Holt Algebra 2
5-7 Solving Quadratic Inequalities
Warm Up1. Graph the inequality y < 2x + 1.
Solve using any method.2. x2 – 16x + 63 = 0
3. 3x2 + 8x = 3
7, 9
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve quadratic inequalities by using tables and graphs.
Solve quadratic inequalities by using algebra.
Objectives
Holt Algebra 2
5-7 Solving Quadratic Inequalities
quadratic inequality in two variables
Vocabulary
Holt Algebra 2
5-7 Solving Quadratic Inequalities
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).
Holt Algebra 2
5-7 Solving Quadratic Inequalities
In Lesson 2-5, you solved linear inequalities in two variables by graphing. You can use a similar procedure to graph quadratic inequalities.
y < ax2 + bx + c y > ax2 + bx + c
y ≤ ax2 + bx + c y ≥ ax2 + bx + c
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Graph y ≥ x2 – 7x + 10.
Example 1: Graphing Quadratic Inequalities in Two Variables
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.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
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.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Example 1 Continued
Check Use a test point to verify the solution region.
y ≥ x2 – 7x + 10
0 ≥ (4)2 –7(4) + 10
0 ≥ 16 – 28 + 10
0 ≥ –2
Try (4, 0).
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Graph the inequality.
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.
Check It Out! Example 1a
y ≥ 2x2 – 5x – 2
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.
Check It Out! Example 1a Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Check Use a test point to verify the solution region.
y < 2x2 – 5x – 2
0 ≥ 2(2)2 – 5(2) – 2
0 ≥ 8 – 10 – 2
0 ≥ –4
Try (2, 0).
Check It Out! Example 1a Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Graph each inequality.
Step 1 Graph the boundary of the related parabola y = –3x2 – 6x – 7 with a dashed curve. Its y-intercept is –7.
Check It Out! Example 1b
y < –3x2 – 6x – 7
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Step 2 Shade below the parabola because the solution consists of y-values less than those on the parabola for corresponding x-values.
Check It Out! Example 1b Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Check Use a test point to verify the solution region.
y < –3x2 – 6x –7
–10 < –3(–2)2 – 6(–2) – 7
–10 < –12 + 12 – 7
–10 < –7
Try (–2, –10).
Check It Out! Example 1b Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
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
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve the inequality by using tables or graphs.
Example 2A: Solving Quadratic Inequalities by Using Tables and Graphs
x2 + 8x + 20 ≥ 5
Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of x for which Y1 ≥ Y2.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Example 2A Continued
The parabola is at or above the line when x is less than or equal to –5 or greater than or equal to –3. So, the solution set is x ≤ –5 or x ≥ –3 or (–∞, –5] U [–3, ∞). The table supports your answer.
–6 –4 –2 0 2 4 6
The number line shows the solution set.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve the inequality by using tables and graph.
Example 2B: Solving Quadratics Inequalities by Using Tables and Graphs
x2 + 8x + 20 < 5
Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of which Y1 < Y2.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Example 2B Continued
The parabola is below the line when x is greater than –5 and less than –3. So, the solution set is –5 < x < –3 or (–5, –3). The table supports your answer.
–6 –4 –2 0 2 4 6
The number line shows the solution set.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve the inequality by using tables and graph.
x2 – x + 5 < 7
Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 – x + 5 and Y2 equal to 7. Identify the values of which Y1 < Y2.
Check It Out! Example 2a
Holt Algebra 2
5-7 Solving Quadratic Inequalities
The parabola is below the line when x is greater than –1 and less than 2. So, the solution set is –1 < x < 2 or (–1, 2). The table supports your answer.
–6 –4 –2 0 2 4 6
Check It Out! Example 2a Continued
The number line shows the solution set.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve the inequality by using tables and graph.
2x2 – 5x + 1 ≥ 1 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to 2x2 – 5x + 1 and Y2 equal to 1. Identify the values of which Y1 ≥ Y2.
Check It Out! Example 2b
Holt Algebra 2
5-7 Solving Quadratic Inequalities
The parabola is at or above the line when x is less than or equal to 0 or greater than or greater than or equal to 2.5. So, the solution set is (–∞, 0] U [2.5, ∞)
–6 –4 –2 0 2 4 6
Check It Out! Example 2b Continued
The number line shows the solution set.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
The number lines showing the solution sets in Example 2 are divided into three distinct regions by the points –5 and –3. These points are called critical values. By finding the critical values, you can solve quadratic inequalities algebraically.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra.
Example 3: Solving Quadratic Equations by Using Algebra
Step 1 Write the related equation.
x2 – 10x + 18 = –3
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Example 3 Continued
Write in standard form.
Step 2 Solve the equation for x to find the critical values.
x2 –10x + 21 = 0
x – 3 = 0 or x – 7 = 0
(x – 3)(x – 7) = 0 Factor.
Zero Product Property.
Solve for x.x = 3 or x = 7
The critical values are 3 and 7. The critical values divide the number line into three intervals: x ≤ 3, 3 ≤ x ≤ 7, x ≥ 7.
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Example 3 Continued
Step 3 Test an x-value in each interval.
(2)2 – 10(2) + 18 ≤ –3
x2 – 10x + 18 ≤ –3
(4)2 – 10(4) + 18 ≤ –3
(8)2 – 10(8) + 18 ≤ –3
Try x = 2.
Try x = 4.
Try x = 8.
–3 –2 –1 0 1 2 3 4 5 6 7 8 9
Critical values
Test points
x
x
Holt Algebra 2
5-7 Solving Quadratic Inequalities
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 –1 0 1 2 3 4 5 6 7 8 9
Example 3 Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve the inequality by using algebra.
Step 1 Write the related equation.
Check It Out! Example 3a
x2 – 6x + 10 ≥ 2
x2 – 6x + 10 = 2
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Write in standard form.
Step 2 Solve the equation for x to find the critical values.
x2 – 6x + 8 = 0
x – 2 = 0 or x – 4 = 0
(x – 2)(x – 4) = 0 Factor.
Zero Product Property.
Solve for x.x = 2 or x = 4
The critical values are 2 and 4. The critical values divide the number line into three intervals: x ≤ 2, 2 ≤ x ≤ 4, x ≥ 4.
Check It Out! Example 3a Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Step 3 Test an x-value in each interval.
(1)2 – 6(1) + 10 ≥ 2
x2 – 6x + 10 ≥ 2
(3)2 – 6(3) + 10 ≥ 2
(5)2 – 6(5) + 10 ≥ 2
Try x = 1.
Try x = 3.
Try x = 5.
Check It Out! Example 3a Continued
x
–3 –2 –1 0 1 2 3 4 5 6 7 8 9
Critical values
Test points
Holt Algebra 2
5-7 Solving Quadratic Inequalities
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 –1 0 1 2 3 4 5 6 7 8 9
Check It Out! Example 3a Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve the inequality by using algebra.
Step 1 Write the related equation.
Check It Out! Example 3b
–2x2 + 3x + 7 < 2
–2x2 + 3x + 7 = 2
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Write in standard form.
Step 2 Solve the equation for x to find the critical values.
–2x2 + 3x + 5 = 0
–2x + 5 = 0 or x + 1 = 0
(–2x + 5)(x + 1) = 0 Factor.
Zero Product Property.
Solve for x.x = 2.5 or x = –1
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.
Check It Out! Example 3b Continued
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Step 3 Test an x-value in each interval.
–2(–2)2 + 3(–2) + 7 < 2
–2(1)2 + 3(1) + 7 < 2
–2(3)2 + 3(3) + 7 < 2
Try x = –2.
Try x = 1.
Try x = 3.
–3 –2 –1 0 1 2 3 4 5 6 7 8 9
Critical values
Test points
Check It Out! Example 3b Continued
x
–2x2 + 3x + 7 < 2
Holt Algebra 2
5-7 Solving Quadratic Inequalities
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 –1 0 1 2 3 4 5 6 7 8 9
Check It Out! Example 3
Holt Algebra 2
5-7 Solving Quadratic Inequalities
A compound inequality such as 12 ≤ x ≤ 28 can be written as {x|x ≥12 U x ≤ 28}, or x ≥ 12 and x ≤ 28. (see Lesson 2-8).
Remember!
Holt Algebra 2
5-7 Solving Quadratic Inequalities
Lesson Quiz: Part I
1. Graph y ≤ x2 + 9x + 14.
Solve each inequality.
2. x2 + 12x + 39 ≥ 12
3. x2 – 24 ≤ 5x
x ≤ –9 or x ≥ –3
–3 ≤ x ≤ 8
Holt Algebra 2
5-7 Solving Quadratic Inequalities
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) = –2x2 + 120x – 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