟑𝟑𝟑 + 𝟏𝟏𝟑𝟑𝟐𝟐 + 𝟒𝟒 + 𝟐𝟐𝒚𝒚 = 𝟎𝟎 𝟐𝟐𝒚𝒚 − 𝟑𝟑𝟑𝟑 = 𝟐𝟐

- Write equations in standardform .

�𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1𝑧𝑧 = 𝑘𝑘1 𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦 + 𝑐𝑐2𝑧𝑧 = 𝑘𝑘2 𝑎𝑎3𝑥𝑥 + 𝑏𝑏3𝑦𝑦 + 𝑐𝑐3𝑧𝑧 = 𝑘𝑘3

�2𝑥𝑥 − 𝑦𝑦 + 3𝑧𝑧 = 1 3𝑥𝑥 + 2𝑦𝑦 + 0𝑧𝑧 = -4 0𝑥𝑥 + 2𝑦𝑦 − 3𝑧𝑧 = 2

- Determine D, Dx, Dy and Dz . Use ‘expansion by minors .’


D = 𝟐𝟐 -1 3 𝟑𝟑 2 0 =𝟎𝟎 2 -3

𝟐𝟐 �2 02 -3� − 𝟑𝟑 �-1 3

2 -3� + 𝟎𝟎 �-1 3

2 0�

= 2[2(-3) – 0(2)] –3[(-1)(-3) – 2∙3] + 0[(-1)∙0–3∙2]

Choose column 1 . = 2(-6) – 3(3 – 6) + 0 = -12 + 9 = -3

Dx = 𝑘𝑘1 𝑏𝑏1 𝑐𝑐1 𝑘𝑘2 𝑏𝑏2 𝑐𝑐2𝑘𝑘3 𝑏𝑏3 𝑐𝑐3

Dx =𝟏𝟏 -1 3 -𝟒𝟒 2 0 =𝟐𝟐 2 -3

𝟏𝟏 �2 02 -3� − (-𝟒𝟒) �-1 3

2 -3� + 𝟐𝟐 �-1 3

2 0�

= 1[2(-3) – 0∙2] – (-4)[(-1)(-3) – 3∙2] + 2[(-1)∙0 – 3∙2]

= -6 + 4(3–6) + 2(-6) = -6 – 12 – 12 = -30

Dy = 𝑎𝑎1 𝑘𝑘1 𝑐𝑐1 𝑎𝑎2 𝑘𝑘2 𝑐𝑐2 𝑎𝑎3 𝑘𝑘3 𝑐𝑐3

Dy = 2 1 𝟑𝟑 3 -4 𝟎𝟎 =0 2 -𝟑𝟑

𝟑𝟑 �3 -40 2

� − 𝟎𝟎 �2 10 2� + (-𝟑𝟑) �

2 13 -4�

Choose column 3 . = 3[3∙2 – (-4)∙0] – 0 + (-3)[2(-4) – 1∙3]

= 3∙6 – 3(-8 – 3) = 18 + 33 = 51

DZ = 𝑎𝑎1 𝑏𝑏1 𝑘𝑘1 𝑎𝑎2 𝑏𝑏2 𝑘𝑘2 𝑎𝑎3 𝑏𝑏3 𝑘𝑘3

DZ =2 -1 1 3 2 -4 = 𝟎𝟎 𝟐𝟐 𝟐𝟐

𝟎𝟎 �-1 12 -4

� − 𝟐𝟐 �2 13 -4� + 𝟐𝟐 �2 -1

3 2�

Choose row 3 . = 0 – 2[2(- 4) –1∙3] + 2[2∙2 – (-1)3] = -2(-8 – 3) + 2(4 + 3) = 22 + 14 = 36

- Solve for x , y and z .



= -30-3

= 10


𝒚𝒚 = 𝐷𝐷𝑦𝑦𝐷𝐷

= 51-3

= -17


z = 36 -3

= -12

Note: It gives the same answers as expansion by diagonals .

+ − + − + − + − +

Tip: Add a coefficient for the missing variable .

+ − + − + − + − +

+ − + − + − + − +

Page 11-7





11-3 MATRICES

Introduction to Matrices

Example

• Array: a set of elements arranged in rows and columns . 𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

, 2 34 5

• Matrix: a rectangular array of elements enclosed in brackets . �3 5 12 4 7� , �

𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

�

• Columns and rows: �𝑎𝑎11 𝑏𝑏12 𝑐𝑐13 𝑎𝑎21 𝑏𝑏22 𝑐𝑐23𝑎𝑎31 𝑏𝑏32 𝑐𝑐33

� �2 1 54 2 45 3 2

�


• Dimensions: A matrix has �𝑚𝑚 rows 𝑛𝑛 columns m × n dimensions �2 3

4 5� , �3 5 12 4 7�

Rows Columns 2×2 2×3

�352� , [2 1 3]

3×1 1×3

• A 3×3 system: Linear System Example

�𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1𝑧𝑧 = 𝑘𝑘1𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦+ 𝑐𝑐2𝑧𝑧 = 𝑘𝑘2𝑎𝑎3𝑥𝑥 + 𝑏𝑏3𝑦𝑦+ 𝑐𝑐3𝑧𝑧 = 𝑘𝑘3

�2𝑥𝑥 + 3𝑦𝑦 + 4𝑧𝑧 = 1 𝑥𝑥 + 2𝑦𝑦 + 3𝑧𝑧 = 4

3𝑥𝑥 + 𝑦𝑦 + 5𝑧𝑧 = 2

• Coefficient matrix: the matrix obtained from the coefficients in a linear system . Coefficient Matrix Example

�𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

� �2 3 41 2 33 1 5

�

• Augmented matrix: the matrix obtained from the coefficients and constant terms in a

linear system . Augmented Matrix Example

�𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

𝑘𝑘1𝑘𝑘2𝑘𝑘3� �

2 3 4 1 2 3 3 1 5

142�

coefficients of 𝑥𝑥 coefficients of y constants (with a vertical line)

coefficients of 𝑧𝑧

Tip: augmented matrix = constants & coefficients

Row 2

Row 3

Row 1

Page 11-8

11-3 MATRICES

Introduction to Matrices

Example

• Array: a set of elements arranged in rows and columns . 𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

, 2 34 5

• Matrix: a rectangular array of elements enclosed in brackets . �3 5 12 4 7� , �

𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

�

• Columns and rows: �𝑎𝑎11 𝑏𝑏12 𝑐𝑐13 𝑎𝑎21 𝑏𝑏22 𝑐𝑐23𝑎𝑎31 𝑏𝑏32 𝑐𝑐33

� �2 1 54 2 45 3 2

�


• Dimensions: A matrix has �𝑚𝑚 rows 𝑛𝑛 columns m × n dimensions �2 3

4 5� , �3 5 12 4 7�

Rows Columns 2×2 2×3

�352� , [2 1 3]

3×1 1×3


�𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1𝑧𝑧 = 𝑘𝑘1𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦+ 𝑐𝑐2𝑧𝑧 = 𝑘𝑘2𝑎𝑎3𝑥𝑥 + 𝑏𝑏3𝑦𝑦+ 𝑐𝑐3𝑧𝑧 = 𝑘𝑘3

�2𝑥𝑥 + 3𝑦𝑦 + 4𝑧𝑧 = 1 𝑥𝑥 + 2𝑦𝑦 + 3𝑧𝑧 = 4

3𝑥𝑥 + 𝑦𝑦 + 5𝑧𝑧 = 2

• Coefficient matrix: the matrix obtained from the coefficients in a linear system . Coefficient Matrix Example

�𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

� �2 3 41 2 33 1 5

�

• Augmented matrix: the matrix obtained from the coefficients and constant terms in a

linear system . Augmented Matrix Example


𝑘𝑘1𝑘𝑘2𝑘𝑘3� �

2 3 4 1 2 3 3 1 5

142�



Tip: augmented matrix = constants & coefficients

Row 2

Row 3

Row 1

Page 11-8





Matrix Addition & Subtraction

• Naming a matrix with a single letter (bold face)

Example: 𝑨𝑨 = �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2

� 𝑩𝑩 = �𝑎𝑎3 𝑏𝑏3𝑎𝑎4 𝑏𝑏4

� 2×2

or A = �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

� B = �𝑎𝑎4 𝑏𝑏4𝑎𝑎5 𝑏𝑏5𝑎𝑎6 𝑏𝑏6

� 3×2

• Matrix equality: two equal matrices have the same dimensions (or size) and the equal

corresponding elements .

Example

The same dimensions: A = 2×2 & B = 2×2

or A = 3×2 & B = 3×2

The equal corresponding elements: A = �𝑥𝑥 4 𝑦𝑦2 5 3� , B = �6 4 3

2 5 √9�

2×3 2×3

A = B only if x = 6 and y = 3

• Requirements for adding/subtracting matrices: only matrices of the same dimensions

can be added or subtracted .

• Add/subtract two matrices of the same dimensions: combine the elements in the

corresponding (or matching) positions .

Matrix Operations Example

matrix addition

A + B


� + �𝑎𝑎3 𝑏𝑏3𝑎𝑎4 𝑏𝑏4

� = �𝑎𝑎1+ 𝑎𝑎3 𝑏𝑏1 + 𝑏𝑏3𝑎𝑎2 + 𝑎𝑎4 𝑏𝑏2 + 𝑏𝑏4

�2×2 2×2

The same dimensions

�1 32 4 � + �3 5

2 1� = �1 + 3 3 + 52 + 2 4 + 1�

= �4 84 5�

matrix subtraction

A – B

�𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

� − �𝑎𝑎4 𝑏𝑏4𝑎𝑎5 𝑏𝑏5𝑎𝑎6 𝑏𝑏6

� = �𝑎𝑎1−𝑎𝑎4 𝑏𝑏1−𝑏𝑏4𝑎𝑎2−𝑎𝑎5 𝑏𝑏2−𝑏𝑏5𝑎𝑎3−𝑎𝑎6 𝑏𝑏3−𝑏𝑏6

�

3×2 3×2

�3 85 76 -9

� − �2 23 -36 4

� = �3 − 2 8 − 25 − 3 7 − (-3)6 − 6 -9 − 4

�

= �1 62 100 -13

�

Page 11-9

Matrix Addition & Subtraction

• Naming a matrix with a single letter (bold face)

Example: 𝑨𝑨 = �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2

� 𝑩𝑩 = �𝑎𝑎3 𝑏𝑏3𝑎𝑎4 𝑏𝑏4

� 2×2

or A = �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

� B = �𝑎𝑎4 𝑏𝑏4𝑎𝑎5 𝑏𝑏5𝑎𝑎6 𝑏𝑏6

� 3×2



Example

The same dimensions: A = 2×2 & B = 2×2

or A = 3×2 & B = 3×2

The equal corresponding elements: A = �𝑥𝑥 4 𝑦𝑦2 5 3� , B = �6 4 3

2 5 √9�

2×3 2×3

A = B only if x = 6 and y = 3

• Requirements for adding/subtracting matrices: only matrices of the same dimensions

can be added or subtracted .

• Add/subtract two matrices of the same dimensions: combine the elements in the

corresponding (or matching) positions .

Matrix Operations Example

matrix addition

A + B




�2×2 2×2

The same dimensions

�1 32 4 � + �3 5

2 1� = �1 + 3 3 + 52 + 2 4 + 1�

= �4 84 5�

matrix subtraction

A – B




�

3×2 3×2

�3 85 76 -9

� − �2 23 -36 4

� = �3 − 2 8 − 25 − 3 7 − (-3)6 − 6 -9 − 4

�

= �1 62 100 -13

�

Page 11-9





Matrix Multiplication

• Scalar matrix multiplication: the product of a scalar (a real number) and a matrix .Scalar Matrix Multiplication Example

k · Ak – scalarA – matrix

k �𝑎𝑎1 𝑏𝑏1𝑎𝑎2 𝑏𝑏2𝑎𝑎3 𝑏𝑏3

� = �𝑘𝑘𝑎𝑎1 𝑘𝑘𝑏𝑏1𝑘𝑘𝑎𝑎2 𝑘𝑘𝑏𝑏2𝑘𝑘𝑎𝑎3 𝑘𝑘𝑏𝑏3

�2�4 -2 𝑦𝑦

3 5 0� = �2 ∙ 4 2(-2) 2 ∙ 𝑦𝑦

2 ∙ 3 2 ∙ 5 2 ∙ 0�

= �8 -4 2𝑦𝑦 6 10 0

�

Tip: Multiply each element by the scalar .

Example: Find the following .

3�2 41 03 -2

� – 1 2�8 40 -22 6

� = �3 ∙ 2 3 ∙ 43 ∙ 1 3 ∙ 03 ∙ 3 3(-2)

� –

⎣⎢⎢⎢⎡

1 2∙ 8 1

2∙ 4

1 2∙ 0 1

2∙ (-2)

1 2∙ 2 1

2∙ 6 ⎦

⎥⎥⎥⎤ = �

6 123 09 -6

� – �4 20 -11 3

�

= �6 − 4 12 − 23 − 0 0 − (-1)9 − 1 -6 − 3

� = �2 103 18 -9

�

• Matrix multiplication: the product of two matrices .

• Requirements for matrix multiplication: the product of two matrices A and B is defined

only when the number of columns of A (1st matrix) is equal to the number of rows of B

(2nd matrix) . Recall: m × n xRows Columns

Requirements for Matrix Multiplication

If A = m1× n1 , B = m2× n2 then A· B is defined only when n1 = m2 .

column for A row for B

Example

Matrix A Matrix B A ∙ B

A = �2 3 -2 11 4 6 4

�

2×4 , n1 = 4B = �

1 422-5

534

� 4×2n1 = m2 , 4 = 4

A·B is defined

A = �2416

� 4×1𝑩𝑩 = [ 3 4 5 7] 1×4 , m2 = 1

n1= m2 , 1 = 1

A·B is defined

𝑨𝑨 = [3 2 1]

1×3 , n1 = 3B = �

4 31 -5�

2×2 , m2 = 2

n1 ≠ m2 , 3 ≠ 2

A·B is not defined

m2 = 4

n1 = 1

Page 11-10

Matrix Multiplication

• Scalar matrix multiplication: the product of a scalar (a real number) and a matrix .Scalar Matrix Multiplication Example




�2�4 -2 𝑦𝑦

3 5 0� = �2 ∙ 4 2(-2) 2 ∙ 𝑦𝑦

2 ∙ 3 2 ∙ 5 2 ∙ 0�

= �8 -4 2𝑦𝑦 6 10 0

�

Tip: Multiply each element by the scalar .

Example: Find the following .

3�2 41 03 -2

� – 1 2�8 40 -22 6

� = �3 ∙ 2 3 ∙ 43 ∙ 1 3 ∙ 03 ∙ 3 3(-2)

� –

⎣⎢⎢⎢⎡

1 2∙ 8 1

2∙ 4

1 2∙ 0 1

2∙ (-2)

1 2∙ 2 1

2∙ 6 ⎦

⎥⎥⎥⎤ = �

6 123 09 -6

� – �4 20 -11 3

�

= �6 − 4 12 − 23 − 0 0 − (-1)9 − 1 -6 − 3

� = �2 103 18 -9

�

• Matrix multiplication: the product of two matrices .

• Requirements for matrix multiplication: the product of two matrices A and B is defined

only when the number of columns of A (1st matrix) is equal to the number of rows of B

(2nd matrix) . Recall: m × n xRows Columns




Example

Matrix A Matrix B A ∙ B

A = �2 3 -2 11 4 6 4

�

2×4 , n1 = 4B = �

1 422-5

534

� 4×2n1 = m2 , 4 = 4

A·B is defined

A = �2416

� 4×1𝑩𝑩 = [ 3 4 5 7] 1×4 , m2 = 1

n1= m2 , 1 = 1

A·B is defined

𝑨𝑨 = [3 2 1]

1×3 , n1 = 3B = �

4 31 -5�

2×2 , m2 = 2

n1 ≠ m2 , 3 ≠ 2

A·B is not defined

m2 = 4

n1 = 1

Page 11-10





• Dimensions of the product for matrix multiplication

Dimensions ofthe Product

If A = (m1× n1) and B = (m2 × n2), then A·B = (m1× n1) (m2 × n2) = (m1 × n2) .

Example:If A = (3 × 2) and B = (2 × 4), Then A·B = (3 × 2) (2 × 4) = (3 × 4) .

Tip: (Rows of A) × (Columns of B)

• Product of 1× n and n × 1 matrices

Matrix Multiplication Example Dimension If A = [ 𝑎𝑎1 𝑎𝑎2 … 𝑎𝑎𝑛𝑛]

B = �

𝑏𝑏1𝑏𝑏2…𝑏𝑏𝑛𝑛

�

then AB = a1b1 + a2b2 + … anbn

A = [ 2 3 1 ]

B = �3-14�

AB = [ 2 3 1 ] �3-14�

= 2∙3 + 3(-1) + 1∙4 = 7

1×3

3×1

1×1

Tip: Multiply each element of the row of the first matrix by the corresponding elements of the column in the second matrix, and then add the products .

• Product of 1× n and n × 2 matrices

Matrix Multiplication Example Dimension If A = [a1, a2, … , an]

B = �

𝑏𝑏11 𝑏𝑏12𝑏𝑏21…𝑏𝑏𝑛𝑛1

𝑏𝑏22…𝑏𝑏𝑛𝑛2

�

then AB = [a1, a2, … , an] �

𝑏𝑏11 𝑏𝑏12𝑏𝑏21…𝑏𝑏𝑛𝑛1

𝑏𝑏22…𝑏𝑏𝑛𝑛2

�

= [𝑎𝑎1𝑏𝑏11 + 𝑎𝑎2𝑏𝑏21 + … + 𝑎𝑎𝑛𝑛𝑏𝑏𝑛𝑛1 𝑎𝑎1𝑏𝑏12 + 𝑎𝑎2𝑏𝑏22 + … + 𝑎𝑎𝑛𝑛𝑏𝑏𝑛𝑛2]

(row of A) × (1st column of B) (row of A) × (2nd column of B)

A = � 4 -1 2 � 1×3

B = �1 -13 22 0

� 3×2

AB = � 4 -1 2 � �1 -1 3 2 2 0

�

= [4∙1+(-1)3+2∙2 4(-1)+(-1)2+2∙0]

= [4-3+4 -4–2] = [5 -6] 1×2

Tips: - Multiply each element of the row of the first matrix by the corresponding elements of each column in the second matrix, and then add the products .

- Double-digit subscripts: Example: b21 - the element in matrix B that is in row 2 and column 1 . bn2 - the element in matrix B that is in row n and column 2 .

• The general caseMatrix Multiplication Example

AB = �𝑎𝑎11 𝑎𝑎12 𝑎𝑎13𝑎𝑎21 𝑎𝑎22 𝑎𝑎23� �

𝑏𝑏11 𝑏𝑏12𝑏𝑏21 𝑏𝑏22𝑏𝑏31 𝑏𝑏32

� A = 2×3 , B = 3×2

(1st row of A) × (1st column of B) (1st row of A) × (2nd column of B)

= �𝑎𝑎11𝑏𝑏11 + 𝑎𝑎12𝑏𝑏21 + 𝑎𝑎13𝑏𝑏31 𝑎𝑎11𝑏𝑏12 + 𝑎𝑎12𝑏𝑏22 + 𝑎𝑎13𝑏𝑏32𝑎𝑎21𝑏𝑏11 + 𝑎𝑎22𝑏𝑏21 + 𝑎𝑎23𝑏𝑏31 𝑎𝑎21𝑏𝑏12 + 𝑎𝑎22𝑏𝑏22 + 𝑎𝑎23𝑏𝑏32

�

(2nd row of A) × (1st column of B) (2nd row of A) × (2nd column of B) 2×2

�1 2 32 0 1� �

2 10 23 -1

� 2×3 , 3×2

= �1 ∙ 2 + 2 ∙ 0 + 3 ∙ 3 1 ∙ 1 + 2 ∙ 2 + 3(-1)2 ∙ 2 + 0 ∙ 0 + 1 ∙ 3 2 ∙ 1 + 0 ∙ 2 + 1(-1)

�

= �11 27 1� 2×2

Page 11-11





Example: Find the products .

1. �2 31 -2� �

3 0-4 4� = �

2 ∙ 3 + 3(-4) 2 ∙ 0 + 3 ∙ 41 ∙ 3 + (-2)(-4) 1 ∙ 0 + (-2) ∙ 4

� = �-6 1211 -8

�

2 × 2 2 × 2 2 × 2

2. �1 -1 2 0 3 0 2 1 2

� �2 0 13 -2 0-2 1 2

�

3 × 3 3 × 3

(1st row of A) × (1st column of B) (1std row of A) × (2nd column of B) 1std row of A) × (3rd column of B)

= �1 ∙ 2 + �-1�3 + 2(-2) 1 ∙ 0 + (-1)(-2) + 2 ∙ 1 1 ∙ 1 + (-1) ∙ 0 + 2 ∙ 2 0 ∙ 2 + 3 ∙ 3 + 0(-2) 0 ∙ 0 + 3(-2) + 0 ∙ 1 0 ∙ 1 + 3 ∙ 0 + 0 ∙ 22 ∙ 2 + 1 ∙ 3 + 2(-2) 2 ∙ 0 + 1(-2) + 2 ∙ 1 2 ∙ 1 + 1 ∙ 0 + 2 ∙ 2

�

(3rd row of A) × (1st column of B) (3rd row of A) × (2nd column of B) (3rd row of A) × (3rd column of B)

= �-5 4 59 -6 03 0 -6

�

3 × 3

(2nd row of A) × (1st column of B)

(2nd row of A) × (3rd column of B)

(2nd row of A) × (2nd column of B)

Page 11-12





Summary – Matrix Multiplication

• Multiply the elements of the first row of the first matrix by the corresponding elements of

each column in the second matrix and add the products .

�𝑎𝑎11 𝑎𝑎12 𝑎𝑎13 𝑎𝑎21 𝑎𝑎22 𝑎𝑎23𝑎𝑎31 𝑎𝑎32 𝑎𝑎33

� �𝑏𝑏11 𝑏𝑏12 𝑏𝑏13 𝑏𝑏21 𝑏𝑏22 𝑏𝑏23𝑏𝑏31 𝑏𝑏32 𝑏𝑏33

� = �𝑐𝑐11 𝑐𝑐12 𝑐𝑐13𝑐𝑐21 𝑐𝑐22 𝑐𝑐23 𝑐𝑐31 𝑐𝑐32 𝑐𝑐33

� 𝑐𝑐11 = 𝑎𝑎11𝑏𝑏11 + 𝑎𝑎12𝑏𝑏21 + 𝑎𝑎13𝑏𝑏31

�𝑎𝑎11 𝑎𝑎12 𝑎𝑎13𝑎𝑎21 𝑎𝑎22 𝑎𝑎23𝑎𝑎31 𝑎𝑎32 𝑎𝑎33

� �𝑏𝑏11 𝑏𝑏12 𝑏𝑏13𝑏𝑏21 𝑏𝑏22 𝑏𝑏23𝑏𝑏31 𝑏𝑏32 𝑏𝑏33

� = �𝑐𝑐11 𝑐𝑐12 𝑐𝑐13 𝑐𝑐21 𝑐𝑐22 𝑐𝑐23𝑐𝑐31 𝑐𝑐32 𝑐𝑐33

� 𝑐𝑐12 = 𝑎𝑎11𝑏𝑏12 + 𝑎𝑎12𝑏𝑏22 + 𝑎𝑎13𝑏𝑏32


� �𝑏𝑏11 𝑏𝑏12 𝑏𝑏13 𝑏𝑏21 𝑏𝑏22 𝑏𝑏23 𝑏𝑏31 𝑏𝑏32 𝑏𝑏33

�= �𝑐𝑐11 𝑐𝑐12 𝑐𝑐13𝑐𝑐21 𝑐𝑐22 𝑐𝑐23𝑐𝑐31 𝑐𝑐32 𝑐𝑐33

� 𝑐𝑐13 = 𝑎𝑎11𝑏𝑏13 + 𝑎𝑎12𝑏𝑏23 + 𝑎𝑎13𝑏𝑏33

• Multiply the elements of the second row of the first matrix by the corresponding elements

of each column in the second matrix and add the products .



� = �𝑐𝑐11 𝑐𝑐12 𝑐𝑐13𝑐𝑐21 𝑐𝑐22 𝑐𝑐23𝑐𝑐31 𝑐𝑐32 𝑐𝑐33

� 𝑐𝑐21 = 𝑎𝑎21𝑏𝑏11 + 𝑎𝑎22𝑏𝑏21 + 𝑎𝑎23𝑏𝑏31




� 𝑐𝑐22 = 𝑎𝑎21𝑏𝑏12 + 𝑎𝑎22𝑏𝑏22 + 𝑎𝑎23𝑏𝑏32




� 𝑐𝑐23 = 𝑎𝑎21𝑏𝑏13 + 𝑎𝑎22𝑏𝑏23 + 𝑎𝑎23𝑏𝑏33

• Multiply the elements of the third row of the first matrix by the corresponding elements of





� 𝑐𝑐31 = 𝑎𝑎31𝑏𝑏11 + 𝑎𝑎32𝑏𝑏21 + 𝑎𝑎33𝑏𝑏31




� 𝑐𝑐32 = 𝑎𝑎31𝑏𝑏12 + 𝑎𝑎32𝑏𝑏22 + 𝑎𝑎33𝑏𝑏32




� 𝑐𝑐33 = 𝑎𝑎31𝑏𝑏13 + 𝑎𝑎32𝑏𝑏23 + 𝑎𝑎33𝑏𝑏33

3 × 3 3 × 3 3 × 3

Page 11-13

Summary – Matrix Multiplication

• Multiply the elements of the first row of the first matrix by the corresponding elements of



� �𝑏𝑏11 𝑏𝑏12 𝑏𝑏13 𝑏𝑏21 𝑏𝑏22 𝑏𝑏23𝑏𝑏31 𝑏𝑏32 𝑏𝑏33

� = �𝑐𝑐11 𝑐𝑐12 𝑐𝑐13𝑐𝑐21 𝑐𝑐22 𝑐𝑐23 𝑐𝑐31 𝑐𝑐32 𝑐𝑐33

� 𝑐𝑐11 = 𝑎𝑎11𝑏𝑏11 + 𝑎𝑎12𝑏𝑏21 + 𝑎𝑎13𝑏𝑏31




� 𝑐𝑐12 = 𝑎𝑎11𝑏𝑏12 + 𝑎𝑎12𝑏𝑏22 + 𝑎𝑎13𝑏𝑏32



�= �𝑐𝑐11 𝑐𝑐12 𝑐𝑐13𝑐𝑐21 𝑐𝑐22 𝑐𝑐23𝑐𝑐31 𝑐𝑐32 𝑐𝑐33

� 𝑐𝑐13 = 𝑎𝑎11𝑏𝑏13 + 𝑎𝑎12𝑏𝑏23 + 𝑎𝑎13𝑏𝑏33

• Multiply the elements of the second row of the first matrix by the corresponding elements

of each column in the second matrix and add the products .




� 𝑐𝑐21 = 𝑎𝑎21𝑏𝑏11 + 𝑎𝑎22𝑏𝑏21 + 𝑎𝑎23𝑏𝑏31




� 𝑐𝑐22 = 𝑎𝑎21𝑏𝑏12 + 𝑎𝑎22𝑏𝑏22 + 𝑎𝑎23𝑏𝑏32




� 𝑐𝑐23 = 𝑎𝑎21𝑏𝑏13 + 𝑎𝑎22𝑏𝑏23 + 𝑎𝑎23𝑏𝑏33

• Multiply the elements of the third row of the first matrix by the corresponding elements of





� 𝑐𝑐31 = 𝑎𝑎31𝑏𝑏11 + 𝑎𝑎32𝑏𝑏21 + 𝑎𝑎33𝑏𝑏31




� 𝑐𝑐32 = 𝑎𝑎31𝑏𝑏12 + 𝑎𝑎32𝑏𝑏22 + 𝑎𝑎33𝑏𝑏32




� 𝑐𝑐33 = 𝑎𝑎31𝑏𝑏13 + 𝑎𝑎32𝑏𝑏23 + 𝑎𝑎33𝑏𝑏33

3 × 3 3 × 3 3 × 3

Page 11-13





11-4 MATRIX INVERSE

Identity Matrix

• Identity matrix I : a square matrix in which all the elements are 0 except the main

diagonal from the top left to the bottom right corner with 1s .

I 2×2 3×3 n×n

identity matrix I I = �1 0

0 1� I = �1 0 00 1 00 0 1

� I = �1 ⋯ 0⋮ 1 ⋮0 ⋯ 1

�

main diagonal

• Identity property: When a square matrix A is multiplied by an identity matrix I, the

result is A .

Identity Property for Matrices A I = I A = A

Tip: Identity property for matrices is same as the identity property for real numbers .

a ∙1 = 1∙ a = a Example: 3∙1 = 1∙ 3 = 3

Example: A = �1 3-2 4� , I = �1 0

0 1�

AI = �1 3-2 4� �

1 00 1�= �

1 ∙ 1 + 3 ∙ 0 1 ∙ 0 + 3 ∙ 1-2 ∙ 1 + 4 ∙ 0 -2 ∙ 0 + 4 ∙ 1 � = �

𝟏𝟏 𝟑𝟑-𝟐𝟐 𝟒𝟒�

IA = �1 00 1� �

1 3-2 4� = �

1 ∙ 1 + 0(-2) 1 ∙ 3 + 0 ∙ 40 ∙ 1 + 1(-2) 0 ∙ 3 + 1 ∙ 4

� = �𝟏𝟏 𝟑𝟑-𝟐𝟐 𝟒𝟒�

The same result for A I and I A.

• Properties of matrix

Property of Multiplication Property of AdditionA, B, and C are matrices

associative property: A(BC) = (AB)C associative property: A + (B + C) = (A + B) + Cdistributive property: A(B + C) = AB + AC (B + C)A = BA + CA commutative property: A + B = B + A

scalar multiplication (k is a constant): k(AB) = (kA)B or A(kB) additive inverse: A + (-A) = 0

multiplicative identity: IA = AI = A additive identity: A + 0 = A

Note: The zero matrix “0” is a matrix that has the same dimension as matrix A but has a “0” for each element .

Page 11-14

11-4 MATRIX INVERSE

Identity Matrix

• Identity matrix I : a square matrix in which all the elements are 0 except the main

diagonal from the top left to the bottom right corner with 1s .

I 2×2 3×3 n×n


0 1� I = �1 0 00 1 00 0 1

� I = �1 ⋯ 0⋮ 1 ⋮0 ⋯ 1

�

main diagonal

• Identity property: When a square matrix A is multiplied by an identity matrix I, the

result is A .

Identity Property for Matrices A I = I A = A

Tip: Identity property for matrices is same as the identity property for real numbers .

a ∙1 = 1∙ a = a Example: 3∙1 = 1∙ 3 = 3

Example: A = �1 3-2 4� , I = �1 0

0 1�

AI = �1 3-2 4� �

1 00 1�= �

1 ∙ 1 + 3 ∙ 0 1 ∙ 0 + 3 ∙ 1-2 ∙ 1 + 4 ∙ 0 -2 ∙ 0 + 4 ∙ 1 � = �

𝟏𝟏 𝟑𝟑-𝟐𝟐 𝟒𝟒�

IA = �1 00 1� �

1 3-2 4� = �

1 ∙ 1 + 0(-2) 1 ∙ 3 + 0 ∙ 40 ∙ 1 + 1(-2) 0 ∙ 3 + 1 ∙ 4

� = �𝟏𝟏 𝟑𝟑-𝟐𝟐 𝟒𝟒�

The same result for A I and I A.

• Properties of matrix

Property of Multiplication Property of AdditionA, B, and C are matrices

associative property: A(BC) = (AB)C associative property: A + (B + C) = (A + B) + Cdistributive property: A(B + C) = AB + AC (B + C)A = BA + CA commutative property: A + B = B + A



Note: The zero matrix “0” is a matrix that has the same dimension as matrix A but has a “0” for each element .

Page 11-14





Inverse of a Matrix

• An inverse matrix (A-1): A-1 is the inverse of a matrix A .

A A-1 = A-1 A = I I - identity matrix

Tip: ordinary algebra: 𝑎𝑎 ∙ 1𝑎𝑎

= 𝑎𝑎𝑎𝑎-1 = 1 𝑎𝑎-1 = 1𝑎𝑎

matrix algebra: 𝑨𝑨 𝑨𝑨-1 = 𝑰𝑰 𝑨𝑨-𝟏𝟏 ≠ 𝟏𝟏𝑨𝑨

Example: A = �2 16 4�, 𝑨𝑨

-1 = �2–12

-3 1�

A A-1 = �2 16 4� �

2–12

-3 1� = �

2 ∙ 2 + 1 ∙ (-3) 2 ∙–12

+ 1 ∙ 1

6 ∙ 2 + 4 ∙ (-3) 6 ∙ �–12� + 4 ∙ 1

�

= �4 − 3 -1 + 1

12 − 12 -3 + 4� = �1 0

0 1� = I , A A-1 = I

• Finding the inverse of a 2×2 matrix A-1

Steps Example: A = �2 16 4�

- Switch the main diagonal elements . A = �𝒂𝒂𝟏𝟏 𝑏𝑏1𝑎𝑎2 𝒃𝒃𝟐𝟐

� �𝒃𝒃𝟐𝟐 𝑏𝑏1𝑎𝑎2 𝒂𝒂𝟏𝟏

� �𝟒𝟒 16 𝟐𝟐�

- Change signs for the remaining elements . � 𝑏𝑏2 -𝑏𝑏1-𝑎𝑎2 𝑎𝑎1

� �4 -1-6 2

�

- Divide the result of the last step by the determinant |𝐴𝐴| . 𝑨𝑨-𝟏𝟏 = �4 -1

-6 2�

�2 16 4

�=

�4 -1-6 2

�

8−6

A-1 = �𝑏𝑏2 -𝑏𝑏1-𝑎𝑎2 𝑎𝑎1

�


= 12� 4 -1-6 2

� =�42

-12

-62

22

� = � 𝟐𝟐-𝟏𝟏𝟐𝟐

-𝟑𝟑 𝟏𝟏�

?- Check: 𝑨𝑨 𝑨𝑨-1 = 𝑰𝑰

𝑨𝑨 𝑨𝑨−1 = �2 16 4

� �2–12

-3 1� = �

2 ∙ 2 + 1 ∙ (-3) 2 ∙ –12 + 1 ∙ 1

6 ∙ 2 + 4 ∙ (-3) 6 ∙ �–12�+ 4 ∙ 1

�

= �4 − 3 -1 + 1

12 − 12 -3 + 4� = �1 0

0 1� = I Correct!

Note: This method can only be used for finding the inverse of a 2 × 2 matrix .

Page 11-15

Inverse of a Matrix



Tip: ordinary algebra: 𝑎𝑎 ∙ 1𝑎𝑎

= 𝑎𝑎𝑎𝑎-1 = 1 𝑎𝑎-1 = 1𝑎𝑎

matrix algebra: 𝑨𝑨 𝑨𝑨-1 = 𝑰𝑰 𝑨𝑨-𝟏𝟏 ≠ 𝟏𝟏𝑨𝑨

Example: A = �2 16 4�, 𝑨𝑨

-1 = �2–12

-3 1�

A A-1 = �2 16 4� �

2–12

-3 1� = �

2 ∙ 2 + 1 ∙ (-3) 2 ∙–12

+ 1 ∙ 1

6 ∙ 2 + 4 ∙ (-3) 6 ∙ �–12� + 4 ∙ 1

�

= �4 − 3 -1 + 1

12 − 12 -3 + 4� = �1 0

0 1� = I , A A-1 = I


Steps Example: A = �2 16 4�



� �𝟒𝟒 16 𝟐𝟐�


� �4 -1-6 2

�

- Divide the result of the last step by the determinant |𝐴𝐴| . 𝑨𝑨-𝟏𝟏 = �4 -1

-6 2�

�2 16 4

�=

�4 -1-6 2

�

8−6

A-1 = �𝑏𝑏2 -𝑏𝑏1-𝑎𝑎2 𝑎𝑎1

�


= 12� 4 -1-6 2

� =�42

-12

-62

22

� = � 𝟐𝟐-𝟏𝟏𝟐𝟐

-𝟑𝟑 𝟏𝟏�

?- Check: 𝑨𝑨 𝑨𝑨-1 = 𝑰𝑰

𝑨𝑨 𝑨𝑨−1 = �2 16 4

� �2–12

-3 1� = �

2 ∙ 2 + 1 ∙ (-3) 2 ∙ –12 + 1 ∙ 1

6 ∙ 2 + 4 ∙ (-3) 6 ∙ �–12�+ 4 ∙ 1

�

= �4 − 3 -1 + 1

12 − 12 -3 + 4� = �1 0

0 1� = I Correct!

Note: This method can only be used for finding the inverse of a 2 × 2 matrix .

Page 11-15





Gauss-Jordan Elimination Method to Find A-1

Gauss-JordanMethod to Find A-1

- Transform matrix A into the identity matrix I. A I - Transform the identity matrix I into inverse matrix A-1 . I A-1

Procedure to use Gaussian-elimination method to find A-1

Steps Example: 𝑨𝑨 = �1 02 3

�

A I

- Write the augmented matrix [A | I] by appending �1 0 1 02 3 0 1�

an identity matrix I on the right of matrix A . Column 1 Column 2

- Use row operations to transform [A | I] to [I | A-1] . �1 0 1 0𝟎𝟎 3 -2 1�

You can:- Switch any two rows .

- Multiply or divide a row by a constant . �1 0 1 0𝟎𝟎 1 –2

3 1

3�

- Add or subtract a row to another row .- Multiply a constant to a row .

Tip: This (row operations) is similar to the elimination method for solving a system of linear equations .

- Determine A-1 . 𝑨𝑨-1 = �𝟏𝟏 𝟎𝟎–𝟐𝟐𝟑𝟑

𝟏𝟏𝟑𝟑�

- Check . A A-1 = I �1 02 3� �

1 0–23

13� = �1 0

0 1�

�1 ∙ 1 + 0 �

–23� 1 ∙ 0 + 0 �1

3�

2 ∙ 1 + 3 �–23� 2 ∙ 0 + 3 �1

3�� = �1 0

0 1�

√

�1 00 1� = �1 0

0 1� Correct!

Note: This method can be used for any n × n matrices .

Row 1

Row 2

-2 × row 1, add to row 2

Get a “1” in the row 2 colunm 2 .

13

× row 2

𝑰𝑰 = �1 00 1�


?

?

?

Page 11-16

Gauss-Jordan Elimination Method to Find A-1


- Transform matrix A into the identity matrix I. A I - Transform the identity matrix I into inverse matrix A-1 . I A-1

Procedure to use Gaussian-elimination method to find A-1

Steps Example: 𝑨𝑨 = �1 02 3

�

A I

- Write the augmented matrix [A | I] by appending �1 0 1 02 3 0 1�

an identity matrix I on the right of matrix A . Column 1 Column 2

- Use row operations to transform [A | I] to [I | A-1] . �1 0 1 0𝟎𝟎 3 -2 1�

You can:- Switch any two rows .

- Multiply or divide a row by a constant . �1 0 1 0𝟎𝟎 1 –2

3 1

3�

- Add or subtract a row to another row .- Multiply a constant to a row .

Tip: This (row operations) is similar to the elimination method for solving a system of linear equations .

- Determine A-1 . 𝑨𝑨-1 = �𝟏𝟏 𝟎𝟎–𝟐𝟐𝟑𝟑

𝟏𝟏𝟑𝟑�

- Check . A A-1 = I �1 02 3� �

1 0–23

13� = �1 0

0 1�

�1 ∙ 1 + 0 �

–23� 1 ∙ 0 + 0 �1

3�

2 ∙ 1 + 3 �–23� 2 ∙ 0 + 3 �1

3�� = �1 0

0 1�

√

�1 00 1� = �1 0

0 1� Correct!

Note: This method can be used for any n × n matrices .

Row 1

Row 2

-2 × row 1, add to row 2


13

× row 2

𝑰𝑰 = �1 00 1�


?

?

?

Page 11-16





Example: Using the Gauss-Jordan method to find A-1 of the following 3 × 3 matrix .

A = �𝟑𝟑 𝟐𝟐 𝟎𝟎𝟏𝟏 -𝟏𝟏 𝟎𝟎𝟎𝟎 𝟓𝟓 𝟏𝟏

� , A-1 = ?

- Write [A | I ] . �3 2 0 1 -1 0 0 5 1

1 0 0 0 1 0 0 0 1

�

- Transform [A | I ] into [I | A-1] .

�3 2 0 1 -1 0 0 5 1

1 0 0 0 1 00 0 1

� �𝟏𝟏 -1 0 3 2 0 0 5 1

0 1 01 0 00 0 1

� �1 -1 0 𝟎𝟎 5 0 0 5 1

0 1 01 -3 00 0 1

�

Switch row 1 & row 2 -3 × row 1 add to row 2

�1 -1 0 0 5 0 5 𝟎𝟎 1

0 1 01 -3 00 5 1

� �1 -1 0 0 𝟏𝟏 0 5 0 1

0 1 015

-35

00 5 1

�

5 × row 1 add to row 3 Row 2 ÷ 5

�1 𝟎𝟎 0 0 1 0 5 0 1

1

5

2

50

1

5

-3

50

0 5 1

�

⎣⎢⎢⎢⎡1 0 0 0 1 0 𝟎𝟎 0 1

𝟏𝟏𝟓𝟓

𝟐𝟐𝟓𝟓

𝟎𝟎𝟏𝟏𝟓𝟓

-𝟑𝟑𝟓𝟓

𝟎𝟎

-𝟏𝟏 𝟑𝟑 𝟏𝟏⎦⎥⎥⎥⎤

Row 2 add to row 1 -5× row 1 add to row 3

- Determine A-1 . A-1 =

⎣⎢⎢⎡𝟏𝟏

𝟓𝟓

𝟐𝟐

𝟓𝟓𝟎𝟎

𝟏𝟏

𝟓𝟓

-𝟑𝟑

𝟓𝟓𝟎𝟎

-𝟏𝟏 𝟑𝟑 𝟏𝟏⎦⎥⎥⎤



Get a “1” in the row 2 colunm 2 . Get a “0” in the row 3 colunm 2 .



Page 11-17





Find Inverse Matrix A-1 – Method II

• Find the inverse of a 3 × 3 matrix – Method II

Example: Find A-1 of the following 3 × 3 matrix .

A = �3 2 01 -1 00 5 1

� A-1 = ?

Steps

• Find the cofactor matrix: determine the cofactor (minors + place signs) of each element .

�3 2 0 1 -1 00 5 1

�

⎣⎢⎢⎢⎡ �-1 0

5 1� - �1 0

0 1� �1 -1

0 5�

- �2 05 1

� �3 00 1

� - �3 20 5

�

�2 0-1 0� - �3 0

1 0� �

3 21 -1� ⎦

⎥⎥⎥⎤

+ − + − + − + − +

cofactor matrix = �-1 -1 5-2 3 -150 0 -5

�

• Determine the transpose of the cofactor matrix AT: exchange all the rows and columns .

AT =

⎣⎢⎢⎢⎡-1 -2 0-1 3 05 -15 -5⎦

⎥⎥⎥⎤

• Calculate the determinant |𝐀𝐀| of the matrix

|𝑨𝑨| = �𝟑𝟑 𝟐𝟐 𝟎𝟎1 -1 00 5 1

� = 3 �-1 05 1

� − 2 �1 00 1� + 0 �1 -1

0 5� = -3 – 2 = -5 Choose row 1 .

• Determine inverse matrix A-1 A-1 = 1|𝐴𝐴| AT

A-1 = 1|𝐴𝐴|

AT = 1-5

⎣⎢⎢⎢⎡-1 -2 0-1 3 05 -15 -5⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡𝟏𝟏𝟓𝟓

𝟐𝟐𝟓𝟓

𝟎𝟎𝟏𝟏𝟓𝟓

-𝟑𝟑𝟓𝟓

𝟎𝟎

-𝟏𝟏 𝟑𝟑 𝟏𝟏⎦⎥⎥⎥⎤

It gives the same result as the Gauss-Jordan method .

Page 11-18

Find Inverse Matrix A-1 – Method II


Example: Find A-1 of the following 3 × 3 matrix .

A = �3 2 01 -1 00 5 1

� A-1 = ?

Steps

• Find the cofactor matrix: determine the cofactor (minors + place signs) of each element .

�3 2 0 1 -1 00 5 1

�

⎣⎢⎢⎢⎡ �-1 0

5 1� - �1 0

0 1� �1 -1

0 5�

- �2 05 1

� �3 00 1

� - �3 20 5

�

�2 0-1 0� - �3 0

1 0� �

3 21 -1� ⎦

⎥⎥⎥⎤

+ − + − + − + − +

cofactor matrix = �-1 -1 5-2 3 -150 0 -5

�

• Determine the transpose of the cofactor matrix AT: exchange all the rows and columns .

AT =

⎣⎢⎢⎢⎡-1 -2 0-1 3 05 -15 -5⎦

⎥⎥⎥⎤

• Calculate the determinant |𝐀𝐀| of the matrix

|𝑨𝑨| = �𝟑𝟑 𝟐𝟐 𝟎𝟎1 -1 00 5 1

� = 3 �-1 05 1

� − 2 �1 00 1� + 0 �1 -1

0 5� = -3 – 2 = -5 Choose row 1 .

• Determine inverse matrix A-1 A-1 = 1|𝐴𝐴| AT

A-1 = 1|𝐴𝐴|

AT = 1-5

⎣⎢⎢⎢⎡-1 -2 0-1 3 05 -15 -5⎦

⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡𝟏𝟏𝟓𝟓

𝟐𝟐𝟓𝟓

𝟎𝟎𝟏𝟏𝟓𝟓

-𝟑𝟑𝟓𝟓

𝟎𝟎

-𝟏𝟏 𝟑𝟑 𝟏𝟏⎦⎥⎥⎥⎤

It gives the same result as the Gauss-Jordan method .

Page 11-18





Solving a Linear System Using the Inverse Matrix

• Write systems of linear equations in matrix formLinear System Matrix Form Example

2×2 system �𝑎𝑎11𝑥𝑥 + 𝑏𝑏12𝑦𝑦 = 𝑐𝑐1𝑎𝑎21𝑥𝑥 + 𝑏𝑏22𝑦𝑦 = 𝑐𝑐2


� �𝑥𝑥𝑦𝑦� = �𝑐𝑐1𝑐𝑐2�

A · X = C

�2𝑥𝑥 + 3𝑦𝑦 = 13𝑥𝑥 − 4𝑦𝑦 = 2

�2 33 -4� �𝑥𝑥𝑦𝑦� = �12�

3×3 system �𝑎𝑎11𝑥𝑥 + 𝑎𝑎12𝑦𝑦+𝑎𝑎13𝑧𝑧 = 𝑐𝑐1𝑎𝑎21𝑥𝑥 + 𝑎𝑎22𝑦𝑦+𝑎𝑎23𝑧𝑧 = 𝑐𝑐2𝑎𝑎31𝑥𝑥 + 𝑎𝑎32𝑦𝑦+𝑎𝑎33𝑧𝑧 = 𝑐𝑐3


� �𝑥𝑥𝑦𝑦𝑧𝑧� = �

𝑐𝑐1𝑐𝑐2𝑐𝑐3�

A · X = C

�3𝑥𝑥 + 2𝑦𝑦 + 𝑧𝑧 = 1

2𝑥𝑥 − 3𝑦𝑦 + 4𝑧𝑧 = 35𝑥𝑥 + 2𝑦𝑦 − 𝑧𝑧 = 2

�3 2 1 2 -3 4 5 2 -1


132�

• Solving a linear system using the inverse matrix

The Matrix Equation of a Linear System

A X = C � 𝑨𝑨 − coefficient matrix𝑪𝑪 − constant matrix 𝑿𝑿 − variable matrix

Solving Using A-1 X = A-1 C A-1− inverse matrix

Note: X = A-1 C , X ≠ C A-1 ∵ AB ≠ BA Multiplication of matrices is not commutative .

Example: Use matrices to solve a 2×2 system . � 𝒙𝒙 − 𝒚𝒚 = 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒚𝒚 = 𝟐𝟐

Steps Example

- Write the system in matrix form . AX = C �1 -12 3

� �𝑥𝑥𝑦𝑦� = �12�- Find A-1 . A X C

- Switch the main diagonal elements .

- Change signs for the remaining elements . A-1 = �

3 1-2 1�

�1 -12 3

�=

�3 1-2 1�

3−(-1)2= 1

5 �

3 1-2 1�

- Divide by the determinant . = �35

15

-25

15

�

- Solve for X: X = A-1C X = A-1C = �35

15

-25

15

� �12� = �35∙1 + 15∙2-25 ∙1 + 15∙2

� = �10�

i .e . �𝑥𝑥𝑦𝑦�=�10� or �𝒙𝒙 = 𝟏𝟏

𝒚𝒚 = 𝟎𝟎 or (𝟏𝟏,𝟎𝟎)

A-1 = � 𝑏𝑏2 -𝑏𝑏1-𝑎𝑎2 𝑎𝑎1

�


A X = C A-1 A X = A-1 C

I X = A-1 C X = A-1 C

Page 11-19

Solving a Linear System Using the Inverse Matrix





A · X = C

�2𝑥𝑥 + 3𝑦𝑦 = 13𝑥𝑥 − 4𝑦𝑦 = 2

�2 33 -4� �𝑥𝑥𝑦𝑦� = �12�

3×3 system �𝑎𝑎11𝑥𝑥 + 𝑎𝑎12𝑦𝑦+𝑎𝑎13𝑧𝑧 = 𝑐𝑐1𝑎𝑎21𝑥𝑥 + 𝑎𝑎22𝑦𝑦+𝑎𝑎23𝑧𝑧 = 𝑐𝑐2𝑎𝑎31𝑥𝑥 + 𝑎𝑎32𝑦𝑦+𝑎𝑎33𝑧𝑧 = 𝑐𝑐3




A · X = C

�3𝑥𝑥 + 2𝑦𝑦 + 𝑧𝑧 = 1

2𝑥𝑥 − 3𝑦𝑦 + 4𝑧𝑧 = 35𝑥𝑥 + 2𝑦𝑦 − 𝑧𝑧 = 2

�3 2 1 2 -3 4 5 2 -1


132�





Note: X = A-1 C , X ≠ C A-1 ∵ AB ≠ BA Multiplication of matrices is not commutative .

Example: Use matrices to solve a 2×2 system . � 𝒙𝒙 − 𝒚𝒚 = 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒚𝒚 = 𝟐𝟐

Steps Example

- Write the system in matrix form . AX = C �1 -12 3

� �𝑥𝑥𝑦𝑦� = �12�- Find A-1 . A X C

- Switch the main diagonal elements .

- Change signs for the remaining elements . A-1 = �

3 1-2 1�

�1 -12 3

�=

�3 1-2 1�

3−(-1)2= 1

5 �

3 1-2 1�

- Divide by the determinant . = �35

15

-25

15

�

- Solve for X: X = A-1C X = A-1C = �35

15

-25

15

� �12� = �35∙1 + 15∙2-25 ∙1 + 15∙2

� = �10�

i .e . �𝑥𝑥𝑦𝑦�=�10� or �𝒙𝒙 = 𝟏𝟏

𝒚𝒚 = 𝟎𝟎 or (𝟏𝟏,𝟎𝟎)

A-1 = � 𝑏𝑏2 -𝑏𝑏1-𝑎𝑎2 𝑎𝑎1

�


A X = C A-1 A X = A-1 C

I X = A-1 C X = A-1 C

Page 11-19





Example: Use matrices to solve a 3×3 system . �𝟐𝟐𝟐𝟐 = 𝒚𝒚 − 𝟑𝟑𝟑𝟑+ 𝟏𝟏 𝟑𝟑𝟐𝟐+ 𝟒𝟒+ 𝟐𝟐𝒚𝒚 = 𝟎𝟎 𝟐𝟐𝒚𝒚 − 𝟑𝟑𝟑𝟑 = 𝟐𝟐

Steps Example

- Write the system in standard form . �2𝑥𝑥 − 𝑦𝑦 + 3𝑧𝑧 = 1 3𝑥𝑥+ 2𝑦𝑦+ 0𝑧𝑧 = -4 0𝑥𝑥+ 2𝑦𝑦 − 3𝑧𝑧 = 2

- Write the system in matrix form . A X = C �2 -1 3 3 2 0 0 2 -3

� �𝑥𝑥𝑦𝑦 𝑧𝑧� = �

1-42�

- Find A-1 . - Find the cofactor matrix .

�2 -1 3 3 2 0 0 2 -3

�

⎣⎢⎢⎢⎢⎡ �

2 02 -3� - �

3 00 -3� �3 2

0 2�

- �-1 32 -3

� �2 30 -3� - �2 -1

0 2�

�-1 32 0

� - �2 33 0

� �2 -13 2

� ⎦⎥⎥⎥⎥⎤

cofactor matrix = �-6 9 6 3 -6 -4 -6 9 7

�

- Determine the transpose of the cofactor matrix AT . AT = �-6 3 -69 -6 96 -4 7

� Exchange rows & columns .

- Calculate the determinant |𝐴𝐴|. |𝐴𝐴| = �2 -1 3 3 2 0 0 2 -3

2 -13 20 2

� Expansion by diagonals .

= 2∙2(-3) + (-1)∙0∙0 + 3∙3∙2 − 3∙2∙0 − 2∙0∙2 – (-1)∙3(-3) = -12 + 18 − 9 = -3 a

- Determine the inverse matrix A-1 . A-1 = 1|𝐴𝐴|

AT = 1-3

�-6 3 -69 -6 96 -4 7

� = �2 -1 2-3 2 -3

-24

3

-7

3

�

- Solve for X . X = A-1C X = �2 -1 2-3 2 -3

-24

3

-7

3 � �

1-4 2� =

⎣⎢⎢⎢⎡ 2 ∙ 1 + (-1)(-4) + 2 ∙ 2(-3) ∙ 1 + 2(-4) + (-3) ∙ 2

(-2) ∙ 1 + 43 (-4) + -7

3 ∙ 2 ⎦⎥⎥⎥⎤

= �2 + 4 + 4-3 − 8 − 6

-2 −16

3−

14

3

�= �10

-17 -12

�

i .e . �𝑥𝑥𝑦𝑦 𝑧𝑧� = �

10-17 -12

� or �𝟐𝟐 = 𝟏𝟏𝟎𝟎 𝒚𝒚 = -𝟏𝟏𝟏𝟏

𝟑𝟑 = -𝟏𝟏𝟐𝟐 or (𝟏𝟏𝟎𝟎, -𝟏𝟏𝟏𝟏,-𝟏𝟏𝟐𝟐)

It gives the same result as using Cramer’s rule for a 3×3 system .

+ − + − + − + − +

Page 11-20





Matrix Inverse on a Graphing Calculator (TI-83 Plus)

• Creating a matrix on a graphing calculator

Example: A = �2 -1 33 2 00 2 -3

�

2nd MATRX Enter the matrix screen .

► ► ENTER Select the ‘Edit’ command .

3 ENTER 3 ENTER Define a 3 × 3 matrix .

2 ENTER -1 ENTER 3 ENTER Enter the numbers of the matrix .

3 ENTER 2 ENTER 0 ENTER

0 ENTER 2 ENTER -3 ENTER

• Finding an inverse matrix on a graphing calculator

Example: A = �2 -1 33 2 00 2 -3

� A-1 = ?

2nd MATRX ► ► ENTER Enter the matrix screen; select the ‘Edit’ command .





2nd QUIT Return to the home screen .

2nd MATRX ENTER Re-enter the matrix screen .

x-1 MATH ENTER ENTER Select the ‘1: ► Frac’ command .

Display:⎣⎢⎢⎢⎡2 -1 2 -3 2 -3-2 4

3-73 ⎦⎥⎥⎥⎤

Display the inverse matrix A-1 .

It gives the same result as manual calculation .

NAMES MATH 𝐄𝐄𝐄𝐄𝐄𝐄𝐄𝐄1: [A]2: [B]

MATRX[A] 𝟑𝟑 𝐱𝐱 𝟑𝟑

[𝐀𝐀]-1 > Frac

Page 11-21

Matrix Inverse on a Graphing Calculator (TI-83 Plus)

• Creating a matrix on a graphing calculator

Example: A = �2 -1 33 2 00 2 -3

�







• Finding an inverse matrix on a graphing calculator

Example: A = �2 -1 33 2 00 2 -3

� A-1 = ?

2nd MATRX ► ► ENTER Enter the matrix screen; select the ‘Edit’ command .






2nd MATRX ENTER Re-enter the matrix screen .

x-1 MATH ENTER ENTER Select the ‘1: ► Frac’ command .

Display:⎣⎢⎢⎢⎡2 -1 2 -3 2 -3-2 4

3-73 ⎦⎥⎥⎥⎤

Display the inverse matrix A-1 .


NAMES MATH 𝐄𝐄𝐄𝐄𝐄𝐄𝐄𝐄1: [A]2: [B]

MATRX[A] 𝟑𝟑 𝐱𝐱 𝟑𝟑

[𝐀𝐀]-1 > Frac

Page 11-21





Solving a System Using Matrices & Graphing Calculator(TI-83 Plus)

Example: Solve a 3×3 system . �𝟐𝟐𝟐𝟐 = 𝒚𝒚− 𝟑𝟑𝟑𝟑+ 𝟏𝟏 𝟑𝟑𝟐𝟐+ 𝟒𝟒+ 𝟐𝟐𝒚𝒚 = 𝟎𝟎 𝟐𝟐𝒚𝒚 − 𝟑𝟑𝟑𝟑 = 𝟐𝟐

Write the system in standard form . �2𝑥𝑥 − 𝑦𝑦+ 3𝑧𝑧 = 1 3𝑥𝑥 + 2𝑦𝑦+ 0𝑧𝑧 = -4 0𝑥𝑥+ 2𝑦𝑦 − 3𝑧𝑧 = 2




2 ENTER -1 ENTER 3 ENTER 1 ENTER Enter the augmented matrix .

3 ENTER 2 ENTER 0 ENTER -4 ENTER �2 -1 3

3 2 0

0 2 -3

1

-4

2

�

0 ENTER 2 ENTER -3 ENTER 2 ENTER


2nd MATRX Re-enter the matrix screen .

► Select the ‘MATH’ command .

▲ ▲ ▲ ▲ ▲ ENTER Select the ‘B:rref (’ command .

2nd MATRX ENTER ENTER

Display: �1 0 0 0 1 0 0 0 1

10-17 -12

�


10-17 -12

� or �𝑥𝑥 = 10 𝑦𝑦 = -17 𝑧𝑧 = -12


NAMES 𝐌𝐌𝐌𝐌𝐌𝐌𝐌𝐌 EDIT1: det (2: T… …

MATRX[A] 𝟑𝟑 𝐱𝐱 𝟒𝟒

𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫 ( NAMES MATH EDIT0: cumSum (A:ref (B:rref (C:rowSwap (D: row+ (… …

Page 11-22

Solving a System Using Matrices & Graphing Calculator(TI-83 Plus)

Example: Solve a 3×3 system . �𝟐𝟐𝟐𝟐 = 𝒚𝒚− 𝟑𝟑𝟑𝟑+ 𝟏𝟏 𝟑𝟑𝟐𝟐+ 𝟒𝟒+ 𝟐𝟐𝒚𝒚 = 𝟎𝟎 𝟐𝟐𝒚𝒚 − 𝟑𝟑𝟑𝟑 = 𝟐𝟐

Write the system in standard form . �2𝑥𝑥 − 𝑦𝑦+ 3𝑧𝑧 = 1 3𝑥𝑥 + 2𝑦𝑦+ 0𝑧𝑧 = -4 0𝑥𝑥+ 2𝑦𝑦 − 3𝑧𝑧 = 2




2 ENTER -1 ENTER 3 ENTER 1 ENTER Enter the augmented matrix .

3 ENTER 2 ENTER 0 ENTER -4 ENTER �2 -1 3

3 2 0

0 2 -3

1

-4

2

�

0 ENTER 2 ENTER -3 ENTER 2 ENTER


2nd MATRX Re-enter the matrix screen .

► Select the ‘MATH’ command .

▲ ▲ ▲ ▲ ▲ ENTER Select the ‘B:rref (’ command .

2nd MATRX ENTER ENTER

Display: �1 0 0 0 1 0 0 0 1

10-17 -12

�


10-17 -12

� or �𝑥𝑥 = 10 𝑦𝑦 = -17 𝑧𝑧 = -12


NAMES 𝐌𝐌𝐌𝐌𝐌𝐌𝐌𝐌 EDIT1: det (2: T… …

MATRX[A] 𝟑𝟑 𝐱𝐱 𝟒𝟒

𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫 ( NAMES MATH EDIT0: cumSum (A:ref (B:rref (C:rowSwap (D: row+ (… …

Page 11-22





Unit 11 Summary

• Evaluate a 2 × 2 determinantDeterminant Evaluation Example



�+

= 𝑎𝑎1𝑏𝑏2 − 𝑏𝑏1𝑎𝑎2 �2 13 4�+

= 2 ∙ 4 − 1 ∙ 3 = 5

• Evaluate a 3 × 3 Determinant – Method I: Using Diagonals3 × 3 Determinant Expansion by Diagonals Example





1 1 3 4 3 2 3 1 2

1 14 33 1

= 1∙3∙2 + 1∙2∙3 + 3∙4∙1-3∙3∙3 − 1∙2∙1 − 1∙4∙2

= 6 + 6 + 12 – 27 – 2 – 8 = -13




+ + − + − + − + − +



• Evaluate a 3 × 3 determinant – Method II: expansion by minors

- Choose any row or column in the determinant .- Multiply each element in the chosen row/column by its cofactor .

Choose this column .





�

+ − + − + − + − +

• Evaluate a determinant that can be expanded by any row or column





(D ≠ 0)




�



constant

Page 11-23

Unit 11 Summary

• Evaluate a 2 × 2 determinantDeterminant Evaluation Example



�+

= 𝑎𝑎1𝑏𝑏2 − 𝑏𝑏1𝑎𝑎2 �2 13 4�+

= 2 ∙ 4 − 1 ∙ 3 = 5

• Evaluate a 3 × 3 Determinant – Method I: Using Diagonals3 × 3 Determinant Expansion by Diagonals Example





1 1 3 4 3 2 3 1 2

1 14 33 1

= 1∙3∙2 + 1∙2∙3 + 3∙4∙1-3∙3∙3 − 1∙2∙1 − 1∙4∙2

= 6 + 6 + 12 – 27 – 2 – 8 = -13




+ + − + − + − + − +



• Evaluate a 3 × 3 determinant – Method II: expansion by minors

- Choose any row or column in the determinant .- Multiply each element in the chosen row/column by its cofactor .

Choose this column .





�

+ − + − + − + − +

• Evaluate a determinant that can be expanded by any row or column





(D ≠ 0)




�



constant

Page 11-23











D ≠ 0




• Matrix: a rectangular array of elements enclosed in brackets .

• Dimensions: A matrix has �𝑚𝑚 rows 𝑛𝑛 columns m × n dimensions

Rows Columns



� 2𝑥𝑥 + 3𝑦𝑦 + 4𝑧𝑧 = 1 𝑥𝑥 + 2𝑦𝑦 + 3𝑧𝑧 = 4

3𝑥𝑥 + 𝑦𝑦 + 5𝑧𝑧 = 2

• Coefficient matrix:

�𝑎𝑎1 𝑏𝑏1 𝑐𝑐1𝑎𝑎2 𝑏𝑏2 𝑐𝑐2 𝑎𝑎3 𝑏𝑏3 𝑐𝑐3

� �2 3 41 2 33 1 5

�

• Augmented matrix:


𝑘𝑘1𝑘𝑘2𝑘𝑘3

� �2 3 4 1 2 3 3 1 5

142�





• Add/subtract two matrices of the same dimensions Matrix Operations Example

matrix addition

A + B




�2×2 2×2

The same dimensions

�1 32 4 � + �3 5

2 1� = �1 + 3 3 + 52 + 2 4 + 1�

= �4 84 5�

matrix subtraction

A – B




�

3×2 3×2

�3 85 76 -9

� − �2 23 -36 4

� = �3 − 2 8 − 25 − 3 7 − (-3)6 − 6 -9 − 4

�

= �1 62 100 -13

�

Page 11-24





• Scalar matrix multiplicationScalar Matrix Multiplication Example




�2�4 -2 𝑦𝑦

3 5 0� = �2 ∙ 4 2(-2) 2 ∙ 𝑦𝑦

2 ∙ 3 2 ∙ 5 2 ∙ 0�

= �8 -4 2𝑦𝑦 6 10 0

�

• Requirements for matrix multiplication




• Dimensions of the product for matrix multiplication

Dimensions ofthe Product

If A = (m1× n1) and B = (m2 × n2), then A·B = (m1× n1) (m2 × n2) = (m1 × n2) .

Example:If A = (3 × 2) and B = (2 × 4), Then A·B = (3 × 2) (2 × 4) = (3 × 4) .

• Matrix multiplicationMatrix Multiplication Example

AB = �𝑎𝑎11 𝑎𝑎12 𝑎𝑎13𝑎𝑎21 𝑎𝑎22 𝑎𝑎23� �

𝑏𝑏11 𝑏𝑏12𝑏𝑏21 𝑏𝑏22𝑏𝑏31 𝑏𝑏32

� A = 2×3 , B = 3×2

(1st row of A) × (1st column of B) (1st row of A) × (2nd column of B)

= �𝑎𝑎11𝑏𝑏11 + 𝑎𝑎12𝑏𝑏21 + 𝑎𝑎13𝑏𝑏31 𝑎𝑎11𝑏𝑏12 + 𝑎𝑎12𝑏𝑏22 + 𝑎𝑎13𝑏𝑏32𝑎𝑎21𝑏𝑏11 + 𝑎𝑎22𝑏𝑏21 + 𝑎𝑎23𝑏𝑏31 𝑎𝑎21𝑏𝑏12 + 𝑎𝑎22𝑏𝑏22 + 𝑎𝑎23𝑏𝑏32

�

(2nd row of A) × (1st column of B) (2nd row of A) × (2nd column of B) 2×2

�1 2 32 0 1� �

2 10 23 -1

� 2×3 , 3×2

= �1 ∙ 2 + 2 ∙ 0 + 3 ∙ 3 1 ∙ 1 + 2 ∙ 2 + 3(-1)2 ∙ 2 + 0 ∙ 0 + 1 ∙ 3 2 ∙ 1 + 0 ∙ 2 + 1(-1)

�

= �11 27 1� 2×2

• Identity matrix II 2×2 3×3 n×n


0 1� I = �1 0 00 1 00 0 1

� I = �1 ⋯ 0⋮ 1 ⋮0 ⋯ 1

�

main diagonal

• Identity property Identity Property for Matrices A I = I A = A

• Properties of matrixProperty of Multiplication Property of Addition

A, B, and C are matricesassociative property: A(BC) = (AB)C associative property: A + (B + C) = (A + B) + Cdistributive property: A(B + C) = AB + AC (B + C)A = BA + CA commutative property: A + B = B + A



Page 11-25










�


�

- Divide the result of the last step by the determinant |𝐴𝐴| .

A-1 = �𝑏𝑏2 -𝑏𝑏1-𝑎𝑎2 𝑎𝑎1

�


?- Check: A A-1 = I

• Gauss-Jordan Method to Find A-1


- Transform matrix A into the identity matrix I. A I

- Transform identity matrix I into inverse matrix A-1 . I A-1


- Find the cofactor matrix (minors + place signs) .

- Determine the transpose of the cofactor matrix AT: exchange all the rows and columns .

- Calculate the determinant |𝐴𝐴| of the matrix.

- Determine inverse matrix A-1 . A-1 = 1|𝐴𝐴| AT





A · X = C

�2𝑥𝑥 + 3𝑦𝑦 = 13𝑥𝑥 − 4𝑦𝑦 = 2

�2 33 -4� �𝑥𝑥𝑦𝑦� = �12�

3×3 system �𝑎𝑎11𝑥𝑥 + 𝑎𝑎12𝑦𝑦+𝑎𝑎13𝑧𝑧 = 𝑐𝑐1𝑎𝑎21𝑥𝑥 + 𝑎𝑎22𝑦𝑦+𝑎𝑎23𝑧𝑧 = 𝑐𝑐2𝑎𝑎31𝑥𝑥 + 𝑏𝑏32𝑦𝑦+𝑎𝑎33𝑧𝑧 = 𝑐𝑐3




A · X = C

�3𝑥𝑥 + 2𝑦𝑦 + 𝑧𝑧 = 1

2𝑥𝑥 − 3𝑦𝑦 + 4𝑧𝑧 = 35𝑥𝑥 + 2𝑦𝑦 − 𝑧𝑧 = 2

�3 2 1 2 -3 4 5 2 -1


132�





Page 11-26





PRACTICE QUIZ

Unit 11 Determinants and Matrices

1. Evaluate the determinant: �2 -1 0 0 3 -22 4 -1

�

2 . Solve the system using Cramer’s rule: �3𝑥𝑥 − 2𝑦𝑦 = -5𝑧𝑧 + 2 4𝑥𝑥 − 7𝑦𝑦 − 𝑧𝑧 = 19 5𝑥𝑥 = 6𝑦𝑦 − 4𝑧𝑧 + 13

3. Find the products: �3 11 -1� �

2 -30 1

�

4. Solve a 2×2 system using matrices: �5𝑥𝑥 + 4𝑦𝑦 = 1 4𝑥𝑥 + 3𝑦𝑦 = -1

5. Solve a 3×3 system using matrices: �3𝑥𝑥 − 2𝑦𝑦+ 5𝑧𝑧 = 2 4𝑥𝑥 − 7𝑦𝑦 − 𝑧𝑧 = 19 5𝑥𝑥 − 6𝑦𝑦+ 4𝑧𝑧 = 13

Page 16

PRACTICE QUIZ

Unit 11 Determinants and Matrices

1. Evaluate the determinant: �2 -1 0 0 3 -22 4 -1

�

2 . Solve the system using Cramer’s rule: �3𝑥𝑥 − 2𝑦𝑦 = -5𝑧𝑧 + 2 4𝑥𝑥 − 7𝑦𝑦 − 𝑧𝑧 = 19 5𝑥𝑥 = 6𝑦𝑦 − 4𝑧𝑧 + 13

3. Find the products: �3 11 -1� �

2 -30 1

�

4. Solve a 2×2 system using matrices: �5𝑥𝑥 + 4𝑦𝑦 = 1 4𝑥𝑥 + 3𝑦𝑦 = -1

5. Solve a 3×3 system using matrices: �3𝑥𝑥 − 2𝑦𝑦+ 5𝑧𝑧 = 2 4𝑥𝑥 − 7𝑦𝑦 − 𝑧𝑧 = 19 5𝑥𝑥 − 6𝑦𝑦+ 4𝑧𝑧 = 13

Page 16




Algebra I & II Key Concepts, Practice, and Quizzes Answers

Answers for Practice Quizzes

Unit 1 Reference Page Number

1. A = { x | x is a number between -5 and 2}A = {-4, -3, -2, -1, 0, 1} (1-1: page 2)

2. a. {-9}b. {7π, √5} (1-1: page 3)

3. a )-2 -2

b. ∙ [ )-1 .5 7 -1 .5 7 (1-1: page 4)

4. a. -3 .3b. 19c. -1 .5d. -3 (1-2: page 6)

5. a. - 0 .008b. m3

c. 1d . -6 (1-3: page 9)

6. 8 (1-4: page 10)

7. a. 7y – 3b. 2(t + 9) (1-4: page 11)

8. a. Inverse property of additionb. Associative property of multiplication (1-4: page 12)

9. a. 5c (ab –5b + 7)b. -6pq + 3prc. 2x² + yd. -x² + 8x -19 (1-5: page 13)

10. a. −8𝑥𝑥6

𝑦𝑦12

b. 𝑦𝑦6

8𝑥𝑥3(1-6: page 16)

11. a. 1.3975 × 105

b. 5.75 × 10–8 (1-6: page 17)

Unit 2

1. a. x = 2 .8b. y ≈ 0 .06c. y ≈ 3 .67 (2-1: pages 26-27)

2. 4x – 5 = 9 + 𝑥𝑥2

, x = 4 (2-2: page 30)

3. 4x = (x + 2) + (x + 4) – 2, 1st = 2, 2nd = 4, 3rd = 6 (2-2: page 32)

4. 2r + 2(r – 10) = 340 , r = 90 km/h , r – 10 = 80 km/h (2-2: page 36)





5. Downstream: t = 𝑑𝑑𝑟𝑟

= 426+12

≈ 0 .11 hr


= 326−12

≈ 0 .21 hr (2-2: page 36)

6. 46x + 66 (3+x) + 86 �12𝑥𝑥 + 2� = 680 ; 2, 5, 3 (2-2: page 37)

7. a. ] -2

(-∞, -2] or {x | x ≤ -2}

b. ) 1

(-∞, 1) or {x | x < 1}

c. [119

�119

, ∞) or {x | x ≥ 119� (2-3: page 43)

8. 76 < 𝑥𝑥+782

< 80, 74 < x < 82 (2-3: page 44)

9. a. Trueb. False (2-3: page 45)

10. a. A = {11, 13, 17}B = {13, 14, 15}A ∪ B = {11, 13, 14, 15, 17}, A ∩ B = {13}

b. A ∪ 𝐵𝐵 {1, 2, 3, 4, 5, 7} A ∩ B {3, 5}

A ∩ 𝐶𝐶 ∅ (2-4: page 46)

11. {x | -5 < 𝑥𝑥 ≤ 1 } or (-5 , 1]

( ] -5 1 (2-4: page 46)

12. a. {-8, 2}

b. {-1, 34� (2-5: pages 49-50)

13. a. {x | -1 ≤ 𝑥𝑥 ≤ 113� or [-1 , 11

3�

b. {x | 𝑥𝑥 < −15

or 𝑥𝑥 > 75�

or (-∞ , −15� ∪ �7

5, ∞) (2-5: pages 51-52)

Unit 3 .

1. a.(3-1: page 60)

b.

(3-1: page 61)

2. a. 52b. -1 (3-2: page 63)

3. a. 5a − 13b. 18 (3-2: page 63)

4. f (2012) ≈ 85 (3-2: page 64)

5. a. {3, -1, 6, -4}b. {4, 6, 3} (3-1: page 66)

∙ (-2, 0)

∙ (0, -2)

∙ (1, 1)





6. {x | x is a real number and x ≠ 5} or (-∞, 5) ∪ (5, ∞)

{x| x is a real number and x ≠ 57� or (-∞, 5

7� ∪ �5

7 , ∞) (3-3: page 67)

7. a. m = 6, b = -15 (3-4: page 68)b. m = 1

2(3-4: page 70)

8. -125 (3-4: page 70)

9.

(3-5: page 72)

10 . m1 = m2 = -72

, L1 ∥ L2 (3-5: page 74)

11 . y = -7x + 17 (3-5: page 76)

12 . a. y = -13𝑥𝑥 − 7

3, L1 ∥ L2

b. y = 3𝑥𝑥 − 9 , L1┴ L2 (3-5: page 77)

13 . f (t) = 20,000 – 1,000t , $15,000

(3-6: page 78)

Unit 4

1. (x, y) = (1, 1)

(4-1: page 84)

2. a. x = 1 , y = 1 (4-1: page 86)b. x = 0 , y = 6 (4-1: pages 87-88)

3. �2𝑙𝑙 + 2𝑤𝑤 = 140 𝑙𝑙 = 4𝑤𝑤 + 10 w = 12m , l = 58m (4-2: page 89)

4. a.

(4-3: page 91)

b.

(4-3: page 92)

f (x)

x0

t

f (t)20000

0

5

∙ (1, 1)

y

∙ (1, 1)x

∙ 4

∙ (1, 3)

x∙ 0

1500010000

x0

y

y

-5

-7

1





5.

The solution set is the region where the shading overlaps . The vertex is (0, 2) . (4-3: page 93)

Unit 5

1. a . f (40) = 2,960 b.

f (30) ≈ 2,500 (5-1: page 99)

2. a. 7𝑥𝑥3 − 2𝑥𝑥2 − 5𝑥𝑥 + 6 = 0b. 4𝑥𝑥3 + 7𝑥𝑥2 − 𝑥𝑥 − 2 = 0 (5-1: page 100)

3. a. 12t 2 – 16t – 3b. u3 + 6u2 + 12u + 8 (5-2: page 103)

4. -b(b + 2) (5-2: page 103)

5. a. (2c – 𝑑𝑑)(2cd + 1) b. 3xy(3x + 𝑦𝑦)(3x – y) c. (x + 1)(x – 3) (5-3: pages 105-106)d . (x – 3)(3x – 8) = 0 e . (t + 1

3) 2 = 0 (5-4: pages 107-108)

6. a. (2x + 3y)(2x – 3y)

b. 2��𝑢𝑢3

+ 𝑣𝑣5� �𝑢𝑢

3– 𝑣𝑣5��

c. (t2 + 4)(t +2)(t – 2)d. (x2 – 2y2)(x4 + 2x2y2 + 4y4) (5-5: pages 110-112)

Unit 6

1. a. 𝑥𝑥12− 1

3

b. b (6-1: page 117)

2. a. (x + 1)(x + 3)

b. 3(𝑦𝑦−1)2

(6-1: pages 118-119)

c. 8𝑥𝑥

(6-2: page 120)

3 . 12x4y2 (6-2: page 122)

4 . 2𝑏𝑏2+3𝑏𝑏+5(𝑏𝑏+2)(𝑏𝑏−2)

(6-2: page 123)

5. 9𝑦𝑦 − 3 − 12𝑦𝑦

(6-3: page 124)

6. a. (2𝑦𝑦 − 1) + 43𝑦𝑦

b. (𝑥𝑥2 + 3𝑥𝑥 + 3) + 6𝑥𝑥−1

(6-3: pages 125-126)

7. (2𝑥𝑥2 + 𝑥𝑥 + 7) + 7𝑥𝑥−2

(6-3: page 127)

8. a. 1+5𝑥𝑥 1+𝑥𝑥2

b. 𝑦𝑦−73𝑦𝑦(𝑦𝑦+3)(𝑦𝑦−2)

(6-4: pages 128-129)

9. a. 𝑦𝑦 = 5 28

(6-5: page 130)b. x = -2 (6-5: page 131)

10. t ≈ 1.71 hr (6-6: page 134)

2x – y ≤ −2

∙ (0, 2)

∙(0, 0)

4x + y > 2





11. $42 (6-6: page 135)

12. r = 5 km/hr (6-6: page 136)

Unit 7

1. a. √17 , √2 b. { x | x ≥ - 2

5� or �- 2

5 , ∞) (7-1: page 143)

2. a. 13

b. 2u (7-1: page 145)

3. -3 (7-1: page 145)

4. a. -3 𝑦𝑦𝑦𝑦3/4

𝑥𝑥𝑥𝑥2/3 𝑧𝑧𝑧𝑧1/5

b. 𝑣𝑣𝑣𝑣3/2

𝑢𝑢𝑢𝑢9/4

c. x9y12

d. 𝑏𝑏𝑏𝑏4

𝑎𝑎𝑎𝑎4

e. 1 (7-1: page 147)

5. a. 2𝑥𝑥𝑥𝑥2 �𝑦𝑦𝑦𝑦3 4 b. √𝑏𝑏𝑏𝑏20

c. √𝑢𝑢𝑢𝑢8𝑣𝑣𝑣𝑣6𝑤𝑤𝑤𝑤912

d. �𝑎𝑎𝑎𝑎3𝑏𝑏𝑏𝑏3

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

4 (7-2: page 148)

6. a. 2𝑥𝑥𝑥𝑥 �7𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 3

b. 2ab √𝑏𝑏𝑏𝑏 4 (7-3: page 150)

7. a. 𝑦𝑦𝑦𝑦�5𝑦𝑦𝑦𝑦 (4√3− 3) (7-4: page 151)

b. 3𝑎𝑎𝑎𝑎 (𝑏𝑏𝑏𝑏 − 4) (7-4: page 152)

c. �5𝑦𝑦𝑦𝑦𝑥𝑥𝑥𝑥2 3

𝑥𝑥𝑥𝑥 (7-5: page 154)

8. a. 𝟑𝟑𝟑𝟑𝟕𝟕𝟕𝟕 (7-6: page 156)

b. 𝑥𝑥𝑥𝑥 = 7 (7-6: page 158)

9. 2329− 14

29𝑖𝑖𝑖𝑖 (7-7: page 163)

Unit 8 1. y = - 5 ± √3 or y ≈ �-3.268

-6.732 (8-1: page 170)

2. x = - 2 ± √7 (8-2: page 171)

3. A =P(1+ r)t , r ≈ 5% (8-2: page 173)

4. a. x = 2 ± √3

b. -32

± 12𝑖𝑖𝑖𝑖 (8-3: pages 175-176)

5. w(w + 30) = 4,000 w = 50m, l = 80m (8-4: page 177)

6. x2 + x2 = 102, x = √50 = 5√2 (8-4: page 178)

7. 11 & -12 (8-4: page 179)

8. 𝑏𝑏𝑏𝑏2 − 4𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 91 > 0: 2 real solutions (8-5: page 181)

9. Let u = m-1, 𝑚𝑚𝑚𝑚 = 17 , m = -1 (8-6: page 184)





10 .

√ × √

x ≤ -4 -4 2 x ≥ 2 { x | -∞ < x ≤ -4 or 2 ≤ x < ∞ } or (-∞, -4] ∪ [2, ∞) (8-6: page 187)

Unit 9

1. Center: (2, -5), radius r = 3 (9-1: page 195)

2.

(9-2: page 202)

3.

(9-3: page 207)

4. 𝑦𝑦2

32− 𝑥𝑥2

12= 1

(9-4: page 210)

5. (𝑥𝑥+3)2

52+ (𝑦𝑦−1)2

22= 1

(9-5: pages 214, 207)

6. (-1, 3), �95

, -135� (9-6: page 216)

Unit 10

1.

(10-1: page 226)

2. f (x) = x3, g (x) = 4 − 3x (10-2: page 232)

3. f -1 (x) = - 𝒙𝒙𝟓𝟓

(10-2: page 234)

x∙

(4, -4)

(0, 12) ∙

(2, 0)

∙ 0

y

∙ (-6, 2) ∙ (4, 2)

∙ (-1, -1)

∙ (-1, 2)

∙ (-1, 5)

(-1, 0) ∙ ∙ (1, 0)

∙ (0, 3)

∙ (0, -3)

∙ (-3, 1) x

0

(-8, 1) ∙ ∙ (2, 1)

∙ (-3, 3)

(-3, -1) ∙

∙ (0, 1)

f(x) = 2xf(x) = 2-x

y

∙ (6, 0)





4. a. x = 4b. x = - 1

3 (10-3: page 237)

5. 3 (10-3: page 239)

6. a. log46 + log4𝑥𝑥 − 5 log4𝑦𝑦 (10-4: page 240)b. 1

2log5𝑎𝑎 + 1

4log5𝑏𝑏 −

14

log5𝑐𝑐 (10-4: page 242)

7. a. ≈ -2.06b. ≈ 1.35 (10-5: page 245)

8. a. 𝑥𝑥 = log8log3

≈ 1.89 (10-5: pages 246-247)

b. 𝑥𝑥 = 935≈ 0.26 (10-6: page 248)

Unit 11

1. 14 (11-1: page 256)

2. D = -5, Dx = -5, Dy = 10, Dz = 5, (1, -2, -1) (11-2: pages 260-261)

3. �6 -8

2 -4� (11-4: page 266)

4. �-79� (11-4: page 273)

5. (1, -2, -1) (11-4: page 274)




Algebra I & II Key Concepts, Practice, and Quizzes Index

AAbsolute value 5, 20Absolute value equation 49, 53Absolute value inequalities 51, 55AC method 109, 114Addend 6, 20Adding complex numbers 161, 166Adding like rational expressions 120, 137Adding radical expressions 151, 165Adding signed numbers 6, 19Adding unlike rational expressions 123, 137Algebraic expression 97, 113Algebraic terms 97, 113Ascending order 98, 113Associative property 12, 19Augmented matrix 262, 278

BBasic mathematic symbols 4, 18Binomial 97, 113Business formulas 33, 35, 55

CChange of base formula 245, 253Circle 195, 217Closed(filled)circle 4,18Closure property 12, 19Coefficient 10,19,113Coefficientmatrix 262,278Cofactor 257, 277Combine like terms 14, 20Commission 35, 55Common logarithm 243, 253Commutative property 12, 19Completing the square 171, 190Complex conjugates 162, 166Complex number system 159, 166Complex rational expression 128, 137Composite function 231, 251Composite number 3Compound inequality 39, 53Compound interest 173, 191

Conjugates 162, 165Consistent & dependent 85, 94Consistent & independent 85, 94Constant 10, 19Converting between exponential

and logarithmic forms 252Co-vertex 205, 219Cramer’s rule 259-260, 277-278Cut point 188, 192

DDegree of a term with one variable 98, 113Degree of a term with more variables 98, 113Degree of a polynomial with

more variables 98, 113Descending order 98, 113Determinant 255, 277Difference 6,20Differenceofcubes 112,115Differenceofsquares 103,115Dimensions of a matrix 262, 278Dimensions of the product 265, 279Discriminant 180, 191Distance Formula 194, 217Distributive property 12, 19Dividend 6, 20Dividing complex numbers 162, 166Dividing rational expressions 119, 137Dividing signed numbers 7, 19Divisor 6, 20Domain 62, 66, 79

EElimination method 87, 94Ellipse 204, 218Empty set (or null set) 45, 55Equation 23, 53Equation-solving strategy 26, 54Equations in quadratic form 183, 192Equations involving decimals 27, 54Equations involving fractions 27, 54Equations of circles 195, 217

Index

Page(s) Page(s)





Equationsofdifferentdegrees 24,53Equations of ellipses 204, 219Equations of hyperbolas 208, 220Equations of parabolas 203, 218Equivalent expressions 13, 20Evaluate a 2 × 2 determinant 255, 277Evaluate a 3 × 3 determinant 256, 277Expansion by diagonals 256, 277Expansion by minors 257, 277Exponent 8 Exponential equation 246, 253Extended number system 159, 166Extraneous solution 156, 165

FFind the LCD 121, 137First-degree equation 24, 53FOIL method 102, 114Formula 28, 53Four quadrants 59, 79Function 62, 79Function notation 63, 79Function transformations 212, 221Function values 63, 79 GGauss-JordanmethodtofindA-1 270, 280General-form conic equation 213, 221Geometry formulas 28, 56Graph of inverse function 229, 251Graph of the logarithmic function 238, 252Greatest / Highest Common Factor 104, 114

HHigher-degree equation 24, 53Horizontal line 73, 80Horizontal-line test 228, 251Hyperbola 208, 219

IIdentity property 12, 19 Identity property for matrices 268, 279Identity matrix 268, 279Imaginary unit 159-160, 166Incomplete quadratic equations 169, 190Inconsistent 85, 94Index of a radical 146, 164

Inequality 39, 53Integers 3, 18Intersection 45, 55Inverse function 229, 251Inverse matrix 269, 280Inverse property 12, 19Irrational numbers 1, 18

KKey or clue words in word problems 11Key to solve an exponential equation 246, 253Key to solve a logarithmic equation 248, 252

LLeadingcoefficient 98,113Leading term of a polynomial 98, 113Least common denominator (LCD) 121Least common multiple (LCM) 121Like radicals 151, 165Like rational expressions 120, 137Like terms 10, 19Linear equation 24, 53Linear equation in two variables 60, 79Linear inequality 90, 94Logarithmdefinition 235,252Logarithm of negative number 235, 252Logarithm of zero 235, 252Logarithmic equation 248, 252Logarithmic function 235, 251

MMatrix 262, 278Matrix addition 263, 278Matrix equality 263, 278Matrix multiplication 264, 279Matrix subtraction 263, 278Methods for solving quadratic equations 174, 191Midpoint formula 194, 217Minuend 6, 20Minor 257Missing terms in long division 126, 138Monomial 97, 113Motion formulas 136, 139Multiplicand 6, 20Multiplier 6, 20Multiplying complex numbers 161, 166Multiplying rational expressions 118, 137





Multiplying signed numbers 6, 19

NNatural numbers 1, 18Natural logarithm 244, 253Negative of the greatest common factor 104, 114Nonlinear equation 61, 79Nonlinear system of equations 215, 221nth root 144, 164nth root to the nth power 146, 164Number line 1

OOne-to-one function 228, 251Open (empty) circle 4, 18Opposite (additive or negative inverse) 7, 20 Opposite of the polynomial 100, 113Order of operations 9, 18Ordered pair 59, 79

PParabola 196, 217Parallel lines 74, 80Percent decrease 33, 56Percent increase 33, 56Perfect square 149Perpendicular lines 74, 80Place signs 277Point-slope equation 75, 80Polynomial 97, 113Polynomial long division 125, 138Power of i 160, 166Powers of roots 146, 165Principle square root 142, 164Prime number 3Procedure for solving

quadratic inequalities 185, 192Procedure to complete the square 171, 190-191Product rule for radicals 149, 165Properties for solving equations 25, 54Properties of absolute value 48, 55Properties of addition 12, 19Properties of multiplication 12, 19Properties of exponential functions 239, 252Properties of linear equations 85, 94

Properties of logarithmic functions 239, 252Properties of matrices 268, 279Properties of zero 7Proportion 135, 139

QQuadratic equation 169, 190Quadratic formula 174, 191Quadratic inequality 185, 192Quotient rule for radicals 149, 165

RRadical (root) 144, 164Radical equation 157, 165Radical expression 148, 165Radical notation 144, 164 Range 66, 79Rate 135, 139Ratio 135, 139Rational (fractional) equation 130, 138Rational exponent notation 144, 164 Rational expression 130, 137Rational function 117, 137Rational inequality 188, 192Rational numbers 1, 18Rationalize the denominator 153, 165Real number system 1, 18Real numbers 1, 18Reflection 250Relation 66, 79Remove parentheses 15, 20Requirements for matrix multiplication 264, 279Root 144, 164Roster notation 2, 18Rules of exponents 16, 20Rules of logarithms 240, 253

SScalar matrix multiplication 264, 279Scientificnotation 17,20Set 2Set-builder notation 2, 18Shifting 212, 221, 224, 250 Simple interest 173, 191





Simplifying a complex rational expression 128, 138

Simplifying radical expressions 148, 165Slope 68-69, 80Slope formula 69, 80Slope-intercept function 68, 80Solve an absolute value equation 49, 55Solving a linear system using the

inverse matrix 273, 280Solving a rational equation 130, 138Solution 23, 53Solution set 23, 53Solutions of the systems of equations 83, 94Squareofdifference 103,115Square of sum 103, 115Square root 142, 164Square root equation 155, 165Steps for solving word problems 30, 54Straight-line equation 76, 80Subset 45, 55Substitution for variable 184, 192Substitution method 86, 94Subtrahend 6, 20Subtracting complex numbers 161, 166Subtracting like rational expressions 120, 137Subtracting radical expressions 151, 165Subtracting signed numbers 6, 19Subtracting unlike rational expressions 123, 137Sum of cubes 112, 115Synthetic division 127, 139System of linear equations 83, 94System of linear inequalities in

two variables 93, 95

T Term 10, 19, 113Test point 91-92, 95Translate words into algebraic expression 11Trinomial 97, 113

UUnion 45, 55Unlike rational expressions 120, 137

VVertical line 73, 80Vertical line test 65, 80

WWork problems 134, 139Writing equation from solutions 182, 192Write systems of linear equations

in matrix form 280Whole numbers 1, 18

Xx-intercept 59, 79x-intercepts of a quadratic equation 170, 190x and y interchanging 226, 251

Yy-intercept 59, 79

ZZero product property 169, 190Zero product property in reverse 182, 192



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