Preview Cambridge IGCSE® and O Level Additional Mathematics Coursebook (Second Edition)

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Chapter 5 Equations, inequalities and graphs This section will show you how to:

solve graphically or algebraically equations of the type ax + b = cx + d

■

solve graphically or algebraically inequalities of the type ax + b > c (c 0 ) ,

■

solve cubic inequalities in the form k (x − a )(x − b )(x − c ) d graphically

SA ■

■ ■

ax + b c ( c > 0 ) and ax + b cx + d

sketch the graphs of cubic polynomials and their moduli, when given in factorised form use substitution to form and solve quadratic equations.

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Cambridge IGCSE and O Level Additional Mathematics

5.1 Solving equations of the type | ax + b | = | cx + d | class discussion

• • •

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Using the fact that a 2 = a 2 and b 2 = b 2 you can say that: a2 − b2 = a 2 − b 2 Using the difference of two squares then gives: a 2 − b 2 = ( a − b )( a + b ) Using the statement above, explain how these three important results can be obtained: (The symbol ⇔ means ‘is equivalent to’.) a = b ⇔ a2 = b2 a > b ⇔ a2 > b2

a < b ⇔ a 2 < b 2 , if b ≠ 0

The next worked example shows you how to solve equations of the form cx + d = ex + f using algebra.

To solve this type of equation you can use the techniques that you learnt in Chapter 1 or you can use the rule: a = b ⇔ a2 = b2

worked example 1

Solve the equation x − 5 = x + 1 using an algebraic method.

Answers

Method 1: x − 5 = x +1 x − 5 = x + 1 or 0=6 or

x − 5 = − (x + 1) 2x = 4 x=2 CHECK: 2 − 5 = 2 + 1 ü The solution is x = 2

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90

Method 2: x − 5 = x +1 ( x − 5 )2 = ( x + 1 )2 x 2 − 10x + 25 = x 2 + 2x + 1 12x = 24 x=2

0 = 6 is false

use a = b ⇔ a 2 = b 2 expand simplify

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Chapter 5: Equations, inequalities and graphs

The equation x − 5 = x + 1 could also have been solved graphically:

y

y = |x + 1|

y = |x – 5|

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6 5 4 3 2 1

–8

–6

–4

–2

O

2

4

6

8

10

12 x

The solution is the x-coordinate where the two graphs intersect.

-----------------------------------------------------------------------------------------------------------------------

worked example 2

Solve the equation 2x + 1 = x − 3 using an algebraic method.

Answers

Method 1 2x + 1 = x − 3 2x + 1 = x − 3 x = –4

2x + 1 = − (x − 3) 3x = 2 2 x = 3 2 2 CHECK: 2 ( −4 ) + 1 = −4 − 3 ü, 2 + 1 = −3 ü 3 3 2 Solution is : x = or x = −4 3 or or

SA

Method 2

2x + 1 = x − 3

use a = b ⇔ a 2 = b 2

( 2x + 1 )2 = ( x − 3 )2

expand

4x 2 + 4x + 1 = x 2 − 6x + 9

simplify

3x + 10x − 8 = 0 2

factorise

(3x − 2)(x + 4 ) = 0 2 x = or x = −4 3

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Cambridge IGCSE and O Level Additional Mathematics

The equation 2x + 1 = x − 3 could also be solved graphically:

y

–6

–4

y = |2x – 5|

y = |x – 3|

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8 7 6 5 4 3 2 1

–2

O

2

4

6

8

10

12 x

From the graph, one of the answers is clearly x = –4. The exact value of the other answer is not so obvious and algebra is needed to find this exact value.

The graph of y = x − 3 can be written as: if

xù3

y = –(x – 3)

if

x<3

The graph of y = 2x + 1 can be written as: 1 y = 2x + 1 if x − 2 1 y = –(2x + 1) if x <− 2 The second answer is found by finding the x value at the point where y = –(x – 3) and y = 2x + 1 intersect.

2x + 1 = –x – 3

2x + 1 = –x + 3

3x = 2 2 x= 3 2 Hence the solution is x = –4 or x = .

3

SA

92

y=x–3

Original material © Cambridge University Press 2017

Chapter 5: Equations, inequalities and graphs

worked example 3 Solve x + 4 + x − 5 = 11.

Answers = 11 = 11 − x − 5 = 11 − x − 5 = x − 5 − 11

subtract x − 5 from both sides split the equation into two parts

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x +4 + x −5 x +4 x +4 x +4

------(1) ------(2)

Using equation (1): x −5 = 7−x split this equation into two parts x–5=7–x or x – 5 = –(7 – x) 2x = 12 or 0 = –2 0 = –2 is false x=6

Using equation (2): x − 5 = x + 15 split this equation into two parts x – 5 = x + 15 or x – 5 = (x + 15) 0 = 20 or 2x = –10 0 = 20 is false x = –5 CHECK: 6 + 4 + 6 − 5 = 11 ü, −5 + 4 + −5 − 5 = 11 ü The solution is x = 6 or x = –5.

Exercise 5.1 1 Solve. a

b

x +5 = x −4

c

2x − 3 = 4 − x

5x + 1 = 1 − 3x

e

1 − 4x = 2 − x

f

1−

3x − 2 = 2x + 5

h

2x − 1 = 2 3 − x

i

2−x = 5

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d

2x − 1 = x

g

2 Solve the simultaneous equations y = x − 5 and y = 8 − x .

3 Solve the equation 6 x + 2 2 + 7 x + 2 − 3 = 0.

4 a

Solve the equation x 2 − 6 x + 8 = 0.

b

Use graphing software to draw the graph of f ( x ) = x 2 − 6 x + 8 .

c

Use your graph in part b to find the range of the function f.

CHALLENGE Q

5 Solve the equation x + 1 + 2x − 3 = 8.

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x = 3x + 2 2

1 x +1 2

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Cambridge IGCSE and O Level Additional Mathematics

CHALLENGE Q

6 Solve the simultaneous equations y = x − 5 and y = 3 − 2x + 2.

CHALLENGE Q

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7 Solve the equation 2 3x + 4 y − 2 + 3 25 − 5x + 2 y = 0.

5.2 Solving modulus inequalities

Two useful properties that can be used when solving modulus inequalities are: a b ⇔ −b a b

and

a b ⇔ a −b or a b

The following examples illustrate the different methods that can be used when solving modulus inequalities.

worked example 4 Solve 2x − 1 < 3.

Answers

Method 1: using algebra 2x − 1 < 3 use a < b ⇔ −b < a < b –3 < 2x – 1 < 3 –2 < 2x < 4 –1 < x < 2

Method 2: using a graph The graphs of y = 2x − 1 and y = 3 intersect at the points A and B. 2x – 1

if

–(2x – 1)

if

1 2 1 x < 2

x

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2x − 1 =

y 5

y = |2x – 1|

4 y=3

3 At A, the line y = –(2x – 1) intersects the line y = 3.

A B –(2x – 1) = 3 2 –2x + 1 = 3 1 2x = –2 x = –1 –2 –1 O 1 2 3x At B, the line y = 2x – 1 intersects the line y = 3.

–1 2x –1 = 3 2x = 4 x =2 To solve the inequality 2x − 1 < 3 you must find where the graph of the function y = 2x − 1 is below the graph of y = 3.

Hence, the solution is –1 < x < 2.

Original material © Cambridge University Press 2017

Chapter 5: Equations, inequalities and graphs

worked example 5 Solve 2x + 3 > 4.

Answers Method 1: using algebra 2x + 3 > 4

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use a > b ⇔ a ( −b or a ) b

2x + 3 < –4 2x < –7 7 x < –__ 2

or or

or

2x + 3 > 4 2x > 1 1 x > __ 2

Method 2: using a graph The graphs of y = 2x + 3 and y = 4 intersect at the points A and B. 2x + 3

if

–(2x + 3)

if

2x + 3 =

1 2 1 x < −1 2

x −1

At A, the line y = –(2x + 3) intersects the line y = 4.

–(2x + 3) = 4 –2x – 3 = 4 2x = –7 7 x = – __ 2 At B, the line y = 2x + 3 intersects the line y = 4.

y 5

A

4 3

y = |2x + 3| B

y=4

2

1

–4 –3 –2 –1 O –1

1x

SA

2x + 3 = 4 2x = 1 1 x = __ 2

To solve the inequality 2x + 3 > 4 you must find where the graph of the function y = 2x + 3 is above the graph of y = 4. 1  7 Hence, the solution is x  − or x  . 2  2

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Cambridge IGCSE and O Level Additional Mathematics

worked example 6 Solve the inequality 2x + 1 3 − x .

Answers

Method 1: using algebra use a b ⇔ a 2 b 2

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2x + 1 3 − x ( 2x + 1 )2 ( 3 − x )2 4x 2 + 4x + 1 9 − 6x + x 2 3x 2 + 10x − 8 0 ( 3x − 2 )( x + 4 ) 0 Critical values are

factorise

2 and –4.

3

y = 3x 2 + 10 x – 8

+

+

–4

2 – 3

–

Hence, x −4 or x

2 .

3

Method 2: using a graph The graphs of y = 2x + 1 and y = 3 − x intersect at the points A and B. 1 2x + 1 if x − 2 2x + 1 = 1 –(2x + 1) if x < − 2

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y

9 8 7 6 5 4 3 2 1

–7 –6 –5 –4 –3 –2 –1 O

y = |2x + 1| y = |3 – x |

1 2 3 4 5 6 7 8 9 x

Original material © Cambridge University Press 2017

Chapter 5: Equations, inequalities and graphs

3−x = x −3 =

x–3 –(x – 3)

if if

x 3 x <3

At A, the line y = –(2x + 1) intersects the line y = –(x – 3).

2x + 1 = x – 3

x = –4

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At B, the line y = 2x + 1 intersects the line y = –(x – 3).

2x + 1 = –(x – 3) 3x = 2 2 x = __ 3 To solve the inequality 2x + 1 3 − x you must find where the graph of the function y = 2x + 1 is above the graph of y = 3 − x .

2 Hence, x −4 or x .

3

Exercise 5.2 y

1

y = |x – 2|

8 7 6 5 4 3 2 1

y = |2x – 10|

1 2 3 4 5 6 7 8 9 10 x

–1 –1

SA

The graphs of y = x − 2 and y = 2x − 10 are shown on the grid.

Write down the set of values of x that satisfy the inequality x − 2 > 2x − 10 .

2 a

b

On the same axes sketch the graphs of y = 3x − 6 and y = 4 − x .

Solve the inequality 3x − 6 4 − x .

3 Solve. a

2x − 3 > 5

b

4 − 5x 9

c

8 − 3x < 2

d

2x − 7 > 3

e

3x + 1 > 8

f

5 − 2x 7

b

5 + x > 7 − 2x

c

x − 2 − 3x 1

c

x > 3x − 2

f

2x < x − 3

4 Solve. a

2x − 3 x − 1

5 Solve. a

2x − 1 3x

b x +1 > x

d

4x + 3 > x

e

x + 3 2x

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Cambridge IGCSE and O Level Additional Mathematics

6 Solve. a

x +1 > x − 4

b x −2 x +5

d

2x + 3 x − 3

e

c

x + 1 3x + 5

1 x −5 2

f

3x − 2 x + 4

b 3 x − 1 < 2x + 1

c

2x − 5 3 2x + 1

x +2 <

7 Solve. a 2 x − 3 > 3x + 1

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8 Solve the inequality x + 2k x − 3k where k is a positive constant.

9 Solve the inequality x + 3k < 4 x − k where k is a positive constant.

CHALLENGE Q

10 Solve 3x + 2 + 3x − 2 8.

5.3 Sketching graphs of cubic polynomials and their moduli

In this section you will learn how to sketch graphs of functions of the form y = k (x − a )(x − b )(x − c ) and their moduli.

To help find the general shape of the curve you need to consider what happens to

• y as x tends to positive infinity (i.e. as x → + ∞) • y as x tends to negative infinity (i.e. as x → – ∞)

worked example 7 a b

Sketch the graph of the function Hence sketch the graph of

.

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98

When sketching graphs of this form you should show the general shape of the curve and all of the axis intercepts.

Answers a

When x = 0, y = –1 × 2 × 1 = –2. ∴ The curve intercepts the y -axis at (0, –2).

When y = 0, (2x – 1)(2 – x)(x + 1) = 0 2x – 1 = 0 2–x =0 x+1=0 1 x = __ x =2 x = –1 2 1 ∴ The curve intercepts the x-axis at  , 0 , (2, 0) and (–1, 0). 2 

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