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CHAPTER 3
Inequalities
3
Learning objectives
After studying this chapter, you should be able to:
■ use sign diagrams to solve inequalities
■ solve inequalities involving rational expressions.
This chapter covers the various techniques for solving inequalities
involving linear rational expressions.
3.1 Introduction
You solve inequalities in a similar way to solving equations.
However, instead of the = sign, you manipulate one of the
following signs:
greater than
greater than or equal to
less than
less than or equal to
You have learned to solve simple linear and quadratic
inequalities in the C1 module. An example is given below to
remind you.
Worked example 3.1
(a) Solve the linear inequality 2(3 5x) 6 2(3x 4).
(b) Solve the quadratic inequality x(x 2) 15.
Solution
(a)
⇒
⇒
⇒
⇒
2(3 5x) 6 2(3x 4)
6 10x 6 6x 8
10x 6x 6 8 6
4x 8
x2
Collect the xs on the left and the
numbers on the right.
Simplify.
Divide by 4, but notice how
the inequality is reversed.
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Inequalities
(b)
⇒
⇒
x(x 2) 15
x2 2x 15 0
(x 5)(x 3) 0
Multiply out, collect terms and
factorise.
Let f(x) (x 5)(x 3).
We introduced the term critical values in C1. The critical
values for this particular expression are x 5 and x 3.
Consider a number line with these critical values marked.
5
3
5 x 3
x 5
x3
The line has been cut into three separate regions:
x 5, 5 x 3 and x 3.
The method consists of choosing a value in each of these
regions and calculating the value of the expression f(x).
You can then draw a sign diagram for f(x) (x 5)(x 3).
5
This is because these values
make each of the brackets in
turn equal to 0.
x 5, choose x 6
f(6) (6 5)(6 3)
1 9 9
5 x 3, choose x 0
f(0) (0 5)(0 3)
5 3 15
x 3, choose x 4
f(4) (4 5)(4 3)
919
3
There are two regions where f(x) 0.
The solution is therefore in two parts: x 5, x 3.
There are important differences between solving equations
and solving inequalities.
1 An inequality will have a range of values as its solution.
2 Whenever you multiply or divide an inequality by a
negative number you must also reverse the inequality
sign.
3.2 Inequalities involving rational
expressions
Firstly, what does the term ‘rational expression’ mean?
Quite simply, a rational expression is one that contains an
x7
algebraic fraction such as .
x2
x7
Consider the equation 3.
x2
You can multiply both sides by (x 2) as this eliminates the
fraction.
⇒ (x 7) 3(x 2) 3x 6
⇒ 13 2x
⇒ x 612
You need to know when the
expression is positive.
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Inequalities
How do you solve an inequality involving a rational expression
x7
such as 3?
x2
One approach is to make sure you are multiplying throughout by a
quantity such as (x 2)2 since a squared quantity is never negative.
Hence, there is no need to reverse the inequality.
3.3 Multiplying both sides by the
square of the denominator
Warning: You cannot simply
multiply both sides by (x 2)
since you don’t know whether
its value is positive or negative.
If it were negative then you
would have to reverse the
inequality sign.
If (x 2) 0 you would keep
the inequality sign the same.
3
Worked example 3.2
x7
Solve the inequality 3.
x2
Solution
x7
3
x2
Multiplying both sides by (x 2)2 gives
(x 7)(x 2)2
3(x 2)2
( x 2)
⇒ (x 7)(x 2) 3(x 2)2
⇒ (x 7)(x 2) 3(x 2)2 0
⇒ (x 2)[(x 7) 3(x 2)] 0
⇒ (x 2)(13 2x) 0
Take all terms to the left-hand
side.
Take (x 2) as a common factor.
Simplify the second bracket.
You then continue in the usual way for solving quadratic
inequalities.
Let
f(x) (x 2)(13 2x)
The critical values are x 2 and x 123.
A sign diagram for (x 2)(13 2x) is given below
13
2
2
The critical values cut the line into three regions:
x 2, 2 x 123 and x 123.
There are two regions which gave a negative value for f(x).
The solution is in two parts: x 2, x 123.
When solving quadratic/cubic/higher order inequalities you
must consider the critical values.
You calculate the value of f(x) in each of the regions on the
number line created by the critical values and produce a
sign diagram.
31
You are trying to solve
(x 2)(13 2x) 0.
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Inequalities
Worked example 3.3
3x 5
Solve the inequality x 3.
1x
Solution
3x 5
x 3
1x
⇒
(3x 5)(1 x) (x 3)(1 x)2
⇒
(3x 5)(1 x) (x 3)(1 x)2 0
⇒
(1 x)[(3x 5) (x 3)(1 x)] 0
⇒
(1 x)(3x 5 x2 4x 3) 0
⇒
(1 x)(x2 x 2) 0
⇒
(1 x)(x 2)(x 1) 0
Let f(x) (1 x)(x 2)(x 1).
The critical values are x 1, x 1, and x 2.
A sign diagram for f(x) is given below:
1
1
f(0) (1 0)(0 2)(0 1)
(1)(2)(1) 2
2
The question asks when the expression is greater than or equal
to zero. There are two regions which gave a positive value for
f(x), so the solution is x 1 or 1 x 2.
EXERCISE 3A
Solve each of the following by multiplying throughout by the
square of the denominator.
x3
1 1
2x
x1
2 2
x4
x5
3 3
x2
1
4 2
x7
6x
5 5
x4
5
6 4
x6
2x 3 1
7 x4
3
x
8 x
x3
x(x 2)
9 3
2x 5
f(2) (1 2)(2 2) (2 1)
(3)(4)(1) 12
x(x 5)
10 2
x4
Notice that the last inequality
must have 1 x since x cannot
equal 1 in the original inequality.
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Inequalities
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3.4 Combining terms into a single
fraction
Another method for solving inequalities of this type is to take all
terms onto one side of the inequality and combine them into a
single algebraic fraction.
Worked example 3.4
4x 5
Find the solution of the inequality 2.
x3
Solution
⇒
⇒
⇒
4x 5
4x 5
2 ⇒ 2 0
x3
x3
4x 5 2(x 3)
0
x3
x3
4x 5 2x 6
0
x3
(2x 11)
0
(x 3)
3
You must have 0 on the righthand side.
Subtract the fractions using a
common denominator.
Take great care with the signs.
Simplify. The brackets are not
strictly necessary here but are
used to emphasise the method.
Again, you find the critical values.
(2x 11)
Let f(x) .
( x 3)
These values make each bracket
in turn equal to 0.
The critical values are x 121 and x 3.
11
2
3
The critical values cut the line into three regions: x 121,
121 < x 3 and x 3.
The sign diagram is shown above:
f(6) 19, f(0) 131, f(4) 19
(2x 11)
You need to find when 0.
( x 3)
There are two regions which gave a positive value for f(x).
The solution is x 121 or x 3.
When solving inequalities involving a rational expression,
you must ensure that the expression is written in an
appropriate form (i.e. factorised/simplified as far as
possible and with 0 on the right-hand side).
You can then find the critical values for the numerator
and denominator.
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Inequalities
Worked example 3.5
x5 x2
Solve the inequality .
x3 x4
Solution
x5 x2
x3 x4
⇒
x5 x2
0
x3 x4
⇒
(x 5)(x 4) (x 2)(x 3)
0
(x 3)(x 4) (x 3)(x 4)
⇒
x2 9x 20
x2 5x 6
0
(x 3)(x 4) (x 3)(x 4)
⇒
x2 9x 20 x2 5x 6
0
(x 3)(x 4)
⇒
(14x 14)
0
(x 3)(x 4)
⇒
14(x 1)
0
(x 3)(x 4)
Finding a few values:
14(x 1)
Let f(x) .
(x 3)(x 4)
The critical values are x 4, x 1 and x 3. A sign diagram
can be drawn:
4
1
3
The critical values cut the line into four regions:
x 4, 4 x 1, 1 x 3 and x 3.
14(x 1)
You need to find when 0.
(x 3)(x 4)
There are two regions which give a negative value for f(x) so the
solution is x 4 or 1 < x < 3.
14(5 1)
f(5) (5 3)(5 4)
5 6
7
(8)(1)
14(0 1)
f(0) (0 3)(0 4)
7
14
6
(3)(4)
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Inequalities
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Worked example 3.6
2x 3
Solve the inequality x 2.
x6
Note that the inequality is not
defined when x 6.
Solution
2x 3
x 2
x6
3
⇒
2x 3
(x 2) 0
x6
⇒
2x 3 (x 2)(x 6)
0
x6
x6
⇒
2x 3 (x2 8x 12)
0
x6
⇒
x2 10x 9
0
( x 6)
⇒
(x2 10x 9)
0
(x 6)
⇒
(x 1)(x 9)
0
(x 6)
(0 1)(0 9)
f(0) (0 6)
9 3
6 2
(x 1)(x 9)
Let f(x) .
(x 6)
The critical values are x 1, x 6 and x 9. A sign diagram is
drawn below:
1
6
9
The critical values suggest four possible intervals to be
considered for the solution:
x 1, 1 x 6, 6 x 9 and x 9.
(x 1)(x 9)
You need to find when 0.
(x 6)
There are two regions which give a positive value for f(x).
The solution is therefore x 1 or 6 x 9.
(7 1)(7 9)
f(7) (7 6)
12
12
1
A very important point here
is that you cannot have x 6 as
part of a solution since the
denominator of f(x) becomes
zero when x 6. This means
that the sign cannot be
attached to the critical value 6.
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Inequalities
EXERCISE 3B
Solve the following inequalities by taking all the terms onto one
side and combining them into a single fraction:
3x 1
1 1
2x 5
7x 2
2 5
x3
5x 2
3 x
x2
4
4 x 1
x2
3
2
5 x1 x2
1
x
6 x4 x2
x
4
7 x3 x
x
x
8 x3 x2
x3 x1
9 x1 x4
x1 x2
10 x3 x5
There are various methods for solving inequalities involving
rational expressions, such as:
1 multiplying throughout by the square of the denominator;
2 combining all the fractions into one single term on one
side of the inequality.
MIXED EXERCISE
Solve inequalities 1–10 by any appropriate method.
5x 4
1 3
x2
x
x
2 x1 x3
6x 5
3 4
x1
3x 4
4 1
2x
6
5 x
x5
5x
6 2 20
(x 3)
x6
7 3
x(x 2)
x3 x5
8 x5 x2
(x 2)(x 3)(x 4)
9 0
(x 1)
2
10 x 4
x5
4x 3
11 Solve the inequality x 3.
x1
[A]
3x 5
12 Solve the inequality 4.
x2
[A]
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Inequalities
37
2x 3
13 A student attempts to solve the inequality 9, and
x2
writes the following statements:
Step 1 2x 3 9(x 2)
Step 2 2x 3 9x 18
Step 3
21 7x
Step 4
x3
(a) Show that, although x 1 satisfies the student’s
solution, it does not satisfy the original inequality.
(b) State, with a reason, the step where the student makes
an error.
(c) Determine the correct solution to the inequality
2x 3
9.
x2
[A]
Key point summary
1 Important differences between solving equations and p30
solving inequalities are:
1 An inequality will have a range of values as its
solution.
2 Whenever you multiply or divide an inequality by a
negative number you must also reverse the
inequality sign.
2 When solving quadratic/cubic/higher order inequalities p31
you need to consider the critical values.
You calculate f(x) in each of the regions of the number
line created by the critical values and produce a sign
diagram.
3 When solving inequalities involving a rational
p33
expression, you must ensure that the expression is
written in an appropriate form (i.e. factorised/simplified
as far as possible and with 0 on the right-hand side).
You can then find the critical values for the numerator
and denominator.
4 There are various methods for solving inequalities
involving rational expressions, such as:
1 multiplying throughout by the square of the
denominator;
2 combining all the fractions into one single term on
one side of the inequality.
p36
3
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Inequalities
Test yourself
What to review
(2x 1)(x 2)
1 Solve the inequality 0
(x 5)
Sections 3.3, 3.4
4x 5
2 Solve the inequality 3.
x2
Sections 3.3, 3.4
x5
3 Solve the inequality x 2.
x7
Sections 3.3, 3.4
x3 x2
4 Solve x6 x2
Sections 3.3, 3.4
ANSWERS
3 1 x 7 or x 9
2 11 x 2
1 5 x 21 or x 2
Test yourself
4 x 6 or 2 x 56
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