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Holt Algebra 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 Solving Quadratic Inequalities Warm Up 1. Graph the inequality y < 2x + 1. Solve using any method. 2. x 2 – 16x + 63 = x 2 + 8x = 3 7, 9

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Holt Algebra Solving Quadratic Inequalities Solve quadratic inequalities by using tables and graphs. Solve quadratic inequalities by using algebra. Objectives

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Holt Algebra Solving Quadratic Inequalities quadratic inequality in two variables Vocabulary

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Holt Algebra 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 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 Solving Quadratic Inequalities Graph y x 2 – 7x 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 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 Solving Quadratic Inequalities Example 1 Continued Check Use a test point to verify the solution region. y x 2 – 7x (4) 2 –7(4) – –2 Try (4, 0).

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Holt Algebra 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 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 Solving Quadratic Inequalities Check Use a test point to verify the solution region. y < 2x 2 – 5x – 2 0 2(2) 2 – 5(2) – – 10 – 2 0 –4 Try (2, 0). Check It Out! Example 1a Continued

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Holt Algebra 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 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 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 < – – 7 –10 < –7 Try (–2, –10). Check It Out! Example 1b Continued

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Holt Algebra 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 Solving Quadratic Inequalities Solve the inequality by using tables or graphs. Example 2A: Solving Quadratic Inequalities by Using Tables and Graphs x 2 + 8x 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 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 – The number line shows the solution set.

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Holt Algebra 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 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 – The number line shows the solution set.

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Holt Algebra 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 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 – Check It Out! Example 2a Continued The number line shows the solution set.

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Holt Algebra Solving Quadratic Inequalities Solve the inequality by using tables and graph. 2x 2 – 5x 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 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 – Check It Out! Example 2b Continued The number line shows the solution set.

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Holt Algebra 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 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 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 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 – Critical values Test points x x

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Holt Algebra 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 – Example 3 Continued

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Holt Algebra Solving Quadratic Inequalities Solve the inequality by using algebra. Step 1 Write the related equation. Check It Out! Example 3a x 2 – 6x x 2 – 6x + 10 = 2

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Holt Algebra 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 Solving Quadratic Inequalities Step 3 Test an x-value in each interval. (1) 2 – 6(1) x 2 – 6x (3) 2 – 6(3) (5) 2 – 6(5) Try x = 1. Try x = 3. Try x = 5. Check It Out! Example 3a Continued x –3 –2 – Critical values Test points

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Holt Algebra 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 – Check It Out! Example 3a Continued

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Holt Algebra 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 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 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 – Critical values Test points Check It Out! Example 3b Continued x –2x 2 + 3x + 7 < 2

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Holt Algebra 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 – Check It Out! Example 3

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Holt Algebra 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 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 x – 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 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 x – 4200.

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Holt Algebra 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 Solving Quadratic Inequalities Solve 3 Write the inequality. –8x x – –8x x – 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 x – 10,200 = 0 –8(x 2 – 75x ) = 0 Example 4 Continued

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Holt Algebra Solving Quadratic Inequalities Solve 3 Use the Quadratic Formula. Simplify. x or x Example 4 Continued

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Holt Algebra Solving Quadratic Inequalities Solve 3 Test an x-value in each of the three regions formed by the critical x-values Critical values Test points Example 4 Continued

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Holt Algebra Solving Quadratic Inequalities Solve 3 –8(25) (25) – –8(45) (45) – –8(50) (50) – Try x = 25. Try x = 45. Try x = Write the solution as an inequality. The solution is approximately x x x Example 4 Continued

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Holt Algebra 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 Solving Quadratic Inequalities Look Back 4 Enter y = –8x x – 4200 into a graphing calculator, and create a table of values. The table shows that integer values of x between and inclusive result in y- values greater than or equal to Example 4 Continued

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Holt Algebra 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 x – 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 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 x – Check It Out! Example 4 Continued

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Holt Algebra 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 Solving Quadratic Inequalities Solve 3 Write the inequality. –25x x – 5000 < 7500 –25x x – 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 x – 12,500 = 0 –25(x 2 – 50x + 500) = 0 Check It Out! Example 4 Continued

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Holt Algebra Solving Quadratic Inequalities Simplify. x or x Use the Quadratic Formula. Solve 3 Check It Out! Example 4 Continued

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Holt Algebra Solving Quadratic Inequalities Test an x-value in each of the three regions formed by the critical x-values Critical values Test points Solve 3 Check It Out! Example 4 Continued

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Holt Algebra Solving Quadratic Inequalities –25(13) (13) – 5000 < < 7500 Try x = 13. Try x = 30. Try x = ,000 < < 7500 Write the solution as an inequality. The solution is approximately x > or x < Because you cannot have a fraction of a person, round each critical value to the appropriate whole number. x –25(30) (30) – 5000 < 7500 –25(37) (37) – 5000 < 7500 Solve 3 Check It Out! Example 4 Continued

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Holt Algebra 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 Solving Quadratic Inequalities Look Back 4 Enter y = –25x x – 5000 into a graphing calculator, and create a table of values. The table shows that integer values of x less than and greater than result in y-values less than Check It Out! Example 4 Continued

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Holt Algebra Solving Quadratic Inequalities Lesson Quiz: Part I 1. Graph y x 2 + 9x Solve each inequality. 2. x x x 2 – 24 5x x –9 or x –3 –3 x 8

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Holt Algebra 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 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

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