lesson 8 – solving quadratic equationslesson 8 – solving quadratic equations - minilesson 327...

Lesson 8 – Solving Quadratic Equations

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We will continue our work with quadratic equations in this lesson and will learn the classic method to solve them…the Quadratic Formula.

We begin by viewing the Quadratic Formula then using it to solve a quadratic equation. Deriving the Quadratic Formula from the standard form quadratic equation follows this.

The remainder of the lesson focuses on using the Quadratic Formula to solve equations, introducing the idea of solutions outside the real number system, then combining all our solution methods together.

Lesson Topics

• Section 8.1: A Case for the Quadratic Formula

• Section 8.2: Solving Quadratic Equations with the Quadratic Formula

• Section 8.3: Complex Numbers

• Section 8.4: Complex Solutions to Quadratic Equations

• Section 8.5: Combining Solution Methods to Solve Any Quadratic Equation


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Lesson 8 – Solving Quadratic Equations - MiniLesson

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Lesson 8 - MiniLesson

Section 8.1 – A Case for the Quadratic Formula

Number and Type of Solutions to a Quadratic Equation

Remember that there are three possible cases for number of solutions to a quadratic equation in standard form as shown below. Graphing just the quadratic part of the equation in each case (i.e. the left side) would give you the accompanying graph. You should always begin your work with any quadratic equation (or function for that matter) by graphing.

CASE 1: One, repeated, real number solution

The parabola touches the x-axis once at the vertex.

Example: !!x

2 − 4x + 4 = 0 Solutions: !!x1 =2, x2 =2

CASE 2: Two unique, real number solutions

The parabola crosses the x-axis at

two unique locations.

Example: !!3x2+ x − 2 =0

Solutions: !!x1 = −1,!!x2 =

23 ≈0.667

CASE 3: No real number solutions but two unique, complex number solutions

The parabola does NOT cross the x-axis.

Example: !!x

2 + x + 1 =0

Solutions: !!x1 = −

12+

32 i , x2 = −

12−

32 i


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How could we determine the solutions for each of the examples listed on the previous page? Case 1: The solutions for the example equation in Case 1 are “nice” numbers (i.e. integers). Factoring as follows can solve the example equation:

!!

x2 − 4x + 4 = 0x −2( ) x −2( ) =0x −2=0

x =2!repeated!solutionx1 =2 and x2 =2

Case 2: One of the solutions in Case 2 is not a “nice” number meaning that factoring is not the preferred method. We could use the Graphing/Intersection Method, which would work, but is there an algebraic method that can be used? Case 3: These solutions do not look like anything we have seen to this point and are definitely not “nice” numbers. Graphing/intersection will not help us since the graph does not cross the x-axis. We will need some other kind of method to solve this kind of equation. The method we will learn that would solve the example equations above is the Quadratic Formula. The Quadratic Formula can be used to solve any quadratic equations written in standard form:

!!x = −b!±! b2!− !4ac

2a

To solve a quadratic equation using the Quadratic Formula:

Step 1: Make sure the quadratic equation is in standard form: !!ax2 + !bx !+ !c = !0 and write down the coefficients a, b and c.

Step 2: Graph the quadratic part of the equation and see how many times the graph crosses the x-axis. Identify the number and type of solutions. If the graph crosses the x-axis, enter 0 as Y2 then find the intersection point(s) and round to the appropriate number of decimal places. Write the x-values down. These will be your approximate solutions.

Step 3: Substitute the coefficient values into the Quadratic Formula Step 4: Simplify your result completely leaving in the form indicated by the

directions.


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Section 8.2 – Solving Quadratic Equations with the Quadratic Formula Problem 1 WORKED EXAMPLE–

Solve Quadratic Equations Using the Quadratic Formula Solve the equation !!3x

2 + x − 2 =0 using the steps as listed on page 318. Leave your solution(s) in exact form and in approximate form rounded to the thousandths place. Step 1: The given equation is in standard form with a = 3, b = 1 and c = !−2 Step 2: Enter !!Y1=3x

2 + x − 2on the graphing calculator and graph on a standard window obtaining the graph below. This graph crosses the x-axis in 2 places meaning there are two unique real number solutions to the original equation. Enter !!Y2=0 and use the Graphing/Intersection Method to obtain the approximate solutions for x as follows:

!!x1 = −1, x2 ≈0.667

Step 3: !!!x = −(1)!±! (1)2!−!4(3)(−2)

2(3) =−1!±! 1!−!(−24)

6 = −1!±! 1+246 = −1!±! 25

6 Step 4: Break up the final fraction into two parts and make computations for x1 and x2 as below noting the complete simplification process:

!!x1 =

−1!− 256 = −1−5

6 = −66 = −1 !!x2 =

−1!+ 256 = −1+5

6 = 46 = 23 ≈0.667

Our final solutions in the requested forms are:

Exact Form Approximate Form to the Thousandths Place

!!x1 = −1 and!!!!x2 =

23 !!x1 = −1, x2 ≈0.667

• !!x1 = −1 is exact to begin with so we use the same form for the approximation.

• We use the symbol ≈ to indicate the approximation for !!x2 . Note this notation is

not needed for !!x1 since exact and approximate forms are equal.


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How to Derive the Quadratic Formula

The Quadratic Formula can be derived directly from a quadratic equation in standard form using a process called completing the square. This process is shown below.

ax2 + bx + c = 0

Start with a quadratic equation in standard form.

!!x2 + bx = − c

Subtract c from both sides.

acx

abx −=+2

Divide both sides by a.

2

2

2

22

44 ab

ac

abx

abx +−=++

Divide the coefficient of x by 2 to get !!b 2a , square it to get !!b

2 4a2 , then add to both sides

2

2

2

2

444

2 ab

aac

abx +−=⎟⎠⎞⎜

⎝⎛ +

Factor the left side. On the right side, get a common denominator of 4a2

!!x + b

2a⎛⎝⎜

⎞⎠⎟

2

= b2 − 4ac4a2

Combine the right side to one fraction, reorder the terms in the numerator, then take square root of both sides

!!x + b

2a = ±b2 − 4ac2a

Simplify the square roots

!!x = − b

2a ±b2 − 4ac2a

Solve for x by subtracting !!b 2a from both sides.

!!x = − b ± b2 − 4ac

2a

Combine over the common denominator to obtain the final form for the Quadratic Formula

Let’s look at more examples using the Quadratic Formula to solve equations.


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Problem 2 MEDIA EXAMPLE– Solve Quadratic Equations Using the Quadratic Formula

Solve the equation !!x

2 − 3 = 3x using the Quadratic Formula using the steps as listed on page 318. Leave your solution(s) in exact form and in approximate form rounded to the thousandths place.


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Step 4 in the previous problems requires skills in simplifying expressions involving radicals. The topic of radicals and radical expressions will be discussed more in a future lesson, but knowing some of the information related to square roots now will help you work through the Quadratic Formula problems in this lesson. Problem 3 WORKED EXAMPLE – Simplifying Square Roots Simplifying Square Roots of Perfect Squares

!

0 =0 1= 1 4 = 2 9 = 3 16 = 4 25 = 536 =6 49= 7 64 = 8 81 = 9 100 = 10121 = 11 144 = 12 169 = 13 196 = 14

Problem 4 MEDIA EXAMPLE – Simplifying Square Roots Simplify each of the following as much as possible. Leave answers in exact form. Simplifying Square Roots of Composite Numbers: a) ! 50 = 18 = 24 = Simplifying Square Roots of Prime Numbers: b) ! 3 = 47 = Simplifying Square Roots with Fractions (I): d)

!502 = 18

6 = 244 =

Simplifying Square Roots with Fractions (II): e)

!2+ 50

2 = 2− 186 =

f)

!8 + 24

4 = −4+ 1612 =


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Problem 5 YOU TRY – Simplifying Square Roots Simplify each of the following as much as possible. Leave final answers in exact form. a) ! 20 =

d)

!3005 =

b) ! 19 =

e)

!4− 84 =

c)

!287 =

f)

!6+ 276 =

Exact vs. Approximate Form

You have seen language asking you to leave answers in either exact form and/or approximated to a given decimal. The number ! 3 is in exact form. If ! 3 is rounded to three decimal places and written as 1.732, then it is in approximate form to the thousandths place. Numbers that don’t need to be rounded to represent them fully (for example the number 12 or the number 0.5) are in exact form.


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Problem 6 MEDIA EXAMPLE – Solve Quadratic Equations Using Quadratic Formula

Solve each quadratic equation by using the Quadratic Formula and the steps illustrated earlier in this lesson. Leave your solution(s) in exact form and in approximate form rounded to two decimals places.

Quadratic Formula: x = !!−b!±! b2!− !4ac

2a

a) Solve –x2 +3x + 10 = 0

b) Solve 2x2 – 4x = 3


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Problem 7 YOU TRY – Solve Quadratic Equations Using Quadratic Formula Solve each quadratic equation by using the Quadratic Formula and the steps illustrated earlier in this lesson. Leave your solution(s) in exact form and in approximate form rounded to the thousandths place. Label solution(s) on your graph.

Quadratic Formula: x = !!−b!±! b2!− !4ac

2a

a) !!x2 − 2x − 5=0

b) !!3x2 = 7x + 20


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Section 8.3 – Complex Numbers Suppose we are asked to solve the quadratic equation x2 = –1. Any real number times itself always gives a positive result. Therefore, there is no real number x such that x2 = –1. Approach this using the Quadratic Formula and see what happens. To solve x2 = –1 using the Quadratic Formula, first write the equation in standard form as:

x2 + 1 = 0. Then, identify the coefficients a, b and c to get: a = 1, b = 0 and c = 1. Substitute these into the Quadratic Formula to get:

!!x = −0 ± 02 − 4(1)(1)

2(1) = ± −42 =

± 4(−1)2 = ± 4 −1

2 = ±2 −12 = ± −1

Again, the number + 1− does not live in the real number system nor does the number 1−− , yet these are the two solutions to the equation x2 + 1 = 0.

The way mathematicians have handled this problem is to define a number system that is an extension of the real number system. This system is called the Complex Number System and it allows equations such as x2 + 1 = 0 to be solved in this system. To do so, a special definition is used and that is the definition that:

i = 1−

With this definition, then, the solutions to x2 + 1 = 0 are just x = i and x = –i which is a lot simpler than the notation with negative numbers under the radical.


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You can see in the image below how the Complex Number System interacts with the other types of numbers we are familiar with.

The Complex Number System

Each set in the image below is a subset of the larger set in which it is contained. Complex numbers contain all the sets of numbers. Natural numbers only contain themselves.

!!

! =Complex!NumbersAll!numbers!of!the!form!a+bi !where!a, b!arereal!numbers,!i = −1, i2 = −1Examples:!3+4i , 2+(−3)i , 0+2i , 3+0i

!

! =Real!NumbersAll!numbers!on!the!real!number!line.Includes!Rational!and!Irrational!Numbers.

Examples:!−4, 13 ,π , 17 , 0.25

!

!=Rational!NumbersRatios!of!integers!with!decimals!that!terminateor!repeat

Examples:!12 , −34 , 0.43

!

!= IntegersIncludes!0,!the!natural!numbers!and!their!negatives.{...−3, −2, −1, 0,1, 2, 3,...}

!

! =Natural!NumbersCounting!Numbers{...1,!2,!3,!4,!5,!...}

When Will We See Complex Solutions? We will see solutions that involve the complex number “i” when we solve quadratic equations whose graphs never cross the x-axis.

Standard Form for a Complex Number

Standard form for a complex number is:

!a + bi

where a and b are real numbers


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Problem 8 WORKED EXAMPLE – Writing Complex Numbers in Exact Form Simplify each of the following leaving your final result in exact a + bi form.

a)

!!

−9 = 9 −1=3 −1=3i

b)

!!− −7 = − 7 −1

= − 7i !

c)

i

i

27

232732

14932493

+=

+=

−+=−+

Problem 9 MEDIA EXAMPLE – Writing Complex Numbers in Exact Form Simplify each of the following leaving your final result in exact a + bi form.

a) !−4 −25 = b) ! 32 = c)

!5−2 −25

10 =

Problem 10 YOU TRY – Writing Complex Numbers in Exact Form Simplify each of the following leaving your final result in exact a + bi form.

a) ! −121 = b) !−4 −3 = c)

!6+ −9

3 =


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Problem 11 WORKED EXAMPLE – Writing Complex Numbers in Approximate Form

Each of the following complex numbers is written in exact a + bi form. Change each of them to an approximate a+bi form rounded to the nearest thousandths place, if possible.

a) !!1− 7i ≈1−2.646i b) !!32+

72 i =1.5+3.5i

Problem 12 MEDIA EXAMPLE – Writing Complex Numbers in Approximate Form

Each of the following complex numbers is written in exact a + bi form. Change each of them to an approximate a+bi form rounded to the nearest thousandths place, if possible.

a) !!3+4 15i = b)

!!37 + −17

⎛⎝⎜

⎞⎠⎟i =

Problem 13 YOU TRY – Writing Complex Numbers in Approximate Form Each of the following complex numbers is written in exact a + bi form. Change each of them to an approximate a+bi form rounded to the nearest thousandths place, if possible.

a) !!1− 11i = b)

!!27 +

37 i


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Section 8.4 – Complex Solutions to Quadratic Equations Work through the following to see how to deal with equations that can only be solved in the Complex Number System. Problem 14 WORKED EXAMPLE –

Solving Quadratic Equations with Complex Solutions Solve 2x2 + x + 1= 0 for x. Leave results in the form of a complex number, a+bi. Include both exact form and approximate form rounded to the thousandths place. The graph below shows that !!y ! = !2x

2 + !x !+ !1! does not cross the x-axis at all. This is an example of our Case 3 possibility and will result in no Real Number solutions but two unique Complex Number Solutions.

To find the solutions, make sure the equation is in standard form (check). Identify the coefficients a = 2, b = 1 and c = 1. Insert these into the Quadratic Formula and simplify as follows:

471

4811

)2(2)1)(2(411 2 −±−=−±−=

−±−=x

Break this into two solutions and use the a+bi form to get

i

i

x

47

41

47

41

47

414

711

+−=

+−=

−+−=

−+−=

and

i

i

x

47

41

47

41

47

414

712

−−=

−−=

−−−=

−−−=

The final exact solutions are ixix

47

41,

47

41

21 −−=+−=

The final approximate solutions are !!x1 ≈ −0.250 + 0.661i ,!!x2 ≈ −0.250 − 0.661i


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Problem 15 MEDIA EXAMPLE – Solving Quadratic Equations with Complex Solutions

Solve x2 + 4x + 8 = 1 for x. Leave results in the form of a complex number, a+bi.


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Problem 16 YOU TRY – Solving Quadratic Equations with Complex Solutions Solve !!2x

2 − 3x = −5 for x. Leave results in the form of a complex number, a+bi.


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Section 8.5 – Combining Solution Methods to Solve any Quadratic Equation We have learned several methods to solve quadratic equations. Which one should you use when? The pros and cons of each method are presented below and the problems in this lesson will provide practice for all solution methods.

Solving Quadratic Equations by Graphing Pros

• Good starting point for all quadratic equations as graphing will provide information about the number and types of solutions

• Not necessary to put the equation in standard form to use this method

Cons • Non-integer solution values will be

approximate, not exact • Window adjustments are often

necessary in order to view necessary parts of the graph

Solving Quadratic Equations by Factoring Pros

• Can be the fastest solution path if you quickly recognize how to factor the quadratic part of your equation

Cons • Equation must be in standard form

to apply this method • Does not work well for non-integer

solutions and large numbers • Can be time consuming if there are

many choices for the Trial and Error Method

Solving Quadratic Equations by Quadratic Formula Pros

• Can be used to solve any quadratic equation.

• Best way to find exact form solutions.

Cons • Equation must be in standard form

to apply this method • Must remember the formula. • Must pay close attention to details

in order to avoid computational errors


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Problem 17 WORKED EXAMPLE – Solving Quadratic Equations Given the quadratic equation x2 – 2x – 3 = 0, solve using the methods indicated below leaving all solutions in exact form. If solutions are complex, leave them in the form a + bi. a) Solve by graphing (if possible). Sketch the graph on a good viewing window (the

vertex, vertical intercept and any horizontal intercepts should appear on the screen). Mark and label the solutions on your graph.

b) Solve by factoring (if possible). Show all steps. Clearly identify your final solutions.

(x+1)(x-3)=0 x+1 = 0 or x – 3 = 0

x = -1 or x = 3 c) Solve using the Quadratic Formula. Clearly identify your final solutions.

!!x =

− −2( )± −2( )2 −4 1( ) −3( )2 1( ) = 2± 4+12

2 = 2± 162 = 2±42

Break up the final fraction into two parts and make computations for x1 and x2 as below:

!!x1 =

2!−42 = −2

2 = −1 !!

x2 =2+42 = 62 =3

Our final solutions in the requested forms are:

Exact Form

!!x1 = −1 and!!!!x2 =3


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Problem 18 MEDIA EXAMPLE – Solving Quadratic Equations Given the quadratic equation!!2x

2 − x = 4 , solve using the methods indicated below, leaving all solutions in exact form and in approximate form rounded to the thousandths place. If solutions are complex, leave them in the form a + bi. a) Solve by graphing (if possible). Sketch the graph on a good viewing window (the

vertex, vertical intercept and any horizontal intercepts should appear on the screen). Mark and label the solutions on your graph.

b) Solve by factoring (if possible). Show all steps. c) Solve using the Quadratic Formula. Show all steps.


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Problem 19 YOU TRY – SOLVING QUADRATIC EQUATIONS Given the quadratic equation !!x

2 + 3x − 7=3 , solve using the methods indicated below leaving all solutions in exact form. If solutions are complex, leave them in the form a + bi. Clearly identify your solutions in all cases. a) Solve by graphing (if possible). Sketch the graph on a good viewing window (the vertex, vertical intercept, and any horizontal intercepts should appear on the screen). Mark and label the solutions on your graph.

b) Solve by factoring (if possible). Show all steps. c) Solve using the Quadratic Formula. Show all steps.


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Problem 20 MEDIA EXAMPLE – Solving Quadratic Functions Given the quadratic equations below, solve for x using an appropriate method of your choice leaving all solutions in exact form and approximate form to the thousandths place. Clearly identify each method that you attempt. If solutions are complex, leave them in the form a + bi.

a) !!x2 −3x −10=0

b) !!x2 − 6x = −10


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Problem 21 YOU TRY – SOLVING QUADRATIC EQUATIONS Given the quadratic equations below, solve for x using an appropriate method of your choice leaving all solutions in exact form and approximate form to the thousandths place. Clearly identify each method that you attempt. If solutions are complex, leave them in the form a + bi. a) !!2x2 +5x −3=0

b) !!x2 +3x = −5

lesson 8 – solving quadratic equationslesson 8 – solving quadratic equations - minilesson 327...

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