172 CHAPTER 1 Equations and Inequalities Step 2 The two numbers -3 and 4 cause the expression x2 - x - 12 to equal 0 and can be used to divide the number line into three intervals, as shown in Figure 12. The expression x2 - x - 12 will take on a value that is either less than 0 or greater than 0 on each of these intervals. We are looking for x-values that make the expression less than 0, so we use open circles at -3 and 4 to indicate that they are not included in the solution set. Interval A (–`, –3) Interval B (–3, 4) Interval C (4, `) 0 –3 4 Figure 12 Use open circles because the inequality symbol does not include equality. -3 and 4 do not satisfy the inequality. EXAMPLE 6 Solving a Quadratic Inequality Solve 2x2 + 5x - 12 Ú 0. SOLUTION Step 1 Find the values of x that satisfy 2x2 + 5x - 12 = 0. 2x2 + 5x - 12 = 0 Corresponding quadratic equation 12x - 321x + 42 = 0 Factor. 2x - 3 = 0 or x + 4 = 0 Zero-factor property x = 3 2 or x = -4 Solve each equation. Step 2 The values 3 2 and -4 cause the expression 2x 2 + 5x - 12 to equal 0 and can be used to form the intervals 1-∞, -42, A -4, 3 2B, and A 3 2, ∞B on the number line, as seen in Figure 14. Interval A (–`, –4) Interval B (–4, ) Interval C ( , `) 0 –4 3 2 3 2 3 2 Figure 14 Use closed circles because the inequality symbol includes equality. -4 and 3 2 satisfy the inequality. Step 3 Choose a test value in each interval to see whether it satisfies the original inequality, x2 - x - 12 60. If the test value makes the statement true, then the entire interval belongs to the solution set. Interval Test Value Is x2 −x −12 *0 True or False? A: 1-∞, -32 -4 1-422 - 1-42 - 12 6 ? 0 8 60 False B: 1-3, 42 0 02 - 0 - 12 6 ? -12 60 True C: 14, ∞2 5 52 - 5 - 12 6 ? 0 8 60 False The values in Interval B make the inequality true. See the graph in Figure 13. The solution set is the interval 1-3, 42. S Now Try Exercise 39. –3 0 4 Figure 13
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