Quadratic Techniques to Solve Polynomial Equations

CCSS: F.IF.4 ; A.APR.3

CCSS: F.IF.4

• For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums.

CCSS: A.APR.3

• Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

Standards for Mathematical Practice

• 1. Make sense of problems and persevere in solving them.

• 2. Reason abstractly and quantitatively.

• 3. Construct viable arguments and critique the reasoning of others.  

• 4. Model with mathematics.

• 5. Use appropriate tools strategically.

• 6. Attend to precision.

• 7. Look for and make use of structure.

• 8. Look for and express regularity in repeated reasoning.

Essential Question:

• How do we use quadratic techniques solving equations?

Objectives

• Solve third and fourth degree equations that contain quadratic factors, and

• Solve other non-quadratic equations that can be written in quadratic form.

Intro

• Some equations are not quadratic but can be written in a form that resembles a quadratic equation. For example, the equation x4 – 20x2 + 64 = 0 can be written as (x2)2 – 20x2 + 64 = 0. Equations that can be written this way are said to be equations in quadratic form.

Key Concept:

• An expression that is quadratic form can be written as:

• au² +bu + c for any numbers a, b, and c, a≠0, where u is some expression in x.

• The expression au² +bu + c is called quadratic form of the original expression.

Once an equation is written in quadratic form, it can be solved by the methods you have already learned to use for solving quadratic equations.

Ex. 1: Solve x4 – 13x2 + 36 = 0

03613 24 xx

0)4)(9( 22 xx

036)(13)( 222 xx

2

02

x

x

0)2)(2)(3)(3( xxxx

2

02

x

x

3

03

x

x

3

03

x

x

The solutions or roots are -3, 3, -2, and 2.

The graph of x4 – 13x2 + 36 = 0 looks like:

The graph of y = x4 – 13x2 + 36 crosses the x-axis 4 times. There will be 4 real solutions.

• Recall that (am)n = amn for any positive number a and any rational numbers n and m. This property of exponents that you learned in chapter 5 is often used when solving equations.

Ex. 2: Solve 086 4

1

2

1

xx

0)4)(2(

08)(6)(

086

4

1

4

1

4

124

1

4

1

2

1

xx

xx

xx

256

4)(

4

0)4(

444

1

4

1

4

1

x

x

x

x

16

)2()(

2

02

444

1

4

1

4

1

x

x

x

x

00

01212

?08124

?08)2(64

08)16(616

086

4

1

2

1

4

1

2

1

xx

check

00

?02424

?08)4(616

08)256(6256

086

4

1

2

1

4

1

2

1

xx

check

Ex. 3: Solve 163

2

t

644

)4(

416

)16()(

16

3

2

16

632

333

2

3

2

ort

t

ort

t

t

1616

?164

?16)64(

?1664

16

2

23

3

2

3

2

t

check

1616

?16)4(

?16)64(

?16)64(

16

2

23

3

2

3

2

t

check

Ex. 4: Solve 087 xx

12

2

2

97

82

16

2

972

817

)1(2

)8)(1(4)7(7(

2

4

08)(7)(

087

2

2

2

x

a

acbbx

xx

xx

64

8)(

822

x

x

x

Ex. 4: Solve 087 xx

00

085664

08)8(764

0864764

087

xx

check

There is no real number x such that is = -1.Since principal root of a number can not be negative, -1 is not a solution. The only solution would be 64.

1x

• Some cubic equations can be solved using the quadratic formula. First a binomial factor must be found.

Ex. 5: Solve 0273 x

3

03

)93)(3(

0272

3

x

x

xxx

x

2

333

2

273

2

3693

)1(2

)9)(1(433

2

4

093

2

2

2

i

a

acbb

xx

Quadratic form

ax2 + bx + c = 0 2x2 – 3x – 5 = 0

This also would be a quadratic form of an equation

0532

0532

0532

3

1

3

2

24

xx

or

xx

or

yy

How would you solve

Use Substitution

0532

0532

0532

3

1

3

2

24

xx

or

xx

or

yy

How would you solve

Use Substitution

Let u = y2

u2 = y40532

0532

2

24

uu

yy

How would you solve

Use Substitution

Let u = y2

u2 = y4

1;2

5

0152

0532 2

uu

uu

uu

How would you solve

Then Use Substitution

Let u = y2

u2 = y4

1;2

5 uu

iyy

yy

yy

;2

10

1;2

5

1;2

5 22

How would you solve

Then Use Substitution

Let

Do both answers

work?

1;2

5 uu

xu

xu

2

1;4

25

1;2

5

xx

xx

How would you solve

Then Use Substitution

Let

Do both answers

work?

1;2

5 uu

xu

xu

2

051312

02

10

2

15

2

25

52

53

2

25

54

253

4

252

0532

xx

How would you solve

Then Use Substitution

Let

1;2

5 uu

3

22

3

1

xu

xu

8

125

2

5

2

5

33

3

1

3

1

x

x

x

1

1

1

3

3

3

1

3

1

x

x

x

How would you solve

Then Use Substitution

Let Do both answers

work?

1;2

5 uu

3

22

3

1

xu

xu

05)1(3)1(2

058

1253

8

1252

3

1

3

2

3

1

3

2