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Solving Quadratic Inequalities
Holt Algebra 2
5-7
Solving Quadratic Inequalities
Example 1: Graphing Quadratic Inequalities in Two
Variables
Graph f(x) ≥ x2 – 7x + 10.
Step 1
Holt Algebra 2
Graph the parabola
f(x) = x2 – 7x + 10
with a solid curve.
x
f(x)
0
10
1
3
2
0
3
-2
3.5
-2.25
4
-2
5
0
6
4
7
10
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Solving Quadratic Inequalities
Example 1 Continued
Step 2
Holt Algebra 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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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 Algebra 2

Try (4, 0).
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Solving Quadratic Inequalities
Check It Out! Example 1b
Graph each inequality.
y < –3x2 – 6x – 7
Step 1
Holt Algebra 2
Graph the boundary
of the related parabola
y = –3x2 – 6x – 7 with
a dashed curve.
x
f(x)
-3
-16
-2
-7
-1
0
0
-7
1
-16
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Solving Quadratic Inequalities
Check It Out! Example 1b Continued
Step 2
Holt Algebra 2
Shade below the
parabola because the
solution consists of
y-values less than
those on the parabola
for corresponding
x-values.
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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 Algebra 2

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Solving Quadratic Inequalities
Quadratic inequalities can also be solved
algebraically. Use factoring or the quadratic
formula to find the zeros and then try a test point
to see if it is inside or outside the solution region.
Holt Algebra 2
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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 Algebra 2
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Solving Quadratic Inequalities
Example 3 Continued
Step 2 Solve the equation for x to find the critical
values.
x2 –10x + 21 = 0
(x – 3)(x – 7) = 0
x – 3 = 0 or x – 7 = 0
x = 3 or x = 7
Write in standard form.
Factor.
Zero Product Property.
Solve for x.
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
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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 Algebra 2
6 7
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 Algebra 2
0 1
2
3 4
5
6 7
8
9
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Solving Quadratic Inequalities
Check It Out! Example 3a
Solve the inequality by using algebra.
x2 – 6x + 8 ≥ 2
Step 1 Write the related equation.
x2 – 6x + 8 = 2
Holt Algebra 2
5-7
Solving Quadratic Inequalities
Check It Out! Example 3a Continued
Step 2 Solve the equation for
values.
x2 – 6x + 6 = 0
6 ± (−6)2 −4(1)(6)
𝑥=
2(1)
6 ± 36 − 24
𝑥=
2
𝑥=
6± 12
6±2 3
𝑥=
=
2
2
6+2 3
6−2 3
𝑎𝑛𝑑 𝑥 =
2
2
𝑥 = 4.732 𝑎𝑛𝑑 𝑥 = 1.268
Holt Algebra 2
x to find the critical
Write in standard form.
Substitute into the
quadratic formula to find
the zeros
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Solving Quadratic Inequalities
Check It Out! Example 3a Continued
Step 3 Test an x-value in the interval.
x2 – 6x + 8 ≥ 2
(2)2 – 6(2) + 8 ≥ 2
Try x = 2.
4 – 12 + 8 ≥ 2
0≥2
Holt Algebra 2

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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 Algebra 2
Solving Quadratic Inequalities
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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 Algebra 2
Solving Quadratic Inequalities
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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 Algebra 2
Solving Quadratic Inequalities
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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 Algebra 2
Write as an equation.
Write in standard form.
Factor out –8 to simplify.
Solving Quadratic Inequalities
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Example 4 Continued
3
Solve
Use the Quadratic
Formula.
Simplify.
x ≈ 26.04 or x ≈ 48.96
Holt 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 Algebra 2
50
60
70
Solving Quadratic Inequalities
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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 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 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 Algebra 2