1. 4.9 Solving Quadratic Inequalities

2. EXAMPLE 4
Solve a quadratic inequality using a table
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 x are between –3 and 2, inclusive.
The solution of the inequality is –3 ≤ x ≤ 2.
ANSWER

3. EXAMPLE 5
Solve a quadratic inequality by graphing
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
x =
– – 4(2)(–4)
2(2)
x = – 4
x or x –1.69

4. EXAMPLE 5
Solve a quadratic inequality by graphing
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.

5. GUIDED PRACTICE
for Examples 4 and 5
Solve the inequality 2x2 + 2x ≤ 3 using a table and using a graph.
ANSWER
–1.8 ≤ x ≤ 0.82

6. Solving a quadratic inequality algebraically-9 -6 -8 -4

7. Solve a quadratic inequality algebraically
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 or x = 5
Zero product property

8. 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:
(–4)2 – 2(–4) = 24 > 15 
12 – 2(1)= –1 >15
62 –2(6) = 24 >15 
The solution is x < –3 or x > 5.
ANSWER

9. Solve the inequality using any method.