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

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.

Holt Algebra 2

5-7 Solving Quadratic Inequalities

y < ax2 + bx + c y > ax2 + bx + c use a dashed line

y ≤ ax2 + bx + c y ≥ ax2 + bx + c use a solid line

If the parabola opens up:

> or shade inside

< or shade outside

If the parabola opens down:

> or shade outside

< or shade inside

To graph Quadratic Inequalities

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

vertex (3.5, -2.25)

roots (2, 0) and (5, 0)

y-intercept (0, 10)

reflection of y-intercept (7, 10)

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

Graph each inequality.

Step 1 Graph the boundary

vertex (-1, -4)

no roots

y-intercept (0, -7)

reflect of y-int (-2, -7)

with a dashed curve.

Example 2

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.

Example 2

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

By finding the critical values,

you can solve quadratic

inequalities algebraically.

Holt Algebra 2

5-7 Solving Quadratic Inequalities

Solve 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

Step 2 Solve for x to find the critical values. x2 –10x + 21 = 0

(x – 3)(x – 7) = 0

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 InequalitiesExample 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

x2 3

6 3

2 3

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.

Example 4

x2 – 6x + 10 ≥ 2

x2 – 6x + 10 = 2

x2 – 6x + 8 = 0

(x – 2)(x – 4) = 0

x = 2 or x = 4

The critical values are 2 and 4.

The 3 intervals are: x ≤ 2, 2 ≤ x ≤ 4, x ≥ 4.

Holt Algebra 2

5-7 Solving Quadratic Inequalities

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.

Example 4

x

–3 –2 –1 0 1 2 3 4 5 6 7 8 9

Critical values

Test points

5 2

1 2

5 2

Holt Algebra 2

5-7 Solving Quadratic Inequalities

Use solid circles for the critical values because the inequality contains them.

Shade the solution regions on the number line.

The solution is x ≤ 2 or x ≥ 4.

–3 –2 –1 0 1 2 3 4 5 6 7 8 9

Example 4

Holt Algebra 2

5-7 Solving Quadratic Inequalities

Example 5: 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

5-7 Solving Quadratic Inequalities

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 5

Holt Algebra 2

5-7 Solving Quadratic Inequalities

x ≈ 26.04 or x ≈ 48.96

Example 5

75 525

2

75 5 21

2

75 5 21

2or

75 5 21

2

2 4

2

b b acx

a

2

175 7 4 15 275

2 1

Holt Algebra 2

5-7 Solving Quadratic Inequalities

10 20 30 40 50 60 70

Critical values

Test points

Example 5

Graph the critical points and test points.

Holt Algebra 2

5-7 Solving Quadratic Inequalities

–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

The solution is approximately 26.04 ≤ x ≤ 48.96.

x

x

Example 5

Holt Algebra 2

5-7 Solving Quadratic Inequalities

For a profit of $6000,

the average price of a helmet needs to be

between $26.04 and $48.96, inclusive.

Example 5

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