holt algebra 2 5-7 solving quadratic inequalities 5-7 solving quadratic inequalities holt algebra 2...

Holt Algebra 2

5-7 Solving Quadratic Inequalities5-7 Solving Quadratic Inequalities

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

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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Solve quadratic inequalities by using tables and graphs.

Solve quadratic inequalities by using algebra.

Objectives

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quadratic inequality in two variables

Vocabulary

Holt Algebra 2


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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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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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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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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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


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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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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y < 2x2 – 5x – 2

0 ≥ 2(2)2 – 5(2) – 2

0 ≥ 8 – 10 – 2

0 ≥ –4

Try (2, 0).


Holt Algebra 2


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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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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y < –3x2 – 6x –7

–10 < –3(–2)2 – 6(–2) – 7

–10 < –12 + 12 – 7

–10 < –7

Try (–2, –10).


Holt Algebra 2


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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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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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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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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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


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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.


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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



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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.


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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



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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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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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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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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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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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Solve the inequality by using algebra.



x2 – 6x + 10 ≥ 2

x2 – 6x + 10 = 2

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x2 – 6x + 8 = 0

x – 2 = 0 or x – 4 = 0

(x – 2)(x – 4) = 0 Factor.


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.


Holt Algebra 2



(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.


x

–3 –2 –1 0 1 2 3 4 5 6 7 8 9

Critical values

Test points

Holt Algebra 2


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


Holt Algebra 2


Solve the inequality by using algebra.



–2x2 + 3x + 7 < 2

–2x2 + 3x + 7 = 2

Holt Algebra 2




–2x2 + 3x + 5 = 0

–2x + 5 = 0 or x + 1 = 0

(–2x + 5)(x + 1) = 0 Factor.


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.


Holt Algebra 2



–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


x

–2x2 + 3x + 7 < 2

Holt Algebra 2


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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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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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?

Holt Algebra 2


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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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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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.


Factor out –8 to simplify.

–8x2 + 600x – 10,200 = 0

–8(x2 – 75x + 1275) = 0

Example 4 Continued

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Solve33

Use the Quadratic Formula.

Simplify.

x ≈ 26.04 or x ≈ 48.96

Example 4 Continued

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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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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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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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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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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

Holt Algebra 2


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

Holt Algebra 2


22 Make a Plan

Write an inequality showing profit less than $7500. Then solve the inequality by using algebra.


Holt Algebra 2


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.


Factor out –25 to simplify.

–25x2 + 1250x – 12,500 = 0

–25(x2 – 50x + 500) = 0


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Simplify.

x ≈ 13.82 or x ≈ 36.18

Use the Quadratic Formula.

Solve33


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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


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–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


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The trip will be rescheduled if the number of people signed up is fewer than 14 people or more than 36 people.

Solve33


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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.


Holt Algebra 2


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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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

holt algebra 2 5-7 solving quadratic inequalities 5-7 solving quadratic inequalities holt algebra 2...

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