Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 1.


Factoring and Applications

Chapter 6


6.7

Solving Quadratic Equations by Factoring


Objectives

1. Solve quadratic equations by factoring.

2. Solve other equations by factoring.

6.7 Solving Quadratic Equations by Factoring


Quadratic EquationsA quadratic equation is an equation that can be written in the form

ax2 + bx + c = 0,where a, b, and c, are real numbers, with a ≠ 0. The given form is called standard form.




Zero-Factor PropertyIf a and b are real numbers and

ab = 0, then a = 0 or b = 0.In words, if the product of two numbers is 0, then at least one of the numbers must be 0. One number must be 0, but both may be 0.



(2y – 3)(y + 1) = 0

Example 1 Solve the equation.



By the zero-factor property, the only way that the product of these two factors can be 0 is if at least one of the factors equals 0.

2y – 3 = 0 or y + 1 = 02y = 3 y = –1

The product is equal to 0.

3

2y

Solve each equation.

The solution set is

3, 1 .

2


x2 – 3x = 4




Subtract 4.

(x – 4)(x + 1) = 0

Zero-factor propertyx – 4 = 0 or x + 1 = 0

x = 4 x = –1

x2 – 3x – 4 = 0

Factor the trinomial.


First, write the equation in standard form.

The solution set is

4, 1 .


6.7 Solving Quadratic Equations by FactoringSolving Quadratic Equations by Factoring

Solving a Quadratic Equation by FactoringStep 1 Write the equation in standard form, that is,

with all terms on one side of the equals sign in descending powers of the variable and 0 on the other side.

Step 2 Factor completely.Step 3 Use the zero-factor property to set each factor

with a variable equal to 0.Step 4 Solve the resulting equations.Step 5 Check each solution in the original equation.


Divide each side by 2.

2x2 + 30 = –16x




Standard form

2(x2 + 8x + 15) = 0

Factor.(x + 3)(x + 5) = 0

Zero-factor propertyx + 3 = 0 or x + 5 = 0

x = –3 x = –5

2x2 + 16x + 30 = 0

Factor out the GCF, 2.

x2 + 8x + 15 = 0



2(–3)2 + 30 = –16(–3)

Example 3 (continued) Solve the equation.



Check that the solution set is {–3, –5} by substituting in the original equation.

18 + 30 = 48

?

48 = 48

? 2(–5)2 + 30 = –16(–5)

50 + 30 = 80

?

80 = 80

?

CAUTIONA common error is to include the common factor 2 as a solution. Only factors containing variables lead to solutions.


y(2y + 5) = 42

Example 4(a) Solve the equation.



2y2 + 5y – 42 = 0

(2y – 7)(y + 6) = 0

2y – 7 = 0 or y + 6 = 02y = 7 y = –6

2y2 + 5y = 42

We need to write this equation in standard form.

7

2y

Multiply.

Subtract 42.

Factor.

Zero-factor property

The solution set is 7

, 6 .2


x2 = 4x

Example 4(b) Solve the equation.



x(x – 4) = 0

x = 0 or x – 4 = 0

x = 4

x2 – 4x = 0 Standard form

Factor.

Zero-factor property

The solution set is {0, 4}.

x = 0


CAUTIONIn Example 4(a), the zero-factor property could not be used tosolve the equation y(2y + 5) = 42 in its given form because of the 42 on the right. The zero-factor property applies only to a product that equals 0.

In Example 4(b), it is tempting to begin by dividing each side of the equation x2 = 4x by x to get x = 4. Note that we do not get the other solution, 0, if we divide by a variable. (We may divide each side of an equation by a nonzero real number, however. For instance, in Example 3, we divided each side by 2.)




x2 + 64 = –16x




Standard form

Factor.(x + 8)2 = 0

Zero-factor propertyx + 8 = 0 or x + 8 = 0

x = – 8 x = – 8

x2 + 16x + 64 = 0


The solution set is {– 8}. This is called a double solution.


NoteNot all quadratic equations can be solved by factoring.





3a3 – 48a = 0

Solving Other Equations by Factoring


3a(a2 – 16) = 0

3a(a + 4)(a – 4) = 0

a = –4

Zero-product property

Factor out 3a.

Factor a2 – 16.

3a = 0 or a + 4 = 0 or a – 4 = 0

a = 0 a = 4

The solution set is {0, – 4, 4}.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 1.

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