1. Solving Quadratic Inequalities
5-7
Solving Quadratic Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2

2. 1. Graph the inequality y < 2x + 1.
Warm Up
1. Graph the inequality y < 2x + 1.
Solve using any method.
2. x2 – 16x + 63 = 0
7, 9
3. 3x2 + 8x = 3

3. Objectives Solve quadratic inequalities by using tables and graphs.
Solve quadratic inequalities by using algebra.

4. Vocabulary
quadratic inequality in two variables

5. Many business profits can be modeled by quadratic functions
Many business profits can be modeled by quadratic functions. To ensure that the profit is above a certain level, financial planners may need to graph and solve quadratic inequalities.
A quadratic inequality in two variables can be written in one of the following forms, where a, b, and c are real numbers and a ≠ 0. Its solution set is a set of ordered pairs (x, y).

6. y < ax2 + bx + c y > ax2 + bx + c
In Lesson 2-5, you solved linear inequalities in two variables by graphing. You can use a similar procedure to graph quadratic inequalities.

7. Example 1: Graphing Quadratic Inequalities in Two Variables
Graph y ≥ x2 – 7x + 10.
Step 1 Graph the boundary of the related parabola y = x2 – 7x + 10 with a solid curve. Its y-intercept is 10, its vertex is (3.5, –2.25), and its x-intercepts are 2 and 5.

8. Example 1 Continued
Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

9. Example 1 Continued
Check Use a test point to verify the solution region.
y ≥ x2 – 7x + 10
0 ≥ (4)2 –7(4) + 10
Try (4, 0).
0 ≥ 16 –
0 ≥ –2 

10. Check It Out! Example 1a
Graph the inequality.
y ≥ 2x2 – 5x – 2
Step 1 Graph the boundary of the related parabola y = 2x2 – 5x – 2 with a solid curve. Its y-intercept is –2, its vertex is (1.3, –5.1), and its x-intercepts are –0.4 and 2.9.

11. Check It Out! Example 1a Continued
Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

12. Check It Out! Example 1a Continued
Check Use a test point to verify the solution region.
y < 2x2 – 5x – 2
0 ≥ 2(2)2 – 5(2) – 2
Try (2, 0).
0 ≥ 8 – 10 – 2
0 ≥ –4 

13. Check It Out! Example 1b
Graph each inequality.
y < –3x2 – 6x – 7
Step 1 Graph the boundary of the related parabola y = –3x2 – 6x – 7 with a dashed curve. Its y-intercept is –7.

14. Check It Out! Example 1b Continued
Step 2 Shade below the parabola because the solution consists of y-values less than those on the parabola for corresponding x-values.

15. Check It Out! Example 1b Continued
Check Use a test point to verify the solution region.
y < –3x2 – 6x –7
–10 < –3(–2)2 – 6(–2) – 7
Try (–2, –10).
–10 < – – 7
–10 < –7 

16. Quadratic inequalities in one variable, such as ax2 + bx + c > 0 (a ≠ 0), have solutions in one variable that are graphed on a number line.
For and statements, both of the conditions must be true. For or statements, at least one of the conditions must be true.
Reading Math

17. Example 3: Solving Quadratic Equations by Using Algebra
Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra.
Step 1 Write the related equation.
x2 – 10x + 18 = –3

18. Example 3 Continued
Step 2 Solve the equation for x to find the critical values.
x2 –10x + 21 = 0
Write in standard form.
(x – 3)(x – 7) = 0
Factor.
x – 3 = 0 or x – 7 = 0
Zero Product Property.
x = 3 or x = 7
Solve for x.
The critical values are 3 and 7. The critical values divide the number line into three intervals: x ≤ 3, 3 ≤ x ≤ 7, x ≥ 7.

19. Step 3 Test an x-value in each interval.
Example 3 Continued
Step 3 Test an x-value in each interval.
–3 –2 –
Critical values
Test points
x2 – 10x + 18 ≤ –3
(2)2 – 10(2) + 18 ≤ –3
Try x = 2. x
(4)2 – 10(4) + 18 ≤ –3 
Try x = 4.
(8)2 – 10(8) + 18 ≤ –3 x
Try x = 8.

20. Example 3 Continued
Shade the solution regions on the number line. Use solid circles for the critical values because the inequality contains them. The solution is 3 ≤ x ≤ 7 or [3, 7].
–3 –2 –

21. Check It Out! Example 3a
Solve the inequality by using algebra.
x2 – 6x + 10 ≥ 2
Step 1 Write the related equation.
x2 – 6x + 10 = 2

22. Check It Out! Example 3a Continued
Step 2 Solve the equation for x to find the critical values.
x2 – 6x + 8 = 0
Write in standard form.
(x – 2)(x – 4) = 0
Factor.
x – 2 = 0 or x – 4 = 0
Zero Product Property.
x = 2 or x = 4
Solve for x.
The critical values are 2 and 4. The critical values divide the number line into three intervals: x ≤ 2, 2 ≤ x ≤ 4, x ≥ 4.

23. Check It Out! Example 3a Continued
Step 3 Test an x-value in each interval.
–3 –2 –
Critical values
Test points
x2 – 6x + 10 ≥ 2
(1)2 – 6(1) + 10 ≥ 2 
Try x = 1.
(3)2 – 6(3) + 10 ≥ 2 x
Try x = 3.
(5)2 – 6(5) + 10 ≥ 2 
Try x = 5.

24. Check It Out! Example 3a Continued
Shade the solution regions on the number line. Use solid circles for the critical values because the inequality contains them. The solution is x ≤ 2 or x ≥ 4.
–3 –2 –

25. Check It Out! Example 3b
Solve the inequality by using algebra.
–2x2 + 3x + 7 < 2
Step 1 Write the related equation.
–2x2 + 3x + 7 = 2

26. Check It Out! Example 3b Continued
Step 2 Solve the equation for x to find the critical values.
–2x2 + 3x + 5 = 0
Write in standard form.
(–2x + 5)(x + 1) = 0
Factor.
–2x + 5 = 0 or x + 1 = 0
Zero Product Property.
x = 2.5 or x = –1
Solve for x.
The critical values are 2.5 and –1. The critical values divide the number line into three intervals: x < –1, –1 < x < 2.5, x > 2.5.

27. Check It Out! Example 3b Continued
Step 3 Test an x-value in each interval.
–3 –2 –
Critical values
Test points
–2x2 + 3x + 7 < 2
–2(–2)2 + 3(–2) + 7 < 2
Try x = –2. 
–2(1)2 + 3(1) + 7 < 2 x
Try x = 1.
–2(3)2 + 3(3) + 7 < 2 
Try x = 3.

28. Check It Out! Example 3
Shade the solution regions on the number line. Use open circles for the critical values because the inequality does not contain or equal to. The solution is x < –1 or x > 2.5.
–3 –2 –

29. Lesson Quiz: Part I
1. Graph y ≤ x2 + 9x + 14.
Solve each inequality.
2. x2 + 12x + 39 ≥ 12
x ≤ –9 or x ≥ –3
3. x2 – 24 ≤ 5x
–3 ≤ x ≤ 8

30. Lesson Quiz: Part II
4. A boat operator wants to offer tours of San Francisco Bay. His profit P for a trip can be modeled by P(x) = –2x x – 788, where x is the cost per ticket. What range of ticket prices will generate a profit of at least $500?
between $14 and $46, inclusive