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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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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
Original material © Cambridge University Press 2017
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.
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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
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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
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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.
Original material © Cambridge University Press 2017
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
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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
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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
Original material © Cambridge University Press 2017