Cambridge International
AS & A Level Mathematics
Pure Mathematics 1
STUDENT’S BOOK: Worked solutions
Helen Ball, Chris Pearce
Series Editor: Dr Adam Boddison
Pure Mathematics 1 International Students Book Title page.indd 1 14/11/17 10:46 pm
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PM
, 1
WORKED SOLUTIONS
Worked solutions
1 Quadratics
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge y
15
1 2x2 – 5x – 3 = 0
10
(2x + 1)(x – 3) = 0
5
1
x=− or x = 3
2 0 x
–25 –20 –15 –10 –5 5 10 15 20 25
–5
2 3x2 + x – 7 = 0
–10
a = 3, b = 1, c = −7
−1 ± 12 − (4)(3)(–7)
x= 3 a x2 – 8x – 5 = (x − 4)2 – 16 − 5
6
= (x − 4)2 − 21
−1 ± 1 + 84
( ) − 94 − 7
x= 2
6 3
b x2 + 3x – 7 = x +
2
−1 ± 85
x=
= (x + ) −
6 3 37
2
x = 1.37 or − 1.70 2 4
c 2x + 3x + 9 = 2 ( x +
2 2)
2 3x 9
3 5x – y = 13 1 2 +
2x + y = 1 2
= 2 ( x + ) −
Add 1 and 2 . 3 9 9
2
+
7x = 14 4 16 2
x = 2, y = −3
= 2 ( x + ) +
3 63
2
4 3x – 5 < 7 4 16
3x < 12
x – 5x + 7 = ( x − ) −
2
5 25
x < 4 4 4 2 +7
2 4
y = (x − ) +
2
–1 0 1 2 3 4 5 6 7 5 3
2 4
Exercise 1.1A At the turning point y has a minimum value
5 5 3
1 a x2 + 4x = (x + 2)2 − 4 i.e. when x = . When x = , y = , the
2 2 4
b 2x2 – 8x = 2(x2 – 4x) coordinates of the turning point.
= 2[(x − 2)2 − 4] y
25
= 2(x – 2)2 – 8
20
c x2 + 8x + 7 = (x + 4)2 – 16 + 7
15
= (x + 4)2 − 9
10
2 x2 – 10x + 11 = (x − 5)2 – 25 + 11 5
= (x − 5)2 − 14
–25 –20 –15 –10 –5
0
5 10 15 20 25
x
–5
1
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, 1 Quadratics
5
2 13
2 (
2 ( x + 2) − = 2 x 2 + 4x + 4 −
13
2 ) 9
5
4
3
4x 2 + 5x − 3 = 4 x 2 + x −
4
= 2x2 + 8x + 8 − 13
( ) − ( 85 ) − 34
2 2
5
= 2x2 + 8x − 5 = 4 x +
8
6
1
a 5x 2 + x − 3 = 5 x 2 +
2
x 3
−
10 5 ( )
( ) − 6473
2
5
( )
1
2
1 3 = 4 x +
= 5 x + − − 8
20 400 5
( ) − 1673
2
5
= 5 ( x + )
1 241
2 =4 x+
− 8
20 400
When 4x2 + 5x – 3 = 0
= 5( x +
20 )
2
1 241
( ) − 1673 = 0
− 2
80 5
4 x+
b 5x – 18 = 5 ( x − 5 ) but can go no further.
8
2 18
2
4(x + ) =
2
5 73
c 5 – 7x – 3x = −3 ( x +
3 3)
7x 5 2 8 16
2 −
( x + 85 ) = 6473
2
= −3 ( x + ) −
7 49 5
2
−
6 36 3
5 73
= −3 ( x + ) −
7 109
2 x+ =±
8 64
6 36
−5 73
= −3( x + ) +
7 109
2 x= ±
8 64
6 12
−5 ± 73
2x – 3x + 11 = 2 ( x − + )
x=
2 3x 11
2 8
7
2 2 2
b b2
10 ax 2 + bx + c = ax + − +c
= 2 ( x − ) −
9 11
2
3 2 a 4a
+
4 16 2 ax2 + bx + c = 0
= 2 ( x − ) +
2
3 79
2
b b2
4 16 ax + − 4a + c = 0
2 a
2 ( x − ) +
3 79
2
b
2
b2 b 2 − 4ac
>0 ax + = 4a − c =
4 16 2 a 4a
Consequently b2 – 4ac < 0 because the curve of
b b 2 − 4ac
this equation will not intersect the x-axis and ax + =±
2 a 4a
so there are no real roots.
y b ± b 2 − 4ac
ax + =
25 2 a 2 a
20
b b 2 − 4ac
ax = − ±
15 2 a 2 a
10
−b ± b 2 − 4ac
5 ax =
2 a
–20 –15 –10 –5
0
5 10 15 20 25
x
−b ± b 2 − 4ac
x=
a ×2 a
8 (1 – 2x) [(x + 4)2 − 1] = (1 – 2x)(x2 + 8x + 16 − 1)
= (1 – 2x) (x2 + 8x + 15) −b ± b 2 − 4ac
x=
= x2 + 8x + 15 – 2x3 – 16x2 – 30x 2a
= 15 – 22x – 15x2 – 2x3
2
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, 1
WORKED SOLUTIONS
Exercise 1.2A b2 – 4ac = 0 so two equal real roots.
c 3x2 + 12 – 5x = 0
1 a x2 + 2x + 13 = 0
b2 – 4ac = (−5)2 – (4)(3)(12)
b2 – 4ac = (2)2 – (4)(1)(13)
= −119
= −48
b2 – 4ac < 0 so no real roots.
b2 – 4ac < 0 so no real roots.
b 2x2 + 3x + 1 = 0 6 2x2 + 5x – c = 0
b2 – 4ac = (3)2 – (4)(2)(1) a No real roots:
=1 b2 – 4ac < 0
b2 – 4ac > 0 so two distinct real roots. (5)2 – (4)(2)(−c) < 0
c x2 + 4x + 4 = 0 25 + 8c < 0
b2 – 4ac = (4)2 – (4)(1)(4) 25
c<−
8
=0
b One repeated real root:
b2 – 4ac = 0 so two equal real roots.
b2 – 4ac = 0
2 2x2 – 5x + 3 = 0
(5)2 –(4)(2)(−c) = 0
b2 – 4ac = (−5)2 – (4)(2)(3)
= 25 − 24 25 + 8c = 0
=1 25
c=−
8
b2 – 4ac > 0 so has two distinct real roots.
c Two distinct real roots:
3 a x2 – 3x – 5 = 0
b2 – 4ac > 0
b2 – 4ac = (−3)2 – (4)(1)(−5)
(5)2 – (4)(2)(−c) > 0
= 29
25 + 8c > 0
b2 – 4ac > 0 so two distinct real roots.
25
b 3x2 – 2x + 7 = 0 c>−
8
b2 – 4ac = (−2)2 – (4)(3)(7)
13 − 3x
= −80 7 x=
3
b2 – 4ac < 0 so no real roots. 13 − 3x
x2 =
c x2 – 12x + 36 = 0 3
b2 – 4ac = (−12)2 – (4)(1)(36) 3x2 = 13 – 3x
=0 3x2 + 3x – 13 = 0
2
b – 4ac = 0 so two equal real roots. b2 – 4ac = 9 + 4 × 3 × 13 = 165 > 0
4 x2 + (2k + 2)x + (5k − 1) = 0 As b2 – 4ac > 0 there are 2 distinct solutions.
b2 – 4ac = 0 for one repeated real roots. 8 x2 – 3x + a2 = 0
2
(2k + 2) – (4)(1)(5k – 1) = 0
No real roots when b2 – 4ac < 0
Expanding and simplifying.
k2 – 3k + 2 = 0 or – k2 + 3k – 2 = 0 b2 – 4ac = 9 – 4a2
Both quadratic equations have the solutions k = 1 9 – 4a2 < 0
or k = 2. 9 < 4a2
When k = 1, x2 + 4x + 4 = 0 and x = −2;
9
when k = 2, x2 + 6x + 9 = 0 and x = −3 a2 >
4
5 a 4 + 2x – 3x2 = 0 −3 3
a< or a >
b2 – 4ac = (2)2 – (4)(–3)(4) 2 2
= 52 9 y = 3 – x meets y = x2 – 5x + 7 when 3 – x = x2 – 5x + 7
2
b – 4ac > 0 so two distinct real roots. ie 0 = x2 – 4x + 4
b 49 + x2 – 14x = 0 b2 – 4ac = 16 – 16 = 0
b2 – 4ac = (−14)2 – (4)(1)(49) One repeated root, so the line touches, but
=0 does not cross the curve.
3
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, 1 Quadratics
10 3x – xy + 1 = 0 iii (x − 3)2 – 9 = 0
3x + 1 x = 0 or 6
y=
x
2 Factorisation:
substitute into 3y – xy + 1 = 0 ›› can only be used on equations that factorise
3( 3xx+ 1 ) − x ( 3xx+ 1 ) + 1 = 0 ›› sometimes spotting factors can be difficult
›› can solve: x2 + 3x + 2 = 0
3( ) − 3x = 0
3x + 1
x ›› cannot solve: x2 − 5x – 7 = 0.
3x2 – 9x – 3 = 0 Quadratic formula:
›› can be used to solve any equation with real roots
b2 – 4ac = 81 + 4 × 3 × 3 = 117 > 0 including ones that don’t factorise
As b2 – 4ac > 0 there are two distinct real roots ›› cumbersome and consequently easy to make a
∴ the curves intersect in two places. mistake
Exercise 1.3A ›› can solve: x2 − 5x – 7 = 0
›› cannot solve: x2 − 5x + 7 = 0.
1 a i x2 – 9 = 0
Completing the square:
(x + 3)(x − 3) = 0
›› can be used to solve any equation with real roots
x = 3 or −3 including ones that don’t factorise
0 ± 0 − ( 4 )(1)( −9) ›› can be cumbersome manipulations if b is odd and
ii x = a > 1, and consequently easy to make a mistake
2
x = 3 or −3 ›› can solve: x2 − 5x – 7 = 0
iii x2 = 9 ›› cannot solve: x2 − 5x + 7 = 0.
x = 3 or −3 3 a (x − 4)2 = 13
b i x2 + 3x + 2 = 0 x = 4 ± 13
(x + 1)(x + 2) = 0 Completing the square. The equation was
x = −1 or −2 already in form of a completed square.
−3 ± 9 − ( 4 )(1)( 2) b 2x2 – 5x – 11 = 0
ii x =
2 5 ± 25 − ( 4 )( 2)( −11)
x=
x = −1 or −2 4
( 32 ) − 14 = 0
2
iii x + 5 ± 113
x=
4
x = −1 or −2 Does not factorise so used the quadratic
formula. Alternatively, could have completed
c i x2 – 5x – 7 = 0
the square but chose not to as a > 1.
Cannot be factorised.
c 3x2 = 5 – 14x
5 ± 25 − ( 4 )(1)( −7 ) 3x2 + 14x – 5 = 0
ii x =
2
(3x − 1)(x + 5) = 0
5 ± 53 1
x= x = or −5
2 3
( 52 ) − 534 = 0 Equation factorises so easy to do this.
2
iii x − Alternatively, could have completed the
square but chose not to as a > 1.
5 ± 53
x=
2 d 4x2 – 16x + 7 = 0
d i x(x − 6) = 0 (2x − 1)(2x − 7) = 0
x = 0 or 6 1 7
x = or
2 2
6 ± 36 − ( 4 )(1)( 0 )
ii x = Equation factorises so easy to do this.
2
Alternatively, could have completed the
x = 0 or 6
square but chose not to as a > 1.
4
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, 1
WORKED SOLUTIONS
4 x2 + bx + c = 0 74x + 33
8 x=
7
Subtract any extra values that completing the
b2 74x + 33
square has produced, in this case− . x2 =
4 7
7x2 – 74x – 33 = 0
( )
2
b b2
x+ − +c =0 (7x + 3) (x – 11) = 0
2 4
The equation can be solved by factorising a
Manipulate the equation to make x the subject.
quadratic expression.
( )
2
b b 2 − 4c a 9x2 + 55x – 56 = 0
x+ = 9
2 4
b2 – 4ac = 552 + 4 × 9 × 56 = 5041
2
b ± b − 4c b 2 − 4ac = 71 so the quadratic can be
x+ =
2 2 factorised.
−b ± b 2 − 4c b (9x – 8)(x + 7) = 0
x=
2 8
x = or x = – 7
5 Cannot be factorised if b2 − 4ac is not a square
a 9
number. Exercise 1.4A
b2 – 4ac = 16 – (4)(3)(−11) = 148
1 a x2 +3x – 4 > 0
148 not a square number so does not
factorise. x2 + 3x – 4 = 0
(x + 4)(x − 1) = 0
b 3x2 + 4x – 11 = 0
x < −4 or x > 1
−4 ± 148
x= b x2 – 6x + 8 0
6
(x − 4)(x − 2) = 0
−2 ± 37
x= 2x4
3
c x2 – 9 0
6 (x − 3)(x − 5) = 8
x2 – 8x + 15 – 8 = 0 (x + 3)(x − 3) 0
x2 – 8x + 7 = 0 x −3 or x 3
(x − 7)(x − 1) = 0 d x2 – 6x < 0
x = 1 or 7 x(x − 6) = 0
7 ax2 + bx + c = 0
0<x<6
(
a x2 +
bx c
a a
+ =0 ) 2 a 2x2 + 7x < −3
2x2 +7x + 3 < 0
( )
2
b b2 c
x+ − 2+ =0 (2x + 1)(x + 3) = 0
2a 4a a
1
( )
b
2
b2 c −3 < x < −
x+ = − 2
2a 4a 2 a b −x2 – 3x + 4 > 0
( )
2
b b2 4ac (−x + 1)(x + 4) = 0
x+ = −
2a 4a 2 4a 2 −4 < x < 1
( ) c x2 > 4
2 2
b b − 4ac
x+ =
2a 4a 2 x2 − 4 > 0
(x − 2)(x + 2) > 0
b ± b 2 − 4ac
x+ = x < −2 or x > 2
2a 2a
d 3x2 5x
−b ± b 2 − 4ac
x= 3x2 – 5x 0
2a
x(3x − 5) 0
5
0x
3
5
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, 1 Quadratics
3 a 6 – 5x – x2 > 0 c 12 + 9x +x2 > 3(x + 1)
(6 + x)(1 − x) = 0 12 + 9x + x2 > 3x + 3
−6 < x < 1 9 + 6x + x2 > 0
and – x + 6 > 0 (3 + x)2 > 0
6>x Inequality true for all values of x.
x<6 d x + 1 > 3 – 4x – 3x2
{−5, −4, −3, −2, −1, 0} 0 > 2 – 5x – 3x2
b 2x2 + 9x – 5 0 3x2 + 5x – 2 > 0
(2x − 1)(x + 5) = 0 (3x – 1)(x + 2) > 0
1
−5 x
2 x < –2 or x 1 [all integers excluding –2, –1
3
or 2x < 5 and 0]
5 6 a x2 + 5x – 6 < 0
x<
2
(x + 6)(x – 1) = 0
5
So x <
2 –6 < x < 1 and x2 + 3x – 4 < 0
{all integers less than 2.5} (x + 4)(x – 1) = 0
c (2x + 1)2 –90 –4 < x < 1
2x + 1 = ±3 So –4 < x < 1
−2 x 1 and 2x < 6 {−3, −2, −1, 0}
x<3 b x2 + 5x < 6 or x2 + 3x < 4
−2 x 1 So –6 < x < 1 or – 4 < x < 1
{−2, −1, 0, 1} {−5, −4, −3, −2, −1, 0}
d 4x2 < 3x c x2 – 6 < −5x and x2 – 4 > −3x
4x2 – 3x < 0 −6 < x < 1 and x < −4 or x > 1
x(4x −3) < 0 So −6 < x < −4
0 < x < 3 or 1 − 4x > 0 {−5}
4 4
1 > 4x d x2 + 5x – 6 > 0 or x2 + 3x – 4 < 0
1
x< x < −6 or x > 1 or −4 < x < 1
4
So x < 3 {all integers excluding −6, −5, −4 and 1}
4
{all integers less than 0.75} 7 x2 – 4x – 3 0, 1 – 2x2 0
4 r < 10 First consider x2 – 4x – 3 0
πr 2 > 250 4 ± 16 + 12
x2 – 4x – 3 = 0 when x = =2± 7
2
r > 8.9
when x = 0 x2 – 4x – 3 = –3
So 8.9 < r < 10
2− 7 x 2+ 7
5 a x2 + 4 < 7x − 2 Now consider 1 – 2x2 0
x2 −7x + 6 < 0 1 −1 1
1 2x2, x 2 , so x
(x − 1)(x − 6) = 0 2 2 2
1<x<6 The range of x values for which x2 – 4x – 3 0,
{2, 3, 4, 5} and 1 – 2x2 0
b 2x2 – 3x – 15 < 2x − 3 1
is 2 − 7 x
2
2x2 – 5x – 12 < 0
(2x + 3)(x − 4) = 0 2
which can be written as 2 − 7 x
3 2
− <x<4
2
{−1, 0, 1, 2, 3}
6
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, 1
WORKED SOLUTIONS
1 − 3x Exercise 1.5A
8 x
5
1 a 2x + y = −4 1
1 − 3x
x2 5x + y = −1 2
5
5x2 + 3x – 1 0 Subtract 1 from 2 .
−3 ± 3 − ( 4 )( 5)(−1)
2 3x = 3
x=
10 x = 1 and y = −6
−3 + 29 b 3x – 2y = 2 1
x=
10 5x + y = 25 2
Or
Multiply 2 by 2.
−3 − 29
x= 10x + 2y = 50 3
10
Add 1 and 3 .
y 13x = 52
4
x = 4 and y = 5
3
c x – 2y = 13 1
2
−x + 3y = −15 2
1
Add 1 and 2 .
–6 –5 –4 –3 –2 –1
0
1 2 3 4 5 6 x y = −2 and x = 9
–1
2 5x + 6y = 6 1
–2
3x – y = 22 2
–3
Multiply 2 by 6.
−3 − 29 5x + 6y = 6 1
So for 5x2 + 3x – 1 0 either x or
10
18x – 6y = 132 3
−3 + 29 Add 1 and 3 .
x
10 23x = 138
9 4x – 7x2 – 8 > 0 x=6
Consider 4x – 7x2 – 8 = 0 Substitute x into 2 .
18 – y = 22
7x2 – 4x + 8 = 0
b2 – 4ac = 16 – 224 = – 208 < 0 y = −4
There are no solutions, ∴ the function does not 3 If there are n unknowns then you need n distinct
cross the x-axis, ie it is all above or below the equations involving the n unknowns.
x-axis. 4 3x – 4y = 3 1
When x = 0, 4x – 7x2 – 8 = – 8 6x + 4y = 3 2
There are no x values for which 4x – 7x2 – 8 > 0
Add 1 and 2 .
10 – 3x2 x – 5 9x = 6
3x2 + x – 5 0 2 1
x= and y = −
−1 ± 1 + 60 −1 ± 61 3 4
3x2 + x – 5 = 0 when x = =
6 6
5 x + y = 50 1
−1 + 61
The higher value is x − 1 = 15(y − 1)
6
x − 15y = −14 2
49 < 61 < 64
Subtract 2 from 1 .
−1 + 49 −1 + 7 6 16y = 64
Consider = = =1
6 6 6 y = 4 and x = 46. Helen was 42 when her
−1 + 61 daughter was born.
∴ >1
6
No, the solution set for – 3x2 x – 5 is not a
subset of x 1.
7
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, 1 Quadratics
6 2x − y = −14 1 1h + 1m = 33.9 C
3y − 2z = 16 2 Multiply A by 2
z−x=3 3 4m + 2p = 67.2 2A
Multiply 1 by 3. 2h + 2p = 42.6 B
6x − 3y = −42 4 Work out 2A – B
3y − 2z = 16 2 4m – 2h = 24.6 D
Add 4 and 2 . Multiply C by 2
6x − 2z = −26 5 2h + 2m = 67.8 2C
Multiply 3 by 2.
Work out D + 2C
2z − 2x = 6 6
Add 5 and 6 . 6m = 92.4
4x = −20 92.4
m= = 15.4
x = −5 6
Substitute x into 3: z = −2 1 maths textbook costs $15.40 (remember
Substitute x into 1: y = 4 to add the ‘0’ to make the money notation
7 A sketch diagram helps see what is required. correct).
y = 2x + 8 Exercise 1.5B
y = 7 – 3x y
1 a x + y = 3 1
2x2 − y = 25 2
Required area
Add 1 and 2 .
2x2 + x − 28 = 0
0 x
x-axis has equation y = 0 (2x − 7)(x + 4) = 0
7
Find the coordinates of the point of x= or −4
2
intersection of the lines y = 2x + 8 and y = 7 – 3x
1
7 – 3x = 2x + 8, so x = –0.2, y = 7.6 y=− or 7
2
The intersections of the lines with the line y = 0 b 2x − y = 20 1
7 x2
are when 0 = 2x + 8, x = –4 and 0 = 7 – 3x, x = + xy = −12 2
3
From 1 .
1
Area of the triangle = × base × height
2 2x − 20 = y
=
1
2 ( )
× 4+
7
3
× 7.6 = 24
1
15
square units Substitute into 2 .
x2 + x(2x − 20) = −12
8 y = 4 – 5x meets y = 7 – 3x when 4 – 5x = 7 – 3x, so, 3x2 − 20x + 12 = 0
x = –1.5, y = 11.5 (–1.5, 11.5)
(3x − 2)(x − 6) = 0
y = 4 – 5x meets y = x + 2 when 4 – 5x = x + 2, so,
( )
2
1 7 1 7 x= or 6
x = ,y = , 3
3 3 3 3
−56
y = 7 – 3x meets y = x + 2 when 7 – 3x = x + 2, so y= or −8
3
x = 1.25, y = 3.25
c y = 4x 1
The coordinates of the vertices are (–1.5, 11.5),
( )
5 − x2 = y 2
1 7
, , and (1.25, 3.25).
3 3 1 = 2
9 Use the information to make 3 equations. 5 − x2 = 4x
Let m be the cost of a maths textbook, h the x2 + 4x − 5 = 0
cost of a history textbook, p the cost of a pen. (x + 5)(x − 1) = 0
2m + 1p = 33.6 A x = −5 or 1
2h + 2p = 42.6 B y = −20 or 4
8
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, 1
WORKED SOLUTIONS
2 a Substitution or elimination: for elimination the 2x − 3(x2 − 7) = 13
first equation would need to be multiplied by 2 3x2 − 2x − 8 = 0
so that y can subsequently be eliminated. (3x + 4)(x − 2) = 0
b S
ubstitution only: neither addition or −4
x= or 2
3
subtraction of the equations will eliminate
a variable. −47 or −3
y=
9
c S
ubstitution only: neither addition or
subtraction of the equations will eliminate So the coordinates of the points of intersection of
3 a
a variable.
2x2 + y = 14 1
(
4 47
)
this line and this curve are − 3 , − 9 and (2, −3)
5 y − x = 1 1
x − 2y = 11 2
x2 + y2 = 64 2
Multiply 1 by 2.
From 1 .
4x2 + 2y = 28 3 y=x+1
Add 2 and 3 . Substitute into 2 .
4x2 + x − 39 = 0 x2 + (x + 1)2 = 64
(4x + 13)(x − 3) = 0 2x2 + 2x − 63 = 0
−2 ± 4 − (4)(2)(−63)
−13 x=
x= or 3 4
4
x = 5.135 or −6.135
−57 y = 6.135 or −5.135
y= or −4
8
So the coordinates of the points of intersection
b xy − x = −4 1 of this line and this circle are (5.135, 6.135) and
x + y = 1 2 (−6.135, −5.135).
From 2 . 6 y − x = 10 1
y=1−x x2 + y2 = 50 2
Substitute into 1 . From 1 .
x(1 − x) − x = −4 y = x + 10
Substitute into 2 .
x − x2 − x = −4
x2 + (x + 10)2 = 50
x = ±2, y = −1 or 3 2x2 + 20x + 50 = 0
c x − y = 10 1 x2 + 10x + 25 = 0
xy = 140 2 (x + 5)2 = 0
x = −5, y = 5
From 1 .
The line is a tangent to the circle at (−5, 5).
x = y + 10
Substitute into 2 . 7 Let the numbers be x and y.
y(y + 10) = 140 Then x + y = 129, x2 + y2 = 8433
y2 + 10y − 140 = 0 The two numbers must satisfy both equations.
x + y = 129 so y = 129 – x
−10 ± 100 − (4)(1)(−140)
y= Substitute into x2 + y2 = 8433
2
x2 + (129 – x)2 = 8433
y = −10 ± 660
2 x2 + 16 641 – 258x + x2 = 8433
y = −5 ± 165 x = 5 ± 165 2x2 – 258x + 8208 = 0
x2 – 129x + 4104 = 0
4 2x − 3y = 13 1 (x – 57)(x – 72) = 0
x2 − y = 7 2 x = 57, x = 72
From 2 .
The two numbers are 57 and 72.
x2 − 7 = y
Substitute into 1 .
9
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Pure Mathematics 1 9780008257736
57736_P001_014.indd 9 15/06/18 6:57 PM
