1.7 Inequalities
Linear Inequalities
Three-Part Inequalities
Quadratic Inequalities
Rational Inequalities
1.7 - 1
Properties of Inequality
For real numbers a, b and c:
1. If a < b, then a + c < b + c,
2. If a < b and if c > 0, then ac < bc,
3. If a < b and if c < 0, then ac > bc.
Replacing < with >, ≤ , or ≥ results in similar properties.
1.7 - 2
Motion Problems
Note Multiplication may be replaced
by division in properties 2 and 3.
Always remember to reverse the
direction of the inequality symbol when
multiplying or dividing by a negative
number.
1.7 - 3
Linear Inequality in One
Variable
A linear inequality in one variable is an
inequality that can be written in the
form
ax + b > 0,
where a and b are real numbers with
a ≠ 0. (Any of the symbols ≥, <, or ≤
may also be used.)
1.7 - 4
Example 1
SOLVING A LINEAR INEQUALITY
Solve −3 x + 5 > −7
Solution
−3 x + 5 > −7
−3 x + 5 − 5 > −7 − 5
−3 x > −12
Don’t forget to
reverse the
symbol here.
−3 x −12
<
−3
−3
Subtract 5.
Divide by –3; reverse
direction of the
inequality symbol
when multiplying or
dividing by a negative
number.
1.7 - 5
Interval
Notation
Open
interval
{x | x > a}
{x | a < x < b}
{x | x < b}
(a, ∞)
(a, b)
(–∞, b)
{x | x ≥ a}
{x | a < x ≤ b}
{x | a ≤ x < b}
{x | x≤ b}
[a, ∞)
(a, b]
[a, b)
(–∞, b]
Other
interval
Graph
(
a
(
a
(
Set
(
Type of
Interval
b
b
[
a
(
a
[
a
]
b
]
)
b
b
1.7 - 6
Set
Interval
Notation
Closed
interval
{x|a ≤ x ≤ b}
[a, b]
Disjoint
interval
{x|x < a or x > b}
(– ∞, a)∪ (b, ∞)
All real
numbers {x|x is a real numbers}
Graph
]
b
[
a
(
Type of
Interval
a
(
b
(– ∞, ∞)
1.7 - 7
Example 2
SOLVING A LINEAR INEQUALITY
Solve 4 – 3x ≤ 7 + 2x. Give the solution
set in interval notation and graph it.
Solution
4 − 3 x ≤ 7 + 2x
4 − 3 x − 4 ≤ 7 + 2x − 4
Subtract 4.
3 x ≤ 3 + 2x
3 x − 2x ≤ 3 + 2x − 2x
Subtract 2x.
−5 x ≤ 3
1.7 - 8
Example 2
SOLVING A LINEAR INEQUALITY
Solve 4 – 3x ≤ 7 + 2x. Give the solution
set in interval notation and graph it.
Solution
−5 x ≤ 3
−5 x 3
≥
−5 x −5
Divide by –5; reverse the
direction of the inequality
symbol.
3
x≥−
5
[
−
3
5
0
 3 
In interval notation the solution set is  − , ∞  .
 5 
1.7 - 9
Example 3
SOLVING A THREE-PART
INEQUALITY
Solve –2 < 5 + 3x < 20.
Solution
−2 < 5 + 3 x < 20
−2 − 5 < 5 + 3 x − 5 < 20 − 5
−7 < 3 x < 15
Divide each
part by 3.
−7 3 x 15
<
<
3
3
3
7
− <x<5
3
Subtract 5 from
each part.
The solution set is
 7 
the interval  − ,5  .
 3 
1.7 - 10
Quadratic Inequalities
A quadratic inequality is an inequality
that can be written in the form
ax 2 + bx + c < 0
for real numbers a, b, and c with a ≠ 0.
(The symbol < can be replaced with >,
≤, or ≥.)
1.7 - 11
Solving a Quadratic Inequality
Step 1 Solve the corresponding quadratic
equation.
Step 2 Identify the intervals determined by
the solutions of the equation.
Step 3 Use a test value from each interval
to determine which intervals form
the solution set.
1.7 - 12
Example 5
SOLVING A QUADRATIC
INEQULITY
Solve x 2 − x − 12 < 0
Solution
Step 1 Find the values of x that satisfy
x2 – x – 12 = 0
x − x − 12 =
0
( x + 3)( x − 4) =
0
2
x +3=
0 or
x−4=
0
Corresponding quadratic
equation
Factor.
Zero-factor
property
1.7 - 13
Example 5
SOLVING A QUADRATIC
INEQULITY
Solve x 2 − x − 12 < 0
Solution
Step 1
x +3=
0 or
x = −3
or
x−4=
0
Zero-factor
property
x=4
1.7 - 14
Example 5
SOLVING A QUADRATIC
INEQULITY
Step 2 The two numbers –3 and 4 divide
the number line into three intervals. If a
value in Interval A makes the polynomial
negative, then all values in Interval A will
make the polynomial negative.
Interval A
Interval B
Interval C
(–∞, – 3)
(– 3, 4)
(4, ∞)
–3
0
4
1.7 - 15
Example 5
SOLVING A QUADRATIC
INEQULITY
Step3 Choose a test value from each interval.
Is x2 –x – 12< 0
Test
Interval
Value
True or False?
–4
(–4)2 – 4(–4) – 12 < 0 ?
A: (– ∞, – 3)
8 < 0 False
B: (– 3, 4)
02 – 0 – 12 < 0 ?
0
–12 < 0 True
52 – 5 – 12 < 0 ?
C: (4, ∞)
5
8 < 0 False
Since the values in Interval B make the inequality
true, the solution set is (– 3, 4).
1.7 - 16
Example 6
SOLVING QUADRATIC
INEQULAITY
Solve 2 x 2 + 5 x − 12 ≥ 0.
Solution
Step 1 Find the values of x that satisfy this
equation.
Corresponding
2
2 x + 5 x − 12 =
0
quadratic equation
(2 x − 3)( x + 4) =
0
2x − 3 =
0
or
x+4=
0
Factor.
Zero-factor
property
1.7 - 17
Example 6
SOLVING QUADRATIC
INEQULAITY
Solve 2 x 2 + 5 x − 12 ≥ 0.
Solution
Step 1 Find the values of x that satisfy this
equation.
2x − 3 =
0
3
x=
2
or
x+4=
0
or
x = −4
Zero-factor
property
1.7 - 18
Example 6
SOLVING QUADRATIC
INEQULAITY
Solve 2 x 2 + 5 x − 12 ≥ 0.
Solution
Step 2 The values form the intervals on
the number line.
Interval A
Interval B
(–∞, – 4)
(– 4, 3/2)
–4
Interval C
(3/2, ∞)
0
3/2
1.7 - 19
Example 6
SOLVING QUADRATIC
INEQULAITY
Step3 Choose a test value from each interval.
Is 2x2 + 5x – 12 ≥ 0
Test
Interval
Value
True or False?
–5
2(–4)2 +5(–5) – 12 ≥ 0 ?
A: (– ∞, – 4)
13 ≥ 0 True
B: (4, 3/2)
2(0)2 +5(0) – 12 ≥ 0 ?
0
–12 ≥ 0 False
2(2)2 + 5(2) – 12 ≥ 0 ?
C: (3/2, ∞)
2
6 ≥ 0 True
The values in Intervals A and B make the inequality true, the
solution set is the union of the intervals. ( −∞, − 4 ) ∪  3 , ∞ 
 8


1.7 - 20