5-7
5-7 Solving
Solving Quadratic
Quadratic Inequalities
Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Holt
McDougal
Algebra 2Algebra
Algebra22
Holt
McDougal
5-7
Solving Quadratic Inequalities
Warm Up
1. Graph the inequality
y < 2x + 1.
Solve using any method.
2. x2 – 16x + 63 = 0
3. 3x2 + 8x = 3
Holt McDougal Algebra 2
7, 9
5-7
Solving Quadratic Inequalities
Objectives
Solve quadratic inequalities by using
tables and graphs.
Solve quadratic inequalities by using
algebra.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Vocabulary
quadratic inequality in two variables
Holt McDougal 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).
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
y < ax2 + bx + c
y > ax2 + bx + c
y ≤ ax2 + bx + c
y ≥ ax2 + bx + c
In Lesson 2-5, you solved linear inequalities in
two variables by graphing. You can use a similar
procedure to graph quadratic inequalities.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Example 1: Graphing Quadratic Inequalities in Two
Variables
Graph y ≥ x2 – 7x + 10.
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.
Holt McDougal Algebra 2
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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 McDougal 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
Holt McDougal Algebra 2

Try (4, 0).
5-7
Solving Quadratic Inequalities
Check It Out! Example 1a
Graph the inequality.
y ≥ 2x2 – 5x – 2
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.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 1a 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 McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 1a Continued
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
Holt McDougal Algebra 2

Try (2, 0).
5-7
Solving Quadratic Inequalities
Check It Out! Example 1b
Graph each inequality.
y < –3x2 – 6x – 7
Step 1
Graph the boundary
of the related parabola
y = –3x2 – 6x – 7 with
a dashed curve. Its
y-intercept is –7.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 1b Continued
Step 2
Shade below the
parabola because the
solution consists of
y-values less than
those on the parabola
for corresponding
x-values.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 1b Continued
Check Use a test point to verify the solution region.
y < –3x2 – 6x –7
–10 < –3(–2)2 – 6(–2) – 7 Try (–2, –10).
–10 < –12 + 12 – 7
–10 < –7
Holt McDougal 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.
Reading Math
For and statements, both of the conditions must
be true. For or statements, at least one of the
conditions must be true.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Example 2A: Solving Quadratic Inequalities by Using
Tables and Graphs
Solve the inequality by using tables or 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.
Holt McDougal 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.
The number line shows the solution set.
–6
Holt McDougal Algebra 2
–4
–2
0
2
4
6
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Solving Quadratic Inequalities
Example 2B: Solving Quadratics Inequalities by Using
Tables and Graphs
Solve the inequality by using tables and graph.
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.
Holt McDougal Algebra 2
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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.
The number line shows the solution set.
–6
Holt McDougal Algebra 2
–4
–2
0
2
4
6
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Solving Quadratic Inequalities
Check It Out! Example 2a
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.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 2a Continued
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.
The number line shows the solution set.
–6
Holt McDougal Algebra 2
–4
–2
0
2
4
6
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Solving Quadratic Inequalities
Check It Out! Example 2b
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.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 2b Continued
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, ∞)
The number line shows the solution set.
–6
Holt McDougal Algebra 2
–4
–2
0
2
4
6
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.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Example 3: Solving Quadratic Equations
by Using Algebra
Solve the inequality x2 – 10x + 18 ≤ –3 by using
algebra.
Step 1 Write the related equation.
x2 – 10x + 18 = –3
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Example 3 Continued
Step 2 Solve the equation for x to find the critical
values.
x2 –10x + 21 = 0
Write in standard form.
(x – 3)(x – 7) = 0
Factor.
Zero Product Property.
Solve for x.
x – 3 = 0 or 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 McDougal Algebra 2
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Solving Quadratic Inequalities
Example 3 Continued
Step 3 Test an x-value in each interval.
Critical values
x2 – 10x + 18 ≤ –3
–3 –2 –1
0 1
2
3 4
5
Test points
(2)2 – 10(2) + 18 ≤ –3 x
Try x = 2.
(4)2 – 10(4) + 18 ≤ –3 
Try x = 4.
(8)2 – 10(8) + 18 ≤ –3 x
Try x = 8.
Holt McDougal Algebra 2
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8
9
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Solving Quadratic Inequalities
Example 3 Continued
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
Holt McDougal Algebra 2
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5
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8
9
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Solving Quadratic Inequalities
Check It Out! Example 3a
Solve the inequality by using algebra.
x2 – 6x + 10 ≥ 2
Step 1 Write the related equation.
x2 – 6x + 10 = 2
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 3a Continued
Step 2 Solve the equation for x to find the critical
values.
x2 – 6x + 8 = 0
(x – 2)(x – 4) = 0
x – 2 = 0 or x – 4 = 0
x = 2 or x = 4
Write in standard form.
Factor.
Zero Product Property.
Solve for x.
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 McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 3a Continued
Step 3 Test an x-value in each interval.
Critical values
x2 – 6x + 10 ≥ 2
–3 –2 –1
0 1
2
3 4
Test points
(1)2 – 6(1) + 10 ≥ 2 
Try x = 1.
(3)2 – 6(3) + 10 ≥ 2 x
Try x = 3.
(5)2 – 6(5) + 10 ≥ 2 
Try x = 5.
Holt McDougal Algebra 2
5
6 7
8
9
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Solving Quadratic Inequalities
Check It Out! Example 3a Continued
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
Holt McDougal Algebra 2
0 1
2
3 4
5
6 7
8
9
5-7
Solving Quadratic Inequalities
Check It Out! Example 3b
Solve the inequality by using algebra.
–2x2 + 3x + 7 < 2
Step 1 Write the related equation.
–2x2 + 3x + 7 = 2
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 3b Continued
Step 2 Solve the equation for x to find the critical
values.
–2x2 + 3x + 5 = 0
Write in standard form.
(–2x + 5)(x + 1) = 0
Factor.
Zero Product Property.
–2x + 5 = 0 or x + 1 = 0
x = 2.5 or x = –1
Solve for x.
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 McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 3b Continued
Step 3 Test an x-value in each interval.
Critical values
–2x2 + 3x + 7 < 2
–3 –2 –1
0 1
2
3 4
5
6 7
Test points
–2(–2)2 + 3(–2) + 7 < 2

Try x = –2.
–2(1)2 + 3(1) + 7 < 2
x
Try x = 1.
–2(3)2 + 3(3) + 7 < 2 
Try x = 3.
Holt McDougal Algebra 2
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9
5-7
Solving Quadratic Inequalities
Check It Out! Example 3
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
Holt McDougal Algebra 2
0 1
2
3 4
5
6 7
8
9
5-7
Solving Quadratic Inequalities
Remember!
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).
Holt McDougal 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?
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Example 4 Continued
1
Understand the Problem
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.
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Example 4 Continued
2
Make a Plan
Write an inequality showing profit greater than
or equal to $6000. Then solve the inequality by
using algebra.
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Example 4 Continued
3
Solve
Write the inequality.
–8x2 + 600x – 4200 ≥ 6000
Find the critical values by solving the related equation.
–8x2 + 600x – 4200 = 6000
–8x2 + 600x – 10,200 = 0
–8(x2 – 75x + 1275) = 0
Holt McDougal Algebra 2
Write as an equation.
Write in standard form.
Factor out –8 to simplify.
Solving Quadratic Inequalities
5-7
Example 4 Continued
3
Solve
Use the Quadratic
Formula.
Simplify.
x ≈ 26.04 or x ≈ 48.96
Holt McDougal Algebra 2
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Solving Quadratic Inequalities
Example 4 Continued
3
Solve
Test an x-value in each of the three regions
formed by the critical x-values.
Critical values
10
20
30
40
Test points
Holt McDougal Algebra 2
50
60
70
Solving Quadratic Inequalities
5-7
Example 4 Continued
3
Solve
–8(25)2 + 600(25) – 4200 ≥ 6000
Try x = 25.
5800 ≥ 6000 x
–8(45)2 + 600(45) – 4200 ≥ 6000
Try x = 45.
6600 ≥ 6000 
Try x = 50.
–8(50)2 + 600(50) – 4200 ≥ 6000
5800 ≥ 6000 x
Write the solution as an inequality. The solution
is approximately 26.04 ≤ x ≤ 48.96.
Holt McDougal Algebra 2
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Solving Quadratic Inequalities
Example 4 Continued
3
Solve
For a profit of $6000, the average price of a
helmet needs to be between $26.04 and $48.96,
inclusive.
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Example 4 Continued
4
Look Back
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 yvalues greater than or equal
to 6000.
Holt McDougal Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 4
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?
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
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) = –25x2 + 1250x – 5000.
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
2
Make a Plan
Write an inequality showing profit less than
$7500. Then solve the inequality by using
algebra.
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
3
Solve
Write the inequality.
–25x2 + 1250x – 5000 < 7500
Find the critical values by solving the related equation.
–25x2 + 1250x – 5000 = 7500 Write as an equation.
–25x2 + 1250x – 12,500 = 0
–25(x2 – 50x + 500) = 0
Holt McDougal Algebra 2
Write in standard form.
Factor out –25 to simplify.
Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
3
Solve
Use the Quadratic
Formula.
Simplify.
x ≈ 13.82 or x ≈ 36.18
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
3
Solve
Test an x-value in each of the three regions
formed by the critical x-values.
Critical values
5
10
15
20
25
30
Test points
Holt McDougal Algebra 2
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Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
3
Solve
–25(13)2 + 1250(13) – 5000 < 7500
7025 < 7500 
–25(30)2 + 1250(30) – 5000 < 7500
10,000 < 7500 x
Try x = 13.
Try x = 30.
Try x = 37.
–25(37)2 + 1250(37) – 5000 < 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.
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
3
Solve
The trip will be rescheduled if the number
of people signed up is fewer than 14
people or more than 36 people.
Holt McDougal Algebra 2
Solving Quadratic Inequalities
5-7
Check It Out! Example 4 Continued
4
Look Back
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 McDougal Algebra 2
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Solving Quadratic Inequalities
Lesson Quiz: Part I
1. Graph y ≤ x2 + 9x + 14.
Solve each inequality.
2. x2 + 12x + 39 ≥ 12
x ≤ –9 or x ≥ –3
3. x2 – 24 ≤ 5x
–3 ≤ x ≤ 8
Holt McDougal Algebra 2
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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
Holt McDougal Algebra 2