The original equation, has two solutions, and Check

these solutions by substituting for xin this equation. Then start overand substi-

tute for x.

CHECK Let Let

✓ True

✓ True

Both and result in true equations, so the solution set is

(b)

Zero-factor property

Checkthese solutions by substituting each one into the original equation. The solution

set isE0, NOW TRY

4

3 F.

y=

4

3

3 y= 4

y=0or 3 y- 4 = 0

y 13 y- 42 = 0

E-3,

1

2 F.

1

• (^32)

7

2

0 = 0

7

2

01 - 72 = 0 11 - 12  0

a

1

2

• 3 ba 2 #

1

2

1 - 3 + 32321 - 32 - 14  0 - 1 b 0

1 x+ 3212 x- 12 = 0 1 x+ 3212 x- 12 = 0

x=

1

2

x=- 3..

1

2

- 3

1

1 x+ 3212 x- 12 = 0, - 3 2.

SECTION 6.5 Solving Quadratic Equations by Factoring 393

NOW TRY
EXERCISE 1
Solve each equation.

(a)

(b)y 14 y- 52 = 0

1 x- 4213 x+ 12 = 0

NOTE The word oras used in Example 1means “one or the other or both.”

Don’t forget that

0 is a solution.

If the polynomial in an equation is not already factored, first make sure that the

equation is in standard form. Then factor.

Solving Quadratic Equations

Solve each equation.

(a)

First, rewrite the equation in standard form by adding 6 to each side.

x 2 - 5 x+ 6 = 0 Add 6.

x^2 - 5 x=- 6

x^2 - 5 x=- 6

EXAMPLE 2

Now factor Find two numbers whose product is 6 and whose sum is

These two numbers are and , so we factor as follows.

Factor.

Zero-factor property

x= 2or x= 3 Solve each equation.

x- 2 = 0or x- 3 = 0

1 x- 221 x- 32 = 0

- 5. - 2 - 3

x^2 - 5 x+6.

Don’tfactor x

out at this step.

NOW TRY ANSWERS

1. (a)E- 31 , 4F (b)E0, 45 F