XEI: EXPRESSIONS, EQUATIONS, AND INEQUALITIES

EXPRESSIONS What is an expression?

An expression is a mathematical statement that consists of terms and represents a value

The difference between expressions and equations is that equations have an equal sign and expressions do not.

If we see an expression with an equal sign, then there is no value on the other side of the equal sign. What are terms?

Terms are parts of an expression separated by a “+” or “-” sign.

EXAMPLES OF EXPRESSIONS

Example 1 – 1 + 1 + 1 – 5 is an expression

Example 2 – (6 + 7 – 10) ∙ 2² is an expression

Example 3 – z – (7 – 6 + x) is an expression

Example 4 – -4x² - 3x + 2 is an expression

Example 5 – x is an expression

EVALUATING EXPRESSIONSWhen we evaluate expressions, we determine the numerical value of the expression.We must use the correct order of operations when we evaluate expressions.Order of Operations:

1. Parentheses2. Exponents3. Multiply/Divide (whichever comes first

from left to right)4. Add/Subtract (whichever comes first

from left to right)

EXAMPLES OF EVALUATING EXPRESSIONS

Example 1 – 1 + 1 + 1 – 5 = ?

-2

Example 2 – (6 + 7 – 10) ∙ 2² = ?

3 · 2²3 · 412

MORE EXAMPLES OF EVALUATING EXPRESSIONS

Example 3 – x + z - = ? when x = -7, y = -6, and z =

4-7 + 4 - substitute variables for values

-7 + 4 – (-1)-3 – (-1) -3 + 1-2

YOU TRY1) ((-) – (-3) · (-7)) ∙ 2 = ?

2)( + m)(9 + n) = ? when m = -2 and n = -10

SIMPLIFYING EXPRESSIONSWhat are like terms?

Like terms are terms that either have the same variable and power or have no variable at all.

Example of like terms – 7x + 2 – 4x – 1 In the expression above, 7x and -4x are like terms, and 2 and -1 are like terms.

What does it mean to “combine” like terms?When like terms are “combined,” they are added together.

COMBINING LIKE TERMSExample 1 – Simplify the following expression

by combining like terms.x + 10 – 4

x + 10 – 4 simplifies to x + 6 since 10 and -4 are like terms

Example 2 – Simplify the following expression by combining like terms.-2b -10 + b – 1-2b -10 + b – 1 simplifies to –b – 11 since -2b and b are like terms, and -10 and -1 are like terms.

YOU TRY1. Simplify x + 3 + 2x + 10 2. Simplify 7x + 4x 3. Simplify 6p – 9p

DISTRIBUTIVE PROPERTYWhat is the Distributive Property?

a(b + c) = a ∙ b + a ∙ c

Examples of Applying the Distributive Property1) 5(9 + 9n) 5 ∙ 9 + 5 ∙ 9n 45 + 45n2) 9(-9m + 1) 9 ∙ -9m + 9 ∙ 1 -81m + 93) -7(6 – 7n) (-7) ∙ 6 – (-7) ∙ 7n -42 – (-49n) -42 + 49n

YOU TRY1. Simplify the

following expression:2(-7v – 1)

2. Simplify the following expression:

-10(n – 1)

SIMPLIFYING EXPRESSIONSExample 1 – Simplify the following expression

by using the Distributive Property and combining like terms:

-3(5x + 10) + 7(-3)∙ 5x + (-3) ∙ 10 + 7 distributive property

-15x + (-30) + 7 -15x + (-23) combine like terms

-15x - 23

MORE YOU TRY1. Simplify the

following expression by using the Distributive Property and combining like terms: -10(1 – x) + 8x

2. Simplify the following expression by using the Distributive Property and combining like terms:

2m + 9(6m + 10)

SOLVING LINEAR EQUATIONS

When we solve linear equations, we are trying to find the numerical value of the variable.

In trying to find the numerical value of the variable, we are isolating the variable by performing inverse operations.

Sometimes, this requires us to combine like terms, use the Distributive Property, or do both.

EXAMPLES OF SOLVING LINEAR EQUATIONS

Example 1 – Solve the following equation:

-5n + 3 + 5 = 13 -5n + 8 = 13 combine like terms

- 8 - 8 inverse operations

-5n = 5 ÷ -5 ÷ -5 inverse

operations

n = -1

MORE EXAMPLES OF SOLVING LINEAR EQUATIONS

Example 2 – Solve the following equation:

3(5a + 5) = 75 15a + 15 = 75 Distributive Property

- 15 - 15 Inverse operations

15a = 60÷ 15 ÷ 15 Inverse operations

a = 4

MORE EXAMPLES OF SOLVING LINEAR

EQUATIONSExample 3 – Solve the following linear

equation:

6x – 2 = 3x + 7 - 3x -3x Inverse operations to move variable to

left side

3x – 2 = 7 + 2 + 2 Inverse operations

3x = 9÷ 3 ÷ 3 Inverse operations

x = 3

MORE EXAMPLES OF SOLVING LINEAR EQUATIONS

Example 4 – Solve the following linear equation:

5 – 4(2 – 4a) = 6a + 27 5 + (-4)(2 – 4a) = 6a + 27 Change subtraction to “adding a negative”

5 + (-4) ∙ 2 – (-4) ∙ 4a = 6a + 27 Distributive Property

5 + (-8) – (-16a) = 6a + 27 5 + (-8) + 16a = 6a + 27 Change “double negative” to addition

-3 + 16a = 6a + 27 Combine like terms

16a – 3 = 6a + 27 Commutative Property to put variable first - 6a - 6a Inverse operations to move variable to left side

10a – 3 = 27 + 3 + 3 Inverse operations

10a = 30 ÷ 10 ÷ 10 Inverse operations

a = 3

YOU TRY1. Solve the following

equation:

-2 – 6(x + 2) = -5x -16

2. Solve the following equation:

6 – 2(3 – 3m) = -12 – 6m

SETTING UP AND SOLVING EQUATIONS

When we set up equations from word problems, we must first identify all of our variables (known and unknown).

Next we must translate the words in the problem to math symbols so that we can derive an equation.

Then we replace all known variables with their numerical value and solve for the unknown variable.

Finally, answer the question in a complete sentence.

EXAMPLE OF SETTING UP AND SOLVING EQUATIONS

Example 1 –

A skating rink charges $125 per hour and $6 per guest in order to rent the rink for a party. If we rented the rink for two hours and 200 people attended the party, then how much did the rink charge us?

Solution –

Let C be the total amount that the rink charges. Let h be the number of hours the rink is rented.Let p be the number of people who attend the party

C = 125h + 6pC = 125(2) + 6(200)C = 250 + 1200C = 1450

The rink charged $1450.

MORE EXAMPLES OF SETTING UP AND SOLVING EQUATIONS

Example 2 –

A large bookshelf holds 50 books, and a small bookshelf holds 20 books. How many books are there if 4 large bookshelves and 6 small bookshelves are filled?

Solution –

Let B be the total amount of books.Let l be the number of large bookshelvesLet s be the number of small bookshelves

B = 50l + 20sB = 50(4) + 20(6)B = 200 + 120B = 320

There are a total of 320 books.

ANOTHER EXAMPLE OF SETTING UP AND SOLVING

EQUATIONSExample 3 –

A hotel charges $80 per night and $10 per family member per night to rent a room. How much would a family of four be charged to rent a room for five nights?

Solution –

Let C be the total cost.Let n be the number of nights for which the room is rented.Let f be the number of family members

C = 80n + 10(f)(n)C = 80(5) + 10(4)(5)C = 400 + 200C = 600

The hotel charged $600 for a family of four to rent a room for five nights.

YOU TRYLebron James scored 27 points last night. He made six 2-point field goals and nine free throws. How many 3-point field goals did he make?

SOLVING “WORK” WORD PROBLEMS

When solving “work” word problems, we are trying to find out how long it takes two or more people to do a job together if we know how long it takes them to do the job individually.

Also, we can find out how long it takes a person to do a job individually if we know how long it takes another person to do the job individually and if we know how long it takes them to do the job together.

EXAMPLE OF SOLVING “WORK” WORD PROBLEMS

Example 1 –

Ijahanna takes 3 hours to do her homework. It takes Emmanuel 4 hours to do his homework. How long would it take them to do their homework if they worked together?

Solution –

Let t be the time it takes both of them to do their homework together.

the amount of work Ijahanna gets done in 1 hour

the amount of work Emmanuel gets done in 1 hour

+ = 1 “1” represents 1 hour(12) + (12) = (12)1 4t + 3t = 12 7t = 12 = t = 1.7

So, it would take Ijahanna and Emmanuel 1.7 hours to do their homework together.

ANOTHER EXAMPLE OF SOLVING “WORK” WORD

PROBLEMSExample 2 –

It takes Adam eight hours to paint a fence. Stephanie can paint the same fence in ten hours. If they worked together how long would it take them?

Solution –

Let t be the amount of time it takes them to paint the fence together.

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YOU TRYWorking alone, it

takes Asanji 14 minutes tosweep a porch. Julio can sweep the sameporch in 8 minutes. Find how long it would take them if they worked together.

Solution –

SOLVING “WORK” WORD PROBLEMS WHEN TRYING TO FIND AN INDIVIDUAL’S TIME

Example 1 –

John can pick forty bushels of apples in 14

hours. One day his friend Pranav helped

him and it only took 7.24 hours. Find how

long it would take Pranav to do it alone.

Solution –

Let x be the amount of time it takes Pranav to do the job alone.

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14.99

6.76 / 101.36

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ANOTHER EXAMPLE OF SOLVING “WORK” WORD PROBLEMS WHEN TRYING TO FIND

AN INDIVIDUAL’S TIME

Example 2 –

Working alone, Perry can tar a roof in nine hours. One day his friend Imani helped him

and it only took 4.24 hours. How long would it take Imani to do it alone?

Solution –

Let x be the time it takes Imani to tar a roof alone.

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8.02

4.76 / 38.16

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YOU TRYWorking together, Amanda and Willie can sweep a porch in 5.32 minutes. Had he done it alone it would have taken Willie 13

minutes. Find how long it would take

Amanda to do it alone.

Solution –

SOLVING DISTANCE-RATE-TIME WORD PROBLEMS

When solving distance-rate-time word problems, we must first know that distance = rate x time or d = r ∙ t

Another word for rate is speed. Also, we must know what we are trying to

find. Then we must identify what we know already

from the problem. Finally, we must set up and solve the

equation.

EXAMPLE OF SOLVING DISTANCE-RATE-TIME WORD

PROBLEM

Example 1 –

Joe left the hospital and traveled toward the recycling plant at an average speed of 27 km/h. Kristin left some time later traveling in the same direction at an average speed of 45 km/h. After traveling for three hours, Kristin caught up with Joe. How long did Joe travel before Kristin caught up?

Solution –

What are we trying to find?

We are trying to find out how many hours Joe traveled before Kristin caught up with him.

What do we already know?

We know Joe’s speed was 27 km/h.

We know Kristin’s speed was 45 km/h.

We know Kristin’s travel time was 3 hours.

We know they both traveled the same distance.

We know d = r ∙ t

How do we set up an equation?

Since they both traveled the same distance, Joe’s “rate x time” must equal Kristin’s “rate x time”

Since we don’t know Joe’s travel time, we’ll let it be t.

27t = 45(3)

27t = 135

t = 5

So, Joe traveled for 5 hours before Kristin caught up with him.

WE TRYLea left the airport and drove toward the lake at an average speed of 30 mph. Nadia left some time later driving in the same direction at an average speed of 75 mph. After driving for two hours Nadia caught up with Lea. How long did Lea drive before Nadia caught up?

Solution –

YOU TRYKayla left the White House and drove toward Georgetown at an average speed of 20 mph. Jennifer left some time later driving in the same direction at an average speed of 25 mph. After driving for four hours, Jennifer caught up with Kayla. Find the number of hours Kayla drove before Jennifer caught up with her.

Solution -

ANOTHER EXAMPLE OF SOLVING DISTANCE-RATE-TIME WORD

PROBLEM

Example 2 –

Gabriella left James’ house and drove toward Uptown. Three hours later, Shayna left driving at 70 mph in an effort to catch up to Gabriella. After driving for two hours, Shayna finally caught up. What was Gabriella’s average speed?

Solution –

What are we trying to find?We are trying to find Gabriella’s average speed or her rate.

What do we already know?We know Shayna’s speed was 70 mph.We know Shayna drove for 2 hours.We know Gabriella drove for 5 hours.We know that d = r ∙ tWe know that they went the same distance since Shayna caught up with Gabriella.

How do we set up an equation?Since they both traveled the same distance, Gabriella’s “rate x time” must equal Shayna’s “rate x time.”Since we don’t know Gabriella’s rate, we will let it be r.

r ∙ 5 = 70(2)5r = 140 r = 28

So, Gabriella drove at an average speed of 28 mph.

WE TRYA passenger plane flew to Jakarta and back. The trip there took seven hours, and the trip back took five hours. The plane averaged 490 mph on the return trip. Find the average speed of the trip there.

Solution –

YOU TRYAn aircraft carrier made a trip to Tahiti and back. The trip there took ten hours and the trip back took 13 hours. The aircraft carrier averaged 20 km/h on the return trip. Find the average speed of the trip there.

Solution –

ANOTHER EXAMPLE OF SOLVING DISTANCE-RATE-TIME

WORD PROBLEM

Example 3 –

Allan left school two hours before Tony. They drove in opposite directions. Tony drove at 25 mph for one hour. After this time, they were 160 miles apart. What was Allan’s speed?

SOLUTION TO PREVIOUS EXAMPLE

What are we trying to find?We are trying to find Allan’s speed or rate.

What do we already know?We know that Tony and Allan drove in opposite directions from school.We know Tony drove for 1 hour.We know Allan drove for 3 hours, since he left 2 hours before Tony.We know Tony’s speed was 25 mph.We know that after 3 hours of driving for Allan and 1 hour of driving for Tony, they were 160 miles apart.

How do we solve?First, we find the distance that Tony drove by multiplying his rate by his time.d = r ∙ td = 25(1)d = 25So, Tony drove 25 miles.

Next, we find the distance Allan drove by subtracting the distance Tony drove from the distance they were apart.160 – 25 = 135

Finally, we find Allan’s rate by using the d = r ∙ t formula.d = r ∙ t 135 = r ∙ 3135 = 3rr = 35So, Allan drove at a speed of 45 mph.

WE TRYA fishing boat left Port 52 and traveled toward Madagascar at an average speed of 10 mph. An aircraft carrier left some time later traveling in the opposite direction with an average speed of 20 mph. After the fishing boat had traveled for four hours, the ships were 120 miles apart. How long did the aircraft carrier travel?

Solution –

YOU TRYRyan left Julio’s house at the same time as Mark. They traveled in opposite directions. Mark traveled at a speed of 65 mph. After one hour, they were 110 miles apart. How fast did Ryan travel?

Solution –

FUNCTIONS A function is how

something works. A function in math is a

relation between a set of inputs and a set of outputs.

The output of a function f corresponding to an input x is denoted by f(x) (read as “f” of “x”).

For example, in the function f(x) = x² the output will be the square of the input. So, f(4) = 4² = 16. Four is the input and 16 is the output.

EVALUATING FUNCTIONSExample 1 –

f(n) = n + 3Find f(0)

Solution –

f(0) = 0 + 3 = 3

Example 2 –

k(a) = a² + 2aFind k(-7)

Solution –

k(-7) = (-7)² + 2(-7) = 49 + (-14) = 35

YOU TRYg(n) = 3n – 4 Find g(9)

f(x) = x² − 4Find f(-10)

PROPERTIES OF EXPONENTS Exponents consist of a base and a power.

For example: 5² 5 is the base and 2 is the power (read as five to the second power).

When we multiply exponents and the bases are the same, then we add the powers (xa ∙ xb = xa+b).

For example: 5² ∙ 5³ = 52+3 = 55

When we raise an exponent to another power, then we multiply the powers ( (xa)b = xa∙b ).

For example: (5²)³ = 52∙3 = 56

When we divide exponents and the bases are the same, then we subtract the powers ( = xa – b ).

For example: = 5-1

MORE PROPERTIES OF EXPONENTS

When a base is raised to a negative power, then it is the same as one over that same base raised to the opposite power (x-a = ).

For example: 5-2 = Any base (except 0) raised to the 0 power is

1(x0 = 1 when x ≠ 0).

For example: 50 = 1 When a number or variable does not show a

power, assume that the power is 1 (x = x1 ).For example: 5 = 51

YOU TRY1. 4² ∙ 4² = 2. 3² ∙ 34 =3. x4 ∙ x = 4. (3²)³ = 5. (3b)4 = 6. (x³)4 = 7. = 8. =

POLYNOMIALSWhat are polynomials?

Polynomials are algebraic expressions defined by their degree and number of terms.

What is the degree of a polynomial?The degree of a polynomial is the highest power of the variable in the polynomial.

What are terms?Terms are parts of an expression separated by a “+” or “-” sign.

EXAMPLES OF POLYNOMIALS

Example 1 –2x² - 3x + 4 is a polynomial2x², -3x, and 4 are the termsSo, this polynomial has 3 terms

Example 2 – x – 9 is a polynomialx and -9 are the termsSo, this polynomial has 2 terms

DEGREES OF POLYNOMIALS

What is the degree of a polynomial?The degree of a polynomial is the highest power of the variable in the polynomial.

*If there is no variable in the polynomial, then the degree is 0.

Example 1 – 2x² - 3x + 4 is a polynomial x² is the exponent with the highest power in the polynomialSo, the degree of the polynomial is 2

MORE EXAMPLES OF DEGREES OF

POLYNOMIALS

Example 3 – 4x³ - 2x² + 5x – 7 is a polynomialx³ is the exponent with highest powerSo, the degree of the polynomial is 3

Example 4 – 10 is a polynomialIt has no variable, so its degree is 0

NAMING POLYNOMIALS

How are polynomials named?Polynomials are named according to their degree and number of terms.

POLYNOMIAL NAMING CHART

Degree Terms

0 Constant n/a

1 Linear Monomial

2 Quadratic Binomial

3 Cubic Trinomial

4 Quartic Polynomial w/ 4 terms

5 Quintic Polynomial w/ 5 terms

EXAMPLES OF NAMING POLYNOMIALS

Example 1 – 2x² - 3x + 4 has a degree of 2 and has 3 termsSo, it is a quadratic trinomial

Example 2 – x – 9 has a degree of 1 and has 2 termsSo, it is a linear binomial

MORE EXAMPLESExample 3 –

4x³ - 2x² + 5x – 7 has a degree of 3 and has 4 termsSo, it is a cubic polynomial with 4 terms

Example 4 – 10 has a degree of 0 and has 1 termSo, it is a constant monomial or often referred to as just a “constant”

YOU TRYName the following polynomials:

1. 8x³ - 272. 2x² - 7x + 103. 5x4. -3x⁴ - 2x² + 1

ADDING AND SUBTRACTING POLYNOMIALS

When we add/subtract polynomials, we can only add/subtract like terms.

*Remember that like terms are terms that have the same variable and power. Align all like terms and

then add or subtract. Make sure final answer

is arranged from highest power to lowest

Example 1 –

4r4 + 3r³ + (r³ - 5r4)

Solution –

4r4 + 3r³+ (-5r4) r³ -1r4 + 4r³

MORE EXAMPLES OF ADDING/SUBTRACTING POLYNOMIALS

Example 2 –

(5 + 5n²) – (5 – 3n²)

Solution –

5 + 5n²− 5 (-3n²) 8n²

Example 3 –

(5x2 – 3x – x4) + (x4 - 5x3 + 5x2)

Solution –

5x2 – 3x – x4

+ 5x2 x4 (-5x3) 10x2 – 3x -5x3

-5x3 + 10x2 – 3x

YOU TRYSimplify the following expression.

(4x + x³ − 3) + (4 – 5x – 5x³)

Simplify the following expression.

(4r² − 3r – 3r4 – 3r³) – (4r³ − 3r² − r4 + 3r)

MULTIPLYING POLYNOMIALS / HOW TO USE AN EXPANSION BOX

Monomial x BinomialExample 1 –

a(b + c)1. Put the first factor on the side and the other factor on the top2. Add what is in the boxes. Combine all like terms if possible.So, a(b + c) = ab + ac

b +c

a a ∙ b a ∙ c

MORE EXAMPLES OF USING AN EXPANSION BOX

Binomial x BinomialExample 2 –

(a + b)(c + d) 1. Put first factor on side and other factor on top.2. Add the boxes and combine like terms if possible.(a + b)(c + d) = ac + ad + bc + bd

c +d

a a ∙ c a ∙ d

+b b ∙ c b ∙ d

REAL EXAMPLESExample 3 –

7(x + 2) = 7x +

14

x +2

7 7x 14

MORE REAL EXAMPLESExample 4 –

(x + 6)(x – 4)

x² -4x + 6x – 24 =

x² + 2x – 24

x -4

x x² -4x

+6 6x -24

YOU TRY1. 2x(3x + 7)2. (x – 5)(4x + 2)3. 3x²(2x² - 3x + 4)

QUADRATIC EXPRESSIONS

What is a quadratic expression?A quadratic expression is a single variable, degree 2 polynomial

Examples of quadratic expressions:1. x² + 10x + 212. x² - 813. 4x² - 254. x² - 7x5. -6x² - 22x -20

SIMPLE QUADRATIC EXPRESSIONS

What are simple quadratic expressions?Simple quadratic expressions are in

the form x² + bx + c

How do we factor simple quadratic expressions?

1. Find the pair of numbers that multiply to produce c and add up to b

2. Place those two numbers in the following binomial factors:

(x + __)(x + __)

EXAMPLES OF FACTORING SIMPLE QUADRATICS

Example 1 –

Factor x² + 10x + 21

1. Find the pair of factors of 21 that add up to 10:

1, 21 1 + 21 = 22 NO-1, -21 -1 + -21 = -22 NO3, 7 3 + 7 = 10 YES-3, -7 -3 + -7 = -10 NO

2. So, x² + 10x + 21 = (x + 3)(x + 7)

MORE EXAMPLES OF FACTORING QUADRATICS

Example 2 –

Factor x² + 7x – 30

1. Find the pair of factors of -30 that add up to 7:

1, -30 1 + -30 = -29 NO-1, 30 -1 + 30 = 29 NO5, -6 5 + -6 = -1 NO-5, 6 -5 + 6 = 1 NO 3, -10 3 + -10 = -7 NO-3, 10 -3 + 10 = 7 YES

2. So, x² + 7x – 30 = (x + -3)(x + 10)

MORE EXAMPLES OF FACTORING

QUADRATICSExample 3 –

Factor x² − 14x + 48

1. Find the pair of factors of 48 that add up to -14:1,48 1 + 48 = 49 NO-1,-48 -1 + -48 = -49 NO2, 24 2 + 24 = 26 NO-2,-24 -2 + -24 = -26 NO3, 16 3 + 16 = 19 NO-3,-16 -3 + -16 = -19 NO4, 12 4 + 12 = 16 NO-4,-12 -4 + -12 = -16 NO6, 8 6 + 8 = 14 NO-6,-8 -6 + -8 = -14 YES

2. So, x² - 14x + 48 = (x + -6)(x + -8)

YOU TRY1. x² + 5x – 242. x² − 8x – 333. x² + 15x + 50

FACTOR OTHER QUADRATIC EXPRESSIONS

What are perfect square trinomials?Perfect square trinomials are produced by

one binomial multiplied by itself.

In x² + bx + c, c is the square of a number and b is 2 times that number.

Example 1 –

x² + 16x + 64 = (x + 8)(x + 8) = (x + 8)²since 64 is 8² and 16 is 2 ∙ 8

MORE EXAMPLES OF PERFECT SQUARE

TRINOMIALS

Example 2 – x² - 6x + 9 = (x + -3)(x + -3) = (x + -3)² since 9 is (-3)² and -6 is 2 ∙ -3

Example 3 – x² + 12x + 36 =

(x + 6)(x + 6) = (x + 6)²

Example 4 – x² -20x + 100 =

(x + -10)(x + -10) = (x + -10)²

YOU TRYFactor the following quadratic expressions:1. x² - 24x + 1442. x² + 10x + 25

DIFFERENCE OF SQUARES

What is difference of squares?Difference of squares is a binomial in which one square is being

subtracted from another.

Factors of difference of squares are two binomials where one is the sum of the square roots of the squares and the other is the

difference of the square roots of the squares.

EXAMPLES OF DIFFERENCE OF SQUARES

Example 1 – x² − 81 = (x + 9)(x – 9)since x² and 81 are perfect squares

Example 2 – x² − 100 = (x + 10)(x – 10)since x² and 100 are perfect squares

Example 3 – 4x² − 49 = (2x + 7)(2x – 7) since 4x² and 49 are perfect squares

YOU TRYFactor the following quadratics:1. x² − 252. 9x² − 36

SOLVING SIMPLE QUADRATIC EQUATIONS

When we solve quadratic equations, we must first make sure that the right side of the equation is 0 and then factor the quadratic expression on the left side of the equation.

Next, we set both factors equal to 0 and then we solve for the variable in both equations.

Example 1 –

Solve the following equation:

x² − 2x – 48 = 0

Solution –

(x – 8)(x + 6) = 0x – 8 = 0 or x + 6 = 0So, x = 8 or x = -6

MORE EXAMPLES OF SOLVING SIMPLE

QUADRATIC EQUATIONSExample 2 –

Solve the following equation:

x² + 10x = -16

Solution –

x² + 10x + 16 = 0 add 16 to both sides to put 0 on right side

(x + 8)(x + 2) = 0 factor left side

x + 8 = 0 or x + 2 = 0 set both factors equal to 0 and solve for x

So, x = -8 or x = -2

MORE EXAMPLES OF SOLVING SIMPLE QUADRATIC EQUATIONS

Example 3 –

Solve the following equation:

x² − x = 20

Solution –

x² − x – 20 = 0 subtract 20 from both sides to make right side equal 0

(x – 5)(x + 4) = 0 factor quadratic on left

x – 5 = 0 or x + 4 = 0 set both factors equal to 0 and then solve for x

So, x = 5 or x = -4

YOU TRYSolve the following equation:

x² − 6x – 40 = 0

Solve the following equation:

x² + 4x = 5