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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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(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) 9 19
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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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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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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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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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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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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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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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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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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