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Holt Algebra 2
5-7 Solving Quadratic Inequalities5-7 Solving Quadratic Inequalities
Holt Algebra 2
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
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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
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve quadratic inequalities by using tables and graphs.
Solve quadratic inequalities by using algebra.
Objectives
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
quadratic inequality in two variables
Vocabulary
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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).
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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
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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.
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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.
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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).
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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.
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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
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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.
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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
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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.
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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.
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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!
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Example 4: Problem-Solving Application
The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = –8x2 + 600x – 4200, where x is the average selling price of a helmet. What range of selling prices will generate a monthly profit of at least $6000?
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
11 Understand the Problem
Example 4 Continued
The answer will be the average price of a helmet required for a profit that is greater than or equal to $6000.
List the important information:
• The profit must be at least $6000.
• The function for the business’s profit is P(x) = –8x2 + 600x – 4200.
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
22 Make a Plan
Write an inequality showing profit greater than or equal to $6000. Then solve the inequality by using algebra.
Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve33
Write the inequality.
–8x2 + 600x – 4200 ≥ 6000
–8x2 + 600x – 4200 = 6000
Find the critical values by solving the related equation.
Write as an equation.
Write in standard form.
Factor out –8 to simplify.
–8x2 + 600x – 10,200 = 0
–8(x2 – 75x + 1275) = 0
Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve33
Use the Quadratic Formula.
Simplify.
x ≈ 26.04 or x ≈ 48.96
Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve33
Test an x-value in each of the three regions formed by the critical x-values.
10 20 30 40 50 60 70
Critical values
Test points
Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve33
–8(25)2 + 600(25) – 4200 ≥ 6000
–8(45)2 + 600(45) – 4200 ≥ 6000
–8(50)2 + 600(50) – 4200 ≥ 6000
5800 ≥ 6000Try x = 25.
Try x = 45.
Try x = 50.
6600 ≥ 6000
5800 ≥ 6000
Write the solution as an inequality. The solution is approximately 26.04 ≤ x ≤ 48.96.
x
x
Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve33
For a profit of $6000, the average price of a helmet needs to be between $26.04 and $48.96, inclusive.
Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Look Back44
Enter y = –8x2 + 600x – 4200 into a graphing calculator, and create a table of values. The table shows that integer values of x between 26.04 and 48.96 inclusive result in y-values greater than or equal to 6000.
Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
A business offers educational tours to Patagonia, a region of South America that includes parts of Chile and Argentina . The profit P for x number of persons is P(x) = –25x2 + 1250x – 5000. The trip will be rescheduled if the profit is less $7500. How many people must have signed up if the trip is rescheduled?
Check It Out! Example 4
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
11 Understand the Problem
The answer will be the number of people signed up for the trip if the profit is less than $7500.
List the important information:
• The profit will be less than $7500.
• The function for the profit is P(x) = –25x2 + 1250x – 5000.
Check It Out! Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
22 Make a Plan
Write an inequality showing profit less than $7500. Then solve the inequality by using algebra.
Check It Out! Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Solve33
Write the inequality.
–25x2 + 1250x – 5000 < 7500
–25x2 + 1250x – 5000 = 7500
Find the critical values by solving the related equation.
Write as an equation.
Write in standard form.
Factor out –25 to simplify.
–25x2 + 1250x – 12,500 = 0
–25(x2 – 50x + 500) = 0
Check It Out! Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Simplify.
x ≈ 13.82 or x ≈ 36.18
Use the Quadratic Formula.
Solve33
Check It Out! Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Test an x-value in each of the three regions formed by the critical x-values.
5 10 15 20 25 30 35
Critical values
Test points
Solve33
Check It Out! Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
–25(13)2 + 1250(13) – 5000 < 75007025 < 7500
Try x = 13.
Try x = 30.
Try x = 37.
10,000 < 7500
7025 < 7500Write the solution as an inequality. The solution is approximately x > 36.18 or x < 13.82. Because you cannot have a fraction of a person, round each critical value to the appropriate whole number.
x
–25(30)2 + 1250(30) – 5000 < 7500
–25(37)2 + 1250(37) – 5000 < 7500
Solve33
Check It Out! Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
The trip will be rescheduled if the number of people signed up is fewer than 14 people or more than 36 people.
Solve33
Check It Out! Example 4 Continued
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Holt Algebra 2
5-7 Solving Quadratic Inequalities
Look Back44
Enter y = –25x2 + 1250x – 5000 into a graphing calculator, and create a table of values. The table shows that integer values of x less than 13.81 and greater than 36.18 result in y-values less than 7500.
Check It Out! Example 4 Continued
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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
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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