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4.9 Solving Quadratic Inequalities. ANSWER. The solution of the inequality is –3 ≤ x ≤ 2. Solve a quadratic inequality using a table. EXAMPLE 4. Solve x 2 + x ≤ 6 using a table. SOLUTION. Rewrite the inequality as x 2 + x – 6 ≤ 0 . Then make a table of values.
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ANSWER The solution of the inequality is –3 ≤ x ≤ 2. Solve a quadratic inequality using a table EXAMPLE 4 Solve x2 + x ≤ 6 using a table. SOLUTION Rewrite the inequality as x2 + x – 6 ≤ 0. Then make a table of values. Notice that x2 + x – 6 ≤ 0 when the values of xare between –3 and 2, inclusive.
–1+ 12– 4(2)(–4) x = –1+33 2(2) x = 4 x 1.19 orx –1.69 Solve a quadratic inequality by graphing EXAMPLE 5 Solve 2x2 + x – 4 ≥ 0 by graphing. SOLUTION The solution consists of the x-values for which the graph of y = 2x2 + x – 4 lies on or above the x-axis. Find the graph’s x-intercepts by letting y = 0 and using the quadratic formula to solve for x. 0 = 2x2 + x – 4
Solve a quadratic inequality by graphing EXAMPLE 5 Sketch a parabola that opens up and has 1.19 and –1.69 as x-intercepts. The graph lies on or above the x-axis to the left of (and including) x = –1.69 and to the right of (and including) x = 1.19. ANSWER The solution of the inequality is approximately x ≤ –1.69 or x ≥ 1.19.
for Examples 4 and 5 GUIDED PRACTICE Solve the inequality 2x2 + 2x ≤ 3using a table and using a graph. ANSWER –1.8 ≤ x ≤ 0.82
Solving a quadratic inequality algebraically -9 -6 0 -8 -4
EXAMPLE 7 Solve a quadratic inequality algebraically Solve x2 – 2x > 15 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with = . x2 – 2x = 15 Write equation that corresponds to original inequality. x2 – 2x – 15 = 0 Write in standard form. (x + 3)(x – 5) = 0 Factor. x = –3 orx = 5 Zero product property
(–4)2– 2(–4) = 24 > 15 62 –2(6) = 24 >15 12 – 2(1)= –1 >15 ANSWER The solution is x < –3 orx > 5. EXAMPLE 7 Solve a quadratic inequality algebraically The numbers –3 and 5 are the critical x-values of the inequality x2 – 2x > 15. Plot –3 and 5 on a number line, using open dots because the values do not satisfy the inequality. The critical x-values partition the number line into three intervals. Test an x-value in each interval to see if it satisfies the inequality. Test x = – 4: Test x = 1: Test x = 6:
