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

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Holt Algebra 2 5-7 Solving Quadratic Inequalities Warm Up 1. Graph the inequality y < 2x + 1. Solve using any method. 2. x 2 – 16x + 63 = 0 3. 3x 2 + 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 ax 2 + bx + c

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Holt Algebra 2 5-7 Solving Quadratic Inequalities Graph y x 2 – 7x + 10. Example 1: Graphing Quadratic Inequalities in Two Variables Step 1 Graph the boundary of the related parabola y = x 2 – 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 x 2 – 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 = 2x 2 – 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 2x 2 – 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 < 2x 2 – 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 = –3x 2 – 6x – 7 with a dashed curve. Its y-intercept is –7. Check It Out! Example 1b y < –3x 2 – 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 < –3x 2 – 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 ax 2 + 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 x 2 + 8x + 20 5 Use a graphing calculator to graph each side of the inequality. Set Y 1 equal to x 2 + 8x + 20 and Y 2 equal to 5. Identify the values of x for which Y 1 Y 2.

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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 x 2 + 8x + 20 < 5 Use a graphing calculator to graph each side of the inequality. Set Y 1 equal to x 2 + 8x + 20 and Y 2 equal to 5. Identify the values of which Y 1 < Y 2.

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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. x 2 – x + 5 < 7 Use a graphing calculator to graph each side of the inequality. Set Y 1 equal to x 2 – x + 5 and Y 2 equal to 7. Identify the values of which Y 1 < Y 2. 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. 2x 2 – 5x + 1 1 Use a graphing calculator to graph each side of the inequality. Set Y 1 equal to 2x 2 – 5x + 1 and Y 2 equal to 1. Identify the values of which Y 1 Y 2. 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 x 2 – 10x + 18 –3 by using algebra. Example 3: Solving Quadratic Equations by Using Algebra Step 1 Write the related equation. x 2 – 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. x 2 –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 x 2 – 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 x 2 – 6x + 10 2 x 2 – 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. x 2 – 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 x 2 – 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 –2x 2 + 3x + 7 < 2 –2x 2 + 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. –2x 2 + 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 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 –2x 2 + 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 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) = –8x 2 + 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 1 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 businesss profit is P(x) = – 8x 2 + 600x – 4200.

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Holt Algebra 2 5-7 Solving Quadratic Inequalities 2 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 Solve 3 Write the inequality. –8x 2 + 600x – 4200 6000 –8x 2 + 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. –8x 2 + 600x – 10,200 = 0 –8(x 2 – 75x + 1275) = 0 Example 4 Continued

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Holt Algebra 2 5-7 Solving Quadratic Inequalities Solve 3 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 Solve 3 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 Solve 3 –8(25) 2 + 600(25) – 4200 6000 –8(45) 2 + 600(45) – 4200 6000 –8(50) 2 + 600(50) – 4200 6000 5800 6000 Try 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 Solve 3 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 Back 4 Enter y = –8x 2 + 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) = –25x 2 + 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 1 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) = –25x 2 + 1250x – 5000. Check It Out! Example 4 Continued

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Holt Algebra 2 5-7 Solving Quadratic Inequalities 2 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 Solve 3 Write the inequality. –25x 2 + 1250x – 5000 < 7500 –25x 2 + 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. –25x 2 + 1250x – 12,500 = 0 –25(x 2 – 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. Solve 3 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 Solve 3 Check It Out! Example 4 Continued

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Holt Algebra 2 5-7 Solving Quadratic Inequalities –25(13) 2 + 1250(13) – 5000 < 7500 7025 < 7500 Try x = 13. Try x = 30. Try x = 37. 10,000 < 7500 7025 < 7500 Write 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 Solve 3 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. Solve 3 Check It Out! Example 4 Continued

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Holt Algebra 2 5-7 Solving Quadratic Inequalities Look Back 4 Enter y = –25x 2 + 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 x 2 + 9x + 14. Solve each inequality. 2. x 2 + 12x + 39 12 3. x 2 – 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) = –2x 2 + 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

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