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2. Equations and
Inequalities
2.6-2.7 Inequalities
Copyright © Cengage Learning. All rights reserved.
Inequalities
An inequality is a statement that two quantities or
expressions are not equal. It may be the case that one
quantity is less than (<), less than or equal to (), greater
than (>), or greater than or equal to () another quantity.
Consider the inequality
2x + 3 > 11,
where x is a variable.
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Inequalities
As illustrated in the following table, certain numbers yield
true statements when substituted for x, and others yield
false statements.
If a true statement is obtained when a number b is
substituted for x, then b is a solution of the inequality.
Thus, x = 5 is a solution of 2x + 3 > 11 since 13 > 11 is
true, but x = 3 is not a solution since 9 > 11 is false.
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Inequalities
To solve an inequality means to find all solutions. Two
inequalities are equivalent if they have exactly the same
solutions.
Most inequalities have an infinite number of solutions. To
illustrate, the solutions of the inequality
2<x<5
consist of every real number x between 2 and 5. We call
this set of numbers an open interval and denote it by
(2, 5).
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Inequalities
The graph of the open interval (2, 5) is the set of all points
on a coordinate line that lie between—but do not
include—the points corresponding to x = 2 and x = 5.
The graph is represented by shading an appropriate part of
the axis, as shown in Figure 1.
Figure 1
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Inequalities
We refer to this process as sketching the graph of the
interval. The numbers 2 and 5 are called the endpoints of
the interval (2, 5).
The parentheses in the notation (2, 5) and in Figure 1 are
used to indicate that the endpoints of the interval are not
included.
If we wish to include an endpoint, we use a bracket instead
of a parenthesis. For example, the solutions of the
inequality 2  x  5 are denoted by [2, 5] and are referred to
as a closed interval.
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Inequalities
The graph of [2, 5] is sketched in Figure 2, where brackets
indicate that endpoints are included. We shall also consider
half-open intervals [a, b) and (a, b] and infinite intervals,
as described in the next chart.
The symbol
(read “infinity”) used for infinite intervals is
merely a notational device and does not represent a real
number.
Figure 2
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Inequalities
Intervals
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Inequalities
Methods for solving inequalities in x are similar to those
used for solving equations.
In particular, we often use properties of inequalities to
replace a given inequality with a list of equivalent
inequalities, ending with an inequality from which solutions
are easily obtained.
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Inequalities
The properties in the following chart can be proved for real
numbers a, b, c, and d.
Properties of Inequalities
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Inequalities
It is important to remember that multiplying or dividing both
sides of an inequality by a negative real number reverses
the inequality sign (see property 4).
Properties similar to those above are true for other
inequalities and for  and .
Thus, if a > b, then a + c > b + c; if a  b and c < 0, then
ac  bc; and so on.
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Example 3 – Solving an inequality
Solve the inequality –6 < 2x – 4 < 2.
Solution:
A real number x is a solution of the given inequality if and
only if it is a solution of both of the inequalities
–6 < 2x – 4 and 2x – 4 < 2.
This first inequality is solved as follows:
–6 < 2x – 4
given
–6 + 4 < (2x – 4) + 4
add 4
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Example 3 – Solution
–2 < 2x
cont’d
simplify
divide by 2
–1 < x
x > –1
simplify
equivalent inequality
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Example 3 – Solution
cont’d
The second inequality is then solved:
2x – 4 < 2
given
2x < 6
add 4
x<3
divide by 2
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Example 3 – Solution
cont’d
Thus, x is a solution of the given inequality if and only if
both
x > –1
and
x < 3;
that is,
–1 < x < 3.
Hence, the solutions are all numbers in the open interval
(–1, 3) sketched in Figure 5.
Figure 5
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Example 3 – Solution
cont’d
An alternative (and shorter) method is to solve both
inequalities simultaneously—that is, solve the continued
inequality:
–6 < 2x – 4 < 2
given
–6 + 4 < 2x < 2 + 4
add 4
–2 < 2x < 6
–1 < x < 3
simplify
divide by 2
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Example 4 – Solving a continued inequality
Solve the continued inequality
Solution:
A number x is a solution of the given inequality if and only if
and
We can either work with each inequality separately or solve
both inequalities simultaneously, as follows (keep in mind
that our goal is to isolate x):
given
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Example 4 – Solution
–10  4 – 3x < 2
–10 – 4  –3x < 2 – 4
–14  –3x < –2
cont’d
multiply by 2
subtract 4
simplify
divide by –3; reverse
the inequality signs
simplify
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Example 4 – Solution
cont’d
equivalent inequality
Thus, the solutions of the inequality are all numbers in the
half-open interval
sketched in Figure 6.
Figure 6
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Example 5 – Solving a rational inequality
Solve the inequality
Solution:
Since the numerator is positive, the fraction is positive if
and only if the denominator, x – 2, is also positive.
Thus, x – 2 > 0 or, equivalently, x > 2, and the solutions are
all numbers in the infinite interval (2, ) sketched in
Figure 7.
Figure 7
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Inequalities
It follows that the solutions of an inequality such as | x| < 3
consist of the coordinates of all points whose distance from
O is less than 3.
This is the open interval (–3, 3) sketched in Figure 10.
Figure 10
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Inequalities
Thus,
|x| < 3 is equivalent to –3 < x < 3.
Similarly, for |x| > 3, the distance between O and a point
with coordinate x is greater than 3; that is,
|x| > 3 is equivalent to x < –3 or x > 3.
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Inequalities
The graph of the solutions to |x| > 3 is sketched in
Figure 11. We often use the union symbol  and write
(
, –3)  (3,
)
to denote all real numbers that are in either (
(3, ).
, –3) or
Figure 11
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Inequalities
The notation
(
, 2)  (2,
)
represents the set of all real numbers except 2.
The intersection symbol  is used to denote the elements
that are common to two sets. For example,
(
, 3)  (–3,
) = (–3, 3),
since the intersection of (
, 3) and (–3, ) consists of
all real numbers x such that both x < 3 and x > –3.
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Inequalities
The preceding discussion may be generalized to obtain the
following properties of absolute values.
In the next example we use property 1 with a = x – 3 and
b = 0.5.
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Example 7 – Solving an inequality containing an absolute value
Solve the inequality |x – 3| < 0.5.
Solution:
|x – 3| < 0.5.
–0.5 < x – 3 < 0.5
–0.5 + 3 < (x – 3) + 3 < 0.5 + 3
2.5 < x < 3.5
given
property 1
isolate x by adding 3
simplify
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Example 7 – Solution
cont’d
Thus, the solutions are the real numbers in the open
interval (2.5, 3.5). The graph is sketched in Figure 12.
Figure 12
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Inequalities
In the next example we use property 2 with a = 2x + 3 and
b = 9.
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Example 8 – Solving an inequality containing an absolute value
Solve the inequality |2x + 3| > 9.
Solution:
given
|2x + 3| > 9
2x + 3 < –9
or
2x + 3 > 9
property 2
2x < –12
or
2x > 6
subtract 3
x < –6
or
x>3
divide by 2
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Example 8 – Solution
cont’d
Consequently, the solutions of the inequality | 2x + 3| > 9
consist of the numbers in (
, 6)  (3, ). The graph is
sketched in Figure 13.
Figure 13
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Inequalities
The trichotomy law states that for any real numbers a and b
exactly one of the following is true:
a > b, a < b, or a = b
Thus, after solving |2x + 3| > 9 in Example 8, we readily
obtain the solutions for |2x + 3| < 9 and
|2x + 3| = 9 —namely, (–6, 3) and {–6, 3}, respectively.
Note that the union of these three sets of solutions is
necessarily the set of real numbers.
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Inequalities
When using the notation a < x < b, we must have a < b.
Thus, it is incorrect to write the solution x < –6 or x > 3 (in
Example 8) as 3 < x < –6.
Another misuse of inequality notation is to write a < x > b,
since when several inequality symbols are used in one
expression, they must point in the same direction.
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2.7
More on Inequalities
Copyright © Cengage Learning. All rights reserved.
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More on Inequalities
To solve an inequality involving polynomials of degree
greater than 1, we shall express each polynomial as a
product of linear factors ax + b and/or irreducible quadratic
factors ax2 + bx + c.
If any such factor is not zero in an interval, then it is either
positive throughout the interval or negative throughout the
interval.
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More on Inequalities
Hence, if we choose any k in the interval and if the factor is
positive (or negative) for x = k, then it is positive (or
negative) throughout the interval.
The value of the factor at x = k is called a test value of the
factor at the test number k. This concept is exhibited in the
following example.
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Example 1 – Solving a quadratic inequality
Solve the inequality 2x2 – x < 3.
Solution:
To use test values, it is essential to have 0 on one side of
the inequality sign.
Thus, we proceed as follows:
2x2 – x < 3
2x2 – x – 3 < 0
(x + 1)(2x – 3) < 0
given
make one side 0
factor
The factors x + 1 and 2x – 3 are zero at –1 and
respectively.
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Example 1 – Solution
cont’d
The corresponding points on a coordinate line (see Figure 1)
determine the nonintersecting intervals
(
, –1), (–1, ) and
Figure 1
We may find the signs of x + 1 and 2x – 3 in each interval by
using a test value taken from each interval.
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Example 1 – Solution
cont’d
To illustrate, if we choose k = – 10 in (
, –1), the values
of both x + 1 and 2x – 3 are negative, and hence they are
negative throughout (
, –1).
A similar procedure for the remaining two intervals gives us
the following sign chart, where the term resulting sign in the
last row refers to the sign obtained by applying laws of
signs to the product of the factors.
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Example 1 – Solution
cont’d
Note that the resulting sign is positive or negative
according to whether the number of negative signs of
factors is even or odd, respectively.
Sometimes it is convenient to represent the signs of x + 1
and 2x – 3 by using a coordinate line and a sign diagram,
of the type illustrated in Figure 2.
Figure 2
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Example 1 – Solution
cont’d
The vertical lines indicate where the factors are zero, and
signs of factors are shown above the coordinate line. The
resulting signs are shown in red.
The solutions of (x + 1)(2x – 3) < 0 are the values of x for
which the product of the factors is negative—that is, where
the resulting sign is negative.
This corresponds to the open interval (–1, ).
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More on Inequalities
The following warning shows this incorrect extension
applied to the inequality in Example 1.
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Example 3 – Using a sign diagram to solve an inequality
Solve the inequality
Solution:
Since 0 is already on the right side of the inequality and the
left side is factored, we may proceed directly to the sign
diagram in Figure 4, where the vertical lines indicate the
zeros (–2, –1, and 3) of the factors.
Figure 4
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Example 3 – Solution
cont’d
The frame around the –1 indicates that –1 makes a factor
in the denominator of the original inequality equal to 0.
Since the quadratic factor x2 + 1 is always positive, it has
no effect on the sign of the quotient and hence may be
omitted from the diagram.
The various signs of the factors can be found using test
values.
Alternatively, we need only remember that as x increases,
the sign of a linear factor ax + b changes from negative to
positive if the coefficient a of x is positive, and the sign
changes from positive to negative if a is negative.
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