Intermediate Algebra (11th edition)

SECTION 9.7 Polynomial and Rational Inequalities 553

Because - 5 satisfies the inequality, allnumbers from Interval A are solutions.

NOW TRY

NOTE If the inequalities in Example 1had used and , the solution sets would

have included the x-values of the intercepts, which make the quadratic expression

equal to 0. They would have been written in interval notation as

and

Square brackets would indicate that the endpoints and 4 are includedin the solu-

tion sets.

Another method for solving a quadratic inequality uses the basic ideas of

Example 1without actually graphing the related quadratic function.

Solving a Quadratic Inequality Using Test Numbers

Solve and graph the solution set of

Solve the quadratic equation by factoring, as in Example 1(a).

or

or

The numbers 4 and divide a number line into Intervals A, B, and C, as shown in

FIGURE 19. Be careful to put the lesser number on the left.

- 3

x= 4 x= - 3

x- 4 = 0 x+ 3 = 0

1 x- 421 x+ 32 = 0

x^2 - x- 12 = 0

x^2 - x- 127 0.

EXAMPLE 2

- 3

1 - q, - 34 ́ 3 4, q 2 3 - 3, 4 4.

Ú ...

NOW TRY
EXERCISE 1
Use the graph to solve each
quadratic inequality.

(a)

(b)x^2 - 3 x- 460

x^2 - 3 x- 470

x

y

0

–4

–1 4

f(x) = x^2 – 3x – 4

(b)

We want values of ythat are less than0. Referring to FIGURE 18(b), we notice from

the graph that x-values between and 4 result in y-values less than 0. Thus, the solu-

tion set of x^2 - x- 126 0,written in interval notation, is 1 - 3, 4 2.

- 3

x^2 - x- 1260

T–3 F 4 T

Interval

C

Interval

B

Interval

A

FIGURE 19

The numbers 4 and are the only numbers that make the quadratic expression

equal to 0. All other numbers make the expression either positive or

negative. The sign of the expression can change from positive to negative or from

negative to positive only at a number that makes it 0. Therefore, if one number in an

interval satisfies the inequality, then all the numbers in that interval will satisfy the

inequality.

To see if the numbers in Interval A satisfy the inequality, choose any number

from Interval A in FIGURE 19(that is, any number less than ). We choose

Substitute this test number for xin the original inequality

Original inequality

Let

Simplify.

187 0 ✓ True

25 + 5 - 127

?

0

1 - 522 - 1 - 52 - 127 x=-5.

?

0

x^2 - x- 127 0

x^2 - x- 127 0.

- 3 - 5.

x^2 - x- 12

- 3

Use parentheses to

avoid sign errors.

NOW TRY ANSWERS

1. (a)
   (b) 1 - 1, 4 2

1 - q, - 12 ́ 1 4, q 2

Notice the similarity between

FIGURE 19and the x-axis with

intercepts and in

FIGURE 18(a).

1 - 3, 0 2 1 4, 0 2