38 Solutions
Chapter 1 More about Equations
Skills Assessment (P.4)
1. From the figure, the point of intersection is
(1 , 2.5).
` The solution of the simultaneous equations
is x = 1, y = 2.5.
2. x y
y x
2 1
3 3
=
=
+
-*
Substituting (ii) into (i), we have
2x = 3x - 3 + 1
x = 2
Substituting x = 2 into (ii), we have
y = 3(2) - 3 = 3
` The solution of the simultaneous equations
is x = 2, y = 3.
3. x y
x y
3
3
4 4
1
- =
+ =*
(i) + (ii) : 5x = 5
x = 1
Substituting x = 1 into (ii), we have
1 + 3y = 1
y = 0
` The solution of the simultaneous equations
is x = 1, y = 0.
4. x y
x y
2 14
4 5 0
- =-
+ =*
(i) # 2 : 4x - 2y = -28 .............................. (iii)
(ii) - (iii) : 7y = 28
y = 4
Substituting y = 4 into (i), we have
2x - 4 = -14
2x = -10
x = -5
` The solution of the simultaneous equations
is x = -5, y = 4.
...................................................... (i)
..................................................... (ii)
................................................... (i)
..................................................... (ii)
................................................. (i)
.................................................. (ii)
5. x2 - 3x - 4 = 0
(x - 4)(x + 1) = 0
` x = 4 or x = 1-
6. x2 - 13x + 40 = 0
(x - 5)(x - 8) = 0
` x = 5 or x = 8
7. x2 - x - 6 = 0
x = 2
( )
( ) ( ) ( )( )
2 1
1 1 4 1 6!- - - - -
= 1 25
2!
= 5
21 !
= 2- or 3
8. 2x2 - 15x - 8 = 0
x = 2
( )
( ) ( ) ( )( )
2 2
15 15 4 2 8!- - - - -
= 15 289
4!
= 17
415 !
= 21
- or 8
9. 9x2 - 12x + 4 = 0
x = 2
( )
( ) ( ) ( )( )
2
12 12 4 9 4
9
!- - - -
= 0
1812 !
= 32
(repeated)
10. From the graph, the x-intercepts are -1 and 3.
` The roots of the equation are -1 and 3.
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Chapter 1: More about Equations 39
Exercise 1A (P.12)
1. (a) From the figure, the points of intersection
of the two graphs are (-3 , 5) and (2 , 5).
` The solutions of the simultaneous
equations are x = -3, y = 5 and
x = 2, y = 5.
(b) From the figure, the points of intersection
of the two graphs are (0 , 2) and (4 , 18).
` The solutions of the simultaneous
equations are x = 0, y = 2 and x = 4,
y = 18.
2. (a) From the figure, the point of intersection
of the two graphs is (-2 , 4).
` The solution of the simultaneous
equations is x = -2, y = 4.
(b) From the figure, the point of intersection
of the two graphs is (3 , -6).
` The solution of the simultaneous
equations is x = 3, y = -6.
3. (a) From the figure, the two graphs have no
points of intersection.
` The simultaneous equations have no
real solutions.
(b) From the figure, the two graphs have no
points of intersection.
` The simultaneous equations have no
real solutions.
4. Add the graph of y = 20 to the figure.
x
y
–1 1
5
10
15
20
25
–5
0
y = 8x2
y = 20
From the figure, the solutions of the simultaneous
equations are x = -1.6, y = 20 and x = 1.6,
y = 20.
5. Add the graph of y = 10 + x to the figure.
y = 10 + x
x -6 -4 -2 0
y 4 6 8 10
x
y
–2–4–6 –1–3–5
5
10
0
y = –x2– 6x
y = 10 + x
From the f igure , the so lu t ions o f the
simultaneous equations are x = -5, y = 5 and
x = -2, y = 8.
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40 Solutions
6. (a) Draw the graph of y = x2 - 3x from
x = -1 to x = 4.
y = x2 - 3x
x -1 0 1 2 3 4
y 4 0 -2 -2 0 4
x
y
–1 1 2 3 4–1
1
2
3
4
–2
–3
–4
0
y = x2– 3x
y = x – 3
(b) Add the graph of y = x - 3 to the figure in
(a).
y = x - 3
x -1 1 3 4
y -4 -2 0 1
From the figure, the solutions of the
simultaneous equations are x = 3, y = 0
and x = 1, y = -2.
7. (a) Draw the graph of y = 1 - 2x2 from
x = -3 to x = 3.
y = 1 - 2x2
x -3 -2 -1 0 1 2 3
y -17 -7 -1 1 -1 -7 -17
x
y
–1–2–3 1 2 3–2
2
4
–4
–6
–8
–10
–12
–14
–16
–18
0
y = 2x – 2y = –1 – 4x
y = 1 – 2x2
(b) (i) Add the graph of y = 2x - 2 to the
figure in (a).
y = 2x - 2
x -3 -1 1 3
y -8 -4 0 4
From the figure, the solutions of
the simultaneous equations are
x = 0.8, y = -0.4 and x = -1.8,
y = -5.6.
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Chapter 1: More about Equations 41
(ii) Add the graph of y = -1 - 4x to the
figure in (a).
y = -1 - 4x
x -1 0 1 3
y 3 -1 -5 -13
From the figure, the solutions of
the simultaneous equations are
x = 2.4, y = -10.6 and x = -0.4,
y = 0.6.
8. (a) No. The graphs drawn by Jacky and Sally
are incomplete. They should draw the
graphs for -2 G x G 5.
(b) By combining the graphs drawn by Jacky
and Sally, the solutions of the simultaneous
equations are x = -1.2, y = 0.8 and
x = 4.2, y = 6.2.
9.
x
y
–3
–4
–1
–2
1
2
0 1 2 3 4 5
y = –x2+ 5x – 4
x + y = 1
y = (x - 1)(4 - x)
= 4x - x2 - 4 + x
= -x2 + 5x - 4
Add the graph of x + y = 1 to the figure.
x + y = 1
x 0 1 3 5
y 1 0 -2 -4
From the figure, the solutions of the simultaneous
equations are x = 1, y = 0 and x = 5, y = -4.
10. (a) From the figure, the two graphs have no
points of intersection.
` The simultaneous equations have no
real solutions.
(b) From the figure, the solutions of the
simultaneous equations are x = -2.7,
y = -8.4 and x = 1.7, y = 0.4.
(c) From the figure, the solution of the
simultaneous equations is x = 1, y = 4.
11.
x
y
–100
–150
–50
100
50
150
0–10–15 –5 5 10 15
y = x2+ 2x – 120
y = x + 10
5x + y = –150
(a) Add the graph of y = x + 10 to the figure.
y = x + 10
x -10 0 10 20
y 0 10 20 30
From the figure, the solutions of the
simultaneous equations are x = 11, y = 20
and x = -12, y = 0.
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42 Solutions
(b) Add the graph of 5x + y = -150 to the
figure.
5x + y = -150
x -20 -10 0 10
y -50 -100 -150 -200
From the figure, the two graphs have no
points of intersection.
` The simultaneous equations have no
real solutions.
12. (a) Draw the graph of y = 4x2 + 8x from
x = -2.5 to x = 0.5.
y = 4x2 + 8x
x -2.5 -2 -1.5 -1 -0.5 0 0.5
y 5 0 -3 -4 -3 0 5
x
y
–0.5–1–1.5–2–2.5 0.5–2
2
4
6
–4
–6
–8
–10
0
y = 4x2+ 8x
4x + y + 9 = 0
(b) Add the graph of 4x + y + 9 = 0 to the
figure in (a).
4x + y + 9 = 0
x -2.5 -2 -1 0
y 1 -1 -5 -9
From the figure, the solution of the
simultaneous equations is x = -1.5,
y = -3.
13. (a) Draw the graph of y = -x2 + x - 2 from
x = -2 to x = 3.
y = -x2 + x - 2
x -2 -1 0 1 2 3
y -8 -4 -2 -2 -4 -8
x
y
–1–2 1 2 3–2
2
4
–4
–6
–8
0
y = –x2+ x – 2
y = 4 – 3x
(b) Add the graph of y = 4 - 3x to the figure in
(a).
y = 4 - 3x
x 0 1 2 3
y 4 1 -2 -5
From the figure, the two graphs have no
points of intersection.
` The simultaneous equations have no
real solutions.
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Chapter 1: More about Equations 43
14. (a) Draw the graph of y = (20 - x)(10 + x)
from x = -20 to x = 30.
y = (20 - x)(10 + x)
x -20 -10 0 10 20 30
y -400 0 200 200 0 -400
x
y
–10–20 10 20 30–100
100
200
300
–200
–300
–400
0
y = (20 – x)(10 + x)
y = 300 – 10x
(b) Add the graph of y = 300 - 10x to the
figure in (a).
y = 300 - 10x
x 0 10 20 30
y 300 200 100 0
From the figure, the solution of the
simultaneous equations is x = 10, y = 200.
Exercise 1B (P.20)
1. y x
y x
2
2=
= +*
Substitute (i) into (ii):
x2 = x + 2
x2 - x - 2 = 0
(x - 2)(x + 1) = 0
` x = 2 or x = -1
Substitute these values of x into (ii):
When x = 2, y = 2 + 2 = 4.
When x = -1, y = -1 + 2 = 1.
` The solutions of the simultaneous equations
are x = 2, y = 4 and x = -1, y = 1.
............................................................. (i)
....................................................... (ii)
2. y x x
y x
2= +
=-*
Substitute (i) into (ii):
x2 + x = -x
x2 + 2x = 0
x(x + 2) = 0
` x = 0 or x = -2
Substitute these values of x into (ii):
When x = 0, y = 0.
When x = -2, y = 2.
` The solutions of the simultaneous equations
are x = 0, y = 0 and x = -2, y = 2.
3. y x x
y x
2 3
1
2= - -
= +*
Substitute (i) into (ii):
x2 - 2x - 3 = x + 1
x2 - 3x - 4 = 0
(x - 4)(x + 1) = 0
` x = 4 or x = -1
Substitute these values of x into (ii):
When x = 4, y = 4 + 1 = 5.
When x = -1, y = -1 + 1 = 0.
` The solutions of the simultaneous equations
are x = 4, y = 5 and x = -1, y = 0.
4. x x
y
y
x
4 0
6 3
2- - =
= -*
Substitute (ii) into (i):
x2 - 4x - (6 - 3x) = 0
x2 - x - 6 = 0
(x - 3)(x + 2) = 0
` x = 3 or x = -2
Substitute these values of x into (ii):
When x = 3, y = 6 - 3(3) = -3.
When x = -2, y = 6 - 3(-2) = 12.
` The so lu t ions o f the s imul taneous
equations are x = 3, y = -3 and x = -2,
y = 12.
...................................................... (i)
........................................................... (ii)
............................................. (i)
....................................................... (ii)
............................................. (i)
.................................................... (ii)
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44 Solutions
5. x y
y x
0
2 8
2- =
- =*
(i) + (ii):
x2 - 2x = 8
x2 - 2x - 8 = 0
(x - 4)(x + 2) = 0
` x = 4 or x = -2
From (i): y = x2 ............................................. (iii)
Substitute these values of x into (iii):
When x = 4, y = 42 = 16.
When x = -2, y = (-2)2 = 4.
` The so lu t ions o f the s imul taneous
equations are x = 4, y = 16 and x = -2,
y = 4.
6. 2 0y x
x y3 2
2- =
=+*
(ii) - (i):
3x + 2x2 = 2
2x2 + 3x - 2 = 0
(2x - 1)(x + 2) = 0
` x = 21
or x = -2
From (ii): y = 2 - 3x .................................... (iii)
Substitute these values of x into (iii):
When x = 21
, y = 2 - 321a k =
21
.
When x = -2, y = 2 - 3(-2) = 8.
` The so lu t ions o f the s imul taneous
equations are x = 21
, y = 21
and x = -2,
y = 8.
...................................................... (i)
.................................................... (ii)
.................................................... (i)
..................................................... (ii)
7. x x
x y
y
6
1
5 0
2- =
- - =
+*
From (ii): y = 5x - 6 .................................... (iii)
Substitute (iii) into (i):
x2 - x = 5x - 6 + 1
x2 - 6x + 5 = 0
(x - 5)(x - 1) = 0
` x = 5 or x = 1
Substitute these values of x into (iii):
When x = 5, y = 5(5) - 6 = 19.
When x = 1, y = 5(1) - 6 = -1.
` The so lu t ions o f the s imul taneous
equations are x = 5, y = 19 and x = 1,
y = -1.
8. x y x
x y
4
020
22
+ - =
+ - =*
From (ii): y = 20 - x .................................... (iii)
Substitute (iii) into (i):
4x + 20 - x - 2 = x2
x2 - 3x - 18 = 0
(x - 6)(x + 3) = 0
` x = 6 or x = -3
Substitute these values of x into (iii):
When x = 6, y = 20 - 6 = 14.
When x = -3, y = 20 - (-3) = 23.
` The solutions of the simultaneous equations
are x = 6, y = 14 and x = -3, y = 23.
9. Rewrite the given equation as:
x y
x x y
1
22
=
=
+
- -*
Substitute (i) into (ii):
x2 - x - 2 = x + 1
x2 - 2x - 3 = 0
(x - 3)(x + 1) = 0
` x = 3 or x = -1
Substitute these values of x into (i):
When x = 3, y = 3 + 1 = 4.
When x = -1, y = -1 + 1 = 0.
` The solutions of the simultaneous equations
are x = 3, y = 4 and x = -1, y = 0.
................................................ (i)
.............................................. (ii)
............................................. (i)
............................................. (ii)
........................................................ (i)
.............................................. (ii)
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Chapter 1: More about Equations 45
10. Rewrite the given equation as:
2 x
y
y
x
15
2 32
=
- = -
- -*
Substitute (i) into (ii):
3 - x = 15 - x2
x2 - x - 12 = 0
(x - 4)(x + 3) = 0
` x = 4 or x = -3
From (i): y = 5 - x ....................................... (iii)
Substitute these values of x into (iii):
When x = 4, y = 5 - 4 = 1.
When x = -3, y = 5 - (-3) = 8.
` The solutions of the simultaneous equations
are x = 4, y = 1 and x = -3, y = 8.
11. Rewrite the given equation as:
x x
x y x
x y 7 30
12 202
=
= + -
+ +
+*
(ii) - (i):
x2 + 7x - 30 - 12x - 20 = 0
x2 - 5x - 50 = 0
(x + 5)(x - 10) = 0
` x = -5 or x = 10
From (i): y = 11x + 20 ................................. (iii)
Substitute these values of x into (iii):
When x = -5, y = 11(-5) + 20 = -35.
When x = 10, y = 11(10) + 20 = 130.
` The solutions of the simultaneous equations
are x = -5, y = -35 and x = 10, y = 130.
12. q p
qp
4
5 02 11
2= -
=+ +*
Substitute (i) into (ii):
5p + 2(p2 - 4) + 11 = 0
2p2 + 5p + 3 = 0
(2p + 3)(p + 1) = 0
` p = -23
or p = -1
Substitute these values of p into (i):
When p = -23
, q = 2
23
-a k - 4 = -47
.
When p = -1, q = (-1)2 - 4 = -3.
` p = -23
, q = -47
or p = -1, q = -3.
................................................. (i)
............................................ (ii)
........................................... (i)
.................................... (ii)
...................................................... (i)
......................................... (ii)
13. ( )( )a b
a b
0
5
1 3
2 =
=+ -
+*
From (i), b = 5 - 2a .................................... (iii)
Substitute (iii) into (ii):
(a + 1)(5 - 2a - 3) = 0
(a + 1)(2 - 2a) = 0
` a = -1 or a = 1
Substitute these values of a into (iii):
When a = -1, b = 5 - 2(-1) = 7.
When a = 1, b = 5 - 2(1) = 3.
` The so lu t ions o f the s imul taneous
equations are a = -1, b = 7 and a = 1,
b = 3.
14. y x
y x
1
2
2= +
=-*
Substitute (i) into (ii):
x2 + 1 = -2x
x2 + 2x + 1 = 0
The discriminant of the equation is:
D = (2)2 - 4(1)(1)
= 0
` There is one solution for x.
i.e. The simultaneous equations have one set of
solution.
15. y x
y x
x3
2 1
2= +
= -*
Substitute (i) into (ii):
3x2 + x = 2x - 1
3x2 - x + 1 = 0
The discriminant of the equation is:
D = (-1)2 - 4(3)(1)
= -11
1 0
` There are no real solutions for x.
i.e. The simultaneous equations have no real
solutions.
..................................................... (i)
........................................ (ii)
....................................................... (i)
........................................................ (ii)
.................................................... (i)
..................................................... (ii)
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46 Solutions
16. x y x
x y 2
12+ = +
+ =*
(i) - (ii):
x2 - x = x + 1 - 2
x2 - 2x + 1 = 0
(x - 1)2 = 0
` x = 1 (repeated)
Substitute x = 1 into (ii):
1 + y = 2
y = 1
` The solution of the simultaneous equations
is x = 1, y = 1.
17. x y x
y x
3
2 10
2+ =
= -*
(i) # 2 - (ii):
6x = 2x2 - x + 10
2x2 - 7x + 10 = 0
The discriminant of the equation is:
D = (-7)2 - 4(2)(10)
= -31
1 0
` There are no real solutions for x.
i.e. The simultaneous equations have no real
solutions.
18. Rewrite the given equation as:
( )
y
y x x x
x2
2 4
4 8
8
- =
- - + =*
From (i): y = 4 + 2x ..................................... (iii)
Substitute (iii) into (ii):
4 + 2x - 2x - x(x + 4) = 8
x2 + 4x + 4 = 0
(x + 2)2 = 0
` x = -2 (repeated)
Substitute x = -2 into (iii):
y = 4 + 2(-2)
= 0
` The solution of the simultaneous equations
is x = -2, y = 0.
................................................ (i)
....................................................... (ii)
..................................................... (i)
................................................... (ii)
................................................... (i)
.................................. (ii)
19. Rewrite the given equation as:
( )x x x
x
y
y
3 2 5 12
6 4 2 12
2- =
- =
-
+*
From (ii), 3x - 2y = 5 .................................. (iii)
Substitute (iii) into (i):
5x - 5x2 = 12
5x2 - 5x + 12 = 0
The discriminant of the equation is:
D = (-5)2 - 4(5)(12)
= -215
1 0
` There are no real solutions for x.
i.e. The simultaneous equations have no real
solutions.
20. Rewrite the given equation as:
x
x y
y
2
4
4
2 2+ =
- + =*
From (ii), x = y + 2 ...................................... (iii)
Substitute (iii) into (i):
(y + 2)2 + y
2 = 4
2y2 + 4y = 0
2y(y + 2) = 0
` y = -2 or y = 0
Substitute these values of y into (iii):
When y = -2, x = -2 + 2 = 0.
When y = 0, x = 0 + 2 = 2.
` The so lu t ions o f the s imul taneous
equations are x = 0, y = -2 and x = 2,
y = 0.
21. Rewrite the given equation as:
x y
x y3
7
7
2 2=
+ =
-*
From (ii), y = 7 - 3x .................................... (iii)
Substitute (iii) into (i):
x2 - (7 - 3x)
2 = 7
4x2 - 21x + 28 = 0
The discriminant of the equation is:
D = (-21)2 - 4(4)(28)
= -7
1 0
` There are no real solutions for x.
i.e. The simultaneous equations have no real
solutions.
.................................. (i)
......................................... (ii)
..................................................... (i)
................................................ (ii)
..................................................... (i)
..................................................... (ii)
01 NSS TM 5A (E)-2P-KJ.indd 46 2010/7/20 4:36:33 PM
Chapter 1: More about Equations 47
22. x y
x y
2 1
3 4 1
2 2=
=
-
+*
From (ii), y = 41
- 43
x ................................. (iii)
Substitute (iii) into (i):
x2 - 2 x
2
41
43
-a k = 1
x2 - 6x + 9 = 0
(x - 3)2 = 0
` x = 3 (repeated)
Substitute x = 3 into (iii):
y = 41
- 43
(3) = -2
` The solution of the simultaneous equations
is x = 3, y = -2.
23. x xy y
x y
4 4
2 0
2 2+ + =
=- -*
From (ii), x = y + 2 ...................................... (iii)
Substitute (iii) into (i):
(y + 2)2 + (y + 2)y + 4y
2 = 4
6y2 + 6y = 0
6y(y + 1) = 0
` y = 0 or y = -1
Substitute these values of y into (iii):
When y = 0, x = 0 + 2 = 2.
When y = -1, x = -1 + 2 = 1.
` The solutions of the simultaneous equations
are x = 2, y = 0 and x = 1, y = -1.
24. x y
x y c
c2 2 2+ =
+ =*
From (ii), y = c - x ...................................... (iii)
Substitute (iii) into (i):
x2 + (c - x)
2 = c
2
2x2 - 2cx = 0
2x(x - c) = 0
` x = c or x = 0
Substitute these values of x into (iii):
When x = c, y = c - c = 0.
When x = 0, y = c - 0 = c.
` The solutions of the simultaneous equations
are x = c, y = 0 and x = 0, y = c.
................................................... (i)
................................................... (ii)
.......................................... (i)
................................................ (ii)
................................................... (i)
....................................................... (ii)
Exercise 1C (P.34)
1. (a) 2x - 3 = x2
2x2 - 3x = 2
2x2 - 3x - 2 = 0
(2x + 1)(x - 2) = 0
` x = 21
- or x = 2
(b) x2
+ x6
= 4
x2 + 12 = 8x
x2 - 8x + 12 = 0
(x - 6)(x - 2) = 0
` x = 6 or x = 2
2. (a) x - x 3
8-
= 1
( )
x
x x
3
3 8
-
- - = 1
x(x - 3) - 8 = x - 3
x2 - 3x - 8 - x + 3 = 0
x2 - 4x - 5 = 0
(x - 5)(x + 1) = 0
` x = 5 or x = 1-
(b) x 1
6+
= 5 - 2x
6 = (5 - 2x)(x + 1)
6 = 5x - 2x2 + 5 - 2x
2x2 - 3x + 1 = 0
(2x - 1)(x - 1) = 0
` x = 21
or x = 1
3. (a) x1
+ x 2
3+
= 2
( )x x
x x2
2 3+
+ + = 2
x + 2 + 3x = 2x(x + 2)
4x + 2 = 2x2 + 4x
x2 - 1 = 0
(x - 1)(x + 1) = 0
` x = 1 or x = 1-
01 NSS TM 5A (E)-2P-KJ.indd 47 2010/7/20 4:36:39 PM
48 Solutions
(b) x1
4-
+ x9
= 1
( )
( )
x x
x x
1
4 9 1
-
+ - = 1
4x + 9(1 - x) = x(1 - x)
4x + 9 - 9x = x - x2
x2 - 6x + 9 = 0
(x - 3)2 = 0
` x = 3 (repeated)
4. (a) Let u = 2x,
then the original equation becomes
u2 - 5u + 4 = 0
(u - 4)(u - 1) = 0
` u = 4 or u = 1
When u = 4, 2x = 4
` x = 2
When u = 1, 2x = 1
` x = 0
(b) Let u = 7x, then u
2 = (7
x)2 = 7
2x.
The original equation becomes
u2 - 56u + 343 = 0
(u - 49)(u - 7) = 0
` u = 49 or u = 7
When u = 49, 7x = 49
` x = 2
When u = 7, 7x = 7
` x = 1
5. (a) Let u = 3x, then u
2 = (3
x)2 = 3
2x.
The original equation becomes
32:u
2 - 10u + 1 = 0
9u2 - 10u + 1 = 0
(9u - 1)(u - 1) = 0
` u = 91
or u = 1
When u = 91
, 3x =
91
` x = 2-
When u = 1, 3x = 1
` x = 0
(b) Let u = 4x, then u
2 = (4
x)2 = (4
2)
x = 16
x.
The original equation becomes
u2 - 20u + 64 = 0
(u - 16)(u - 4) = 0
` u = 16 or u = 4
When u = 16, 4x = 16
` x = 2
When u = 4, 4x = 4
` x = 1
6. (a) log x + log (x - 3) = log 4
log x(x - 3) = log 4
` x(x - 3) = 4
x2 - 3x - 4 = 0
(x - 4)(x + 1) = 0
` x = 4 or x = -1 (rejected)
(b) log (x - 4) + log (x + 4) = log 9
log (x - 4)(x + 4) = log 9
` (x - 4)(x + 4) = 9
x2 - 16 - 9 = 0
x2 - 25 = 0
(x - 5)(x + 5) = 0
` x = 5 or x = -5 (rejected)
7. (a) log (x + 3) + log (x + 12) = 1
log (x + 3)(x + 12) = log 10
` (x + 3)(x + 12) = 10
x2 + 15x + 36 = 10
x2 + 15x + 26 = 0
(x + 2)(x + 13) = 0
` x = 2- or x = -13 (rejected)
(b) log2 (x + 1) + log2 (x - 1) = 3
log2 (x + 1)(x - 1) = log28
` (x + 1)(x - 1) = 8
x2 - 1 - 8 = 0
x2 - 9 = 0
(x - 3)(x + 3) = 0
` x = 3 or x = -3 (rejected)
01 NSS TM 5A (E)-2P-KJ.indd 48 2010/7/20 4:36:44 PM
Chapter 1: More about Equations 49
8. (a) Let u = sin i, then u2 = sin
2i.
The original equation becomes
2u2 - 1 = 0
u2 =
12
` u = !22
When u = 22
, sin i = 22
` i = 45c, 135c
When u = -22
, sin i = -22
` i = 225c, 315c
(b) Let u = tan i, then u2 = tan
2i.
The original equation becomes
3u2 - 1 = 0
u2 =
31
` u = !33
When u = 33
, tan i = 33
` i = 30c, 210c
When u = -33
, tan i = -33
` i = 150c, 330c
9. (a) Let u = cos i, then u2 = cos
2i.
The original equation becomes
2u2 - 3 u = 0
u(2u - 3 ) = 0
` u = 0 or u = 23
When u = 0, cos i = 0
` i = 90c, 270c
When u = 23
, cos i = 23
` i = 30c, 330c
(b) Let u = sin i, then u2 = sin
2i.
The original equation becomes
u + u2 = 0
u(u + 1) = 0
` u = 0 or u = -1
When u = 0, sin i = 0
` i = 0c, 180c
When u = -1, sin i = -1
` i = 270c
10. (a) Let u = tan i, then u2 = tan
2i.
The original equation becomes
u2 + 2u + 1 = 0
(u + 1)2 = 0
` u = -1 (repeated)
When u = -1, tan i = -1
` i = 135c, 315c
(b) Let u = cos i, then u2 = cos
2i.
The original equation becomes
2u2 - u - 1 = 0
(u - 1)(2u + 1) = 0
` u = 1 or u = -21
When u = 1, cos i = 1
` i = 0c
When u = -21
, cos i = -21
` i = 120c, 240c
11. No. The value of u obtained is the solution for
the quadratic equation in one unknown instead
of the solution for the original exponential
equation.
When u = 51
, 5x =
51
` x = 1-
When u = 1, 5x = 1
` x = 0
01 NSS TM 5A (E)-2P-KJ.indd 49 2010/7/20 4:36:48 PM
50 Solutions
12. When solving the equation, we cannot simplify
the equation by dividing both sides by the
common factor which involves the variable. This
will lose part of the solution.
sin i = 2sin2i
sin i(1 - 2sin i) = 0
sin i = 0 or sin i = 21
` i = 0c, 30c, 150c, 180c
13. Let the two integers be x and x + 1, then
x1
+ x 1
1+
= 127
( )x x
x x1
1+
+ + =
127
12(x + 1 + x) = 7x(x + 1)
24x + 12 = 7x2 + 7x
7x2 - 17x - 12 = 0
(x - 3)(7x + 4) = 0
` x = 3 or x = -74
(rejected)
` The two integers are 3 and 4.
14. Let the two numbers be x and x + 2, then
log x + log (x + 2) = log 35
log x(x + 2) = log 35
` x(x + 2) = 35
x2 + 2x - 35 = 0
(x - 5)(x + 7) = 0
` x = 5 or x = -7 (rejected)
` The two numbers are 5 and 7.
15. (a) 1x2
+` j(x + 6) = 1
( ) ( )
xx x2 6+ +
= 1
(2 + x)(x + 6) = x
2x + x2 + 12 + 6x = x
x2 + 7x + 12 = 0
(x + 4)(x + 3) = 0
` x = 4- or x = 3-
(b) x
x3+
+ 2x 9
6
- =
x 31-
- 1
2
( )
x
x x
9
3 6
-
- + = 2
2)(
x
x x
9
93
-
-+ -
x(x - 3) + 6 = x + 3 - x2 + 9
x2 - 2x - 3 = 0
(x + 1)(x - 3) = 0
` x = 1- or x = 3 (rejected)
(c) xx
3 53
-+
+ 1 = ( )
x
x
1
3 2
-
-
x
x x3 53 3 5
-+ + -
= ( )
x
x
1
3 2
-
-
xx
3 524
--
= ( )
x
x
1
3 2
-
-
(4x - 2)(x - 1) = 3(3x - 5)(x - 2)
4x2 - 6x + 2 = 9x
2 - 33x + 30
5x2 - 27x + 28 = 0
(5x - 7)(x - 4) = 0
` x = 57
or x = 4
16. (a) Let u = 4x, then the original equation
becomes
u - 3 = u10
u2 - 3u = 10
u2 - 3u - 10 = 0
(u + 2)(u - 5) = 0
` u = -2 or u = 5
When u = -2, 4x = -2 (rejected)
When u = 5, 4x = 5
log 4x = log 5
x log 4 = log 5
x = log
log
4
5
= .1 16, cor. to 2 d.p.
(b) Let u = 5x, then the original equation
becomes
2u + u1
= 3
2u2 + 1 = 3u
2u2 - 3u + 1 = 0
(u - 1)(2u - 1) = 0
` u = 1 or u = 21
When u = 1, 5x = 1
` x = 0
01 NSS TM 5A (E)-2P-KJ.indd 50 2010/7/20 4:36:56 PM
Chapter 1: More about Equations 51
When u = 21
, 5x =
21
log 5x = log
21
x log 5 = log 21
x = log
log
521
= .0 43- , cor. to 2 d.p.
(c) Let u = 2x, then the original equation
becomes
82
u1` j + u
1 = 7
2u
8 + u
1 = 7
8 + u = 7u2
7u2 - u - 8 = 0
(u + 1)(7u - 8) = 0
` u = -1 or u = 78
When u = -1, 2x = -1 (rejected)
When u = 78
, 2x =
78
log 2x = log
78
x log 2 = log 78
x = log
log
278
= .0 19, cor. to 2 d.p.
17. (a) log (x - 4) - log (3x - 10) = log x1` j
log xx
3 104
--
= log x1` j
` xx
3 104
--
= x1
x(x - 4) = 3x - 10
x2 - 7x + 10 = 0
(x - 2)(x - 5) = 0
` x = 2 (rejected) or x = 5
(b) log (x2 - 9) - log (x - 5) = log (2x - 1)
log 2
xx
59
--
= log (2x - 1)
` 2
xx
59
--
= 2x - 1
x2 - 9 = (2x - 1)(x - 5)
x2 - 11x + 14 = 0
x = 2
( )
( ) ( )( )
2 1
11 11 4 1 14! - -
= 11 65
2+
or 6511
2-
(rejected)
= .9 53, cor. to 2 d.p.
(c) log3 (4 + x2) - log3 (7 - x) - 2 = 0
log3 2
xx
74
-+
- log3 9 = 0
log3 2
xx
74
-+
= log3 9
` 2
xx
74
-+
= 9
4 + x2 = 9(7 - x)
x2 + 9x - 59 = 0
x = 2
( )
( )( )
2 1
9 9 4 1 59!- - -
= 9
2317- +
or 2
9 317- -
= .4 40 or .13 40- , cor. to 2 d.p.
18. (a) sin2i =
23
cos i
1 - cos2i =
23
cos i
2 - 2 cos2i = 3 cos i
2 cos2i + 3 cos i - 2 = 0
(cos i + 2)(2 cos i - 1) = 0
cos i + 2 = 0 or 2 cos i - 1 = 0
cos i = -2 (rejected) or cos i = 21
` i = 60c, 300c
01 NSS TM 5A (E)-2P-KJ.indd 51 2010/7/20 4:37:05 PM
52 Solutions
(b) 2 cos2i + 5 sin i = 5
2(1 - sin2i) + 5 sin i = 5
2 - 2 sin2i + 5 sin i - 5 = 0
2 sin2i - 5 sin i + 3 = 0
(sin i - 1)(2 sin i - 3) = 0
sin i - 1 = 0 or 2 sin i - 3 = 0
sin i = 1 or sin i = 23
(rejected)
` i = 90c
(c) 3 sin2i + 8 cos i - 7 = 0
3(1 - cos2i) + 8 cos i - 7 = 0
3 - 3 cos2i + 8 cos i - 7 = 0
3 cos2i - 8 cos i + 4 = 0
(cos i - 2)(3 cos i - 2) = 0
cos i - 2 = 0 or 3 cos i - 2 = 0
cos i = 2 (rejected) or cos i = 32
` i = 48.19c, 311.81c, cor. to 2 d.p.
19. (a) 2 cos i = tan i
2 cos i = cossin
i
i
2 cos2i = sin i
2(1 - sin2i) = sin i
2 sin2i + sini - 2 = 0
` sin i = 2
( )
( )( )
2 2
1 1 4 2 2!- - -
= 4
1 17- + or
41 17- -
(rejected)
` i = 51.33c, 128.67c, cor. to 2 d.p.
(b) tan
3i
+ sin i = 0
sincos3
i
i + sin i = 0
3 cos i + sin2i = 0
3 cos i + (1 - cos2i) = 0
cos2i - 3 cos i - 1 = 0
` cos i = 2
( )( )( )
( )
4 1 13 3
2 1
! - --
= 3 13
2+
(rejected) or 2
3 13-
` i = 107.62c, 252.38c , cor. to 2 d.p.
(c) sin i tan i + cos i = tan
4i
2
cossin
i
i + cos i =
sincos4
i
i
2 2
coscos
sini
i i+ =
sincos4
i
i
cos
1i
= sincos4
i
i
4 cos2i = sin i
4(1 - sin2i) = sin i
4 sin2i + sin i - 4 = 0
` sin i = 2
( )( )
( )
1 1 4 4 4
2 4
!- - -
= 8
1 56- + or
1 658
- - (rejected)
` i = 61.98c, 118.02c, cor. to 2 d.p.
20. Let Jack’s original cycling speed be x km/h.
From the question,
x30
- x 330+
= 0.5
( )
( )
x x
x x
3
30 3 30
+
+ - = 0.5
( )x x 3
90+
= 0.5
0.5x(x + 3) = 90
0.5x2 + 1.5x - 90 = 0
x2 + 3x - 180 = 0
(x + 15)(x - 12) = 0
x = -15 (rejected) or x = 12
` Jack’s original cycling speed is 12 km/h.
21. (a) Let the original number of glassware bought
by the students be x.
From the question,
(x - 2) 3x48
+` j - 48 = 22
( ) ( )
xx x2 48 3- +
= 70
(x - 2)(48 + 3x) = 70x
48x - 96 + 3x2 - 6x - 70x = 0
3x2 - 28x - 96 = 0
(x - 12)(3x + 8) = 0
x = 12 or x = -38
(rejected)
` The original number of glassware bought
by the students was 12.
01 NSS TM 5A (E)-2P-KJ.indd 52 2010/7/20 4:37:12 PM
Chapter 1: More about Equations 53
(b) The selling price of each piece of
glassware
= $ 31248
+a k
= $7
22. From the question,
20 000(1 - 36%)t - 11 000(1 - 20%)
t = 3 000
20 # 0.64t - 11 # 0.8
t - 3 = 0
Let u = 0.8t, then u
2 = (0.8
t)2 = (0.8
2)
t = 0.64
t.
The original equation becomes
20u2 - 11u - 3 = 0
(4u - 3)(5u + 1) = 0
` u = 43
or u = -51
When u = 43
, 0.8t =
43
log 0.8t = log
43
t log 0.8 = log 43
t = .log
log
0 843
= 1.29, cor. to 2 d.p.
When u = -51
, 0.8t = -
51
(rejected)
` The value of t is 1.29.
23. From the question, R = 25.
` 25 = 2tan
tan
1
80
i
i
+
25(1 + tan2i) = 80 tan i
25 + 25 tan2i - 80 tan i = 0
5 tan2i - 16 tan i + 5 = 0
tan i = 2
) ( )( )(
( )
16 4 5 516
2 5
! --
= 16 156
10+
or 10
16 156-
` i = 70.66c or 19.34c, cor. to 2 d.p.
` The minimum value of i is 19.34c.
Supp. Exercise 1 (P.39)
1. (a) From the figure, the points of intersection
of the two graphs are (-1.8 , -0.6) and
(0.3 , 3.6).
` The solutions of the simultaneous
equations are x = -1.8, y = -0.6 and
x = 0.3, y = 3.6.
(b) From the figure, the point of intersection
of the two graphs is (-1.5 , 6).
` The solution of the simultaneous
equations is x = -1.5, y = 6.
2.
x
y
2
–2
–4
4
6
8
10
0 21 3–2 –1–4 –3
y = x2+ 2x – 3
y = –xy = 2x + 1
(a) Add the graph of y = 2x + 1 to the figure.
y = 2x + 1
x -3 -1 0 3
y -5 -1 1 7
From the figure, the solutions of the
simultaneous equations are x = -2,
y = -3 and x = 2, y = 5.
01 NSS TM 5A (E)-2P-KJ.indd 53 2010/7/20 4:37:16 PM
54 Solutions
(b) Add the graph of y = -x to the figure.
y = -x
x -4 -2 0 2
y 4 2 0 -2
From the figure, the solutions of the
simultaneous equations are x = -3.8,
y = 3.8 and x = 0.8, y = -0.8.
3. y x
y x
1
3
2= +
= -*
Substitute (i) into (ii):
x2 + 1 = 3 - x
x2 + x + 1 - 3 = 0
x2 + x - 2 = 0
(x - 1)(x + 2) = 0
` x = 1 or x = -2
Substitute these values of x into (ii):
When x = 1, y = 3 - 1 = 2.
When x = -2, y = 3 - (-2) = 5.
` The solutions of the simultaneous equations
are x = 1, y = 2 and x = -2, y = 5.
4. y x
x y
x
3
2 3
1
2=
- =
- +*
(i) + (ii):
3x = x2 - 2x + 3 + 1
x2 - 5x + 4 = 0
(x - 4)(x - 1) = 0
` x = 4 or x = 1
From (ii), y = 3x - 1 .................................... (iii)
Substitute these values of x into (iii):
When x = 4, y = 3(4) - 1 = 11.
When x = 1, y = 3(1) - 1 = 2.
` The solutions of the simultaneous equations
are x = 4, y = 11 and x = 1, y = 2.
....................................................... (i)
....................................................... (ii)
.............................................. (i)
..................................................... (ii)
5. y
xy
x
12
4
=
= +*
Substitute (ii) into (i):
x(x + 4) = 12
x2 + 4x - 12 = 0
(x - 2)(x + 6) = 0
` x = 2 or x = -6
Substitute these values of x into (ii):
When x = 2, y = 2 + 4 = 6.
When x = -6, y = -6 + 4 = -2.
` The solutions of the simultaneous equations
are x = 2, y = 6 and x = -6, y = -2.
6. ( )y
x y
x x
3 2
2
0
=
- + =
+*
(i) + (ii):
3x + 2 = x(x + 2)
3x + 2 = x2 + 2x
x2 - x - 2 = 0
(x + 1)(x - 2) = 0
` x = -1 or x = 2
Substitute these values of x into (i):
When x = -1, y = (-1)(-1 + 2) = -1.
When x = 2, y = 2(2 + 2) = 8.
` The solutions of the simultaneous equations
are x = -1, y = -1 and x = 2, y = 8.
7. x2
= 4 - x8
x2
= xx4 8-
x2 = 2(4x - 8)
x2 - 8x + 16 = 0
(x - 4)2 = 0
` x = 4 (repeated)
8. x 1
1+
+ x2
= 3
( )
( )
x x
x x
1
2 1
+
+ + = 3
x + 2x + 2 = 3x2 + 3x
3x2 - 2 = 0
x2 =
32
` x = .0 82 or x = .0 82- , cor. to 2 d.p.
.......................................................... (i)
....................................................... (ii)
.................................................. (i)
.............................................. (ii)
01 NSS TM 5A (E)-2P-KJ.indd 54 2010/7/20 4:37:20 PM
Chapter 1: More about Equations 55
9. 2x2 - 4
= 8x
2x2 - 4
= 23x
` x2 - 4 = 3x
x2 - 3x - 4 = 0
(x + 1)(x - 4) = 0
` x = 1- or x = 4
10. Let u = 5x, then the original equation becomes
u2 - 7u + 6 = 0
(u - 1)(u - 6) = 0
` u = 1 or u = 6
When u = 1, 5x = 1
` x = 0
When u = 6, 5x = 6
log 5x = log 6
x = log
log
5
6
= .1 11, cor. to 2 d.p.
11. Let u = 2x, then u
2 = (2
x)2 = (2
2)
x = 4
x.
The original equation becomes
u2 - 5u + 4 = 0
(u - 4)(u - 1) = 0
` u = 4 or u = 1
When u = 4, 2x = 4
` x = 2
When u = 1, 2x = 1
` x = 0
12. log (x + 2) + log x = 1
log (x + 2)(x) = log 10
` x2 + 2x = 10
x2 + 2x - 10 = 0
x = 2
( )
( )( )
2 1
2 2 4 1 10!- - -
= 2 44
2!-
= -1 + 11 or -1 - 11 (rejected)
= .2 32, cor. to 2 d.p.
13. (log x)2 - log x
2 = 3
(log x)2 - 2 log x = 3
(log x)2 - 2 log x - 3 = 0
(log x + 1)(log x - 3) = 0
` log x = -1 or log x = 3
` x = 10-1
or x = 103
= .0 1 = 1000
14. 4 cos2i = 3cos i
cos i(4 cos i - 3) = 0
cos i = 0 or cos i = 43
` i = 90c, 270c or
i = 41.41c, 318.59c, cor. to 2 d.p.
15. 6 sin2i - sin i - 1 = 0
(2 sin i - 1)(3 sin i + 1) = 0
2 sin i - 1 = 0 or 3 sin i + 1 = 0
sin i = 21
or sin i = -31
` i = 30c, 150c or
i = 199.47c, 340.53c, cor. to 2 d.p.
16. (a) 1x
x y
y
2 5
02
=
+ =
+*
(b) x y
x y
2 5
102
+ =
+ =*
From (ii), x = 10 - y2 .......................... (iii)
Substitute (iii) into (i):
2(10 - y2) + y = 5
20 - 2y2 + y - 5 = 0
2y2 - y - 15 = 0
(y - 3)(2y + 5) = 0
` y = 3 or y = -25
(rejected)
Substitute y = 3 into (iii):
x = 10 - 32
= 1
............................................. (i)
........................................... (ii)
01 NSS TM 5A (E)-2P-KJ.indd 55 2010/7/20 4:37:26 PM
56 Solutions
17. Let the denominator of the original fraction be
x, then
x 27 2
++
- x7
= 201
( )
( )
x x
x x
2
9 7 2
+
- + =
201
2x x
x
2
2 14
+
- =
201
x2 + 2x = 20(2x - 14)
x2 - 38x + 280 = 0
(x - 10)(x - 28) = 0
` x = 10 or x = 28
When x = 10,
the original fraction is 107
.
When x = 28,
the original fraction is 728
(rejected).
` The original fraction is 107
.
18. (a) The original width
= cmx224
The new width
= cm cmor1x x224
1225
+-
` aj k
(b) From the question,
(x - 1) 1x224
+` j = 224 + 1
( ) ( )
xx x1 224- +
= 225
224x - 224 + x2 - x = 225x
x2 - 2x - 224 = 0
(x - 16)(x + 14) = 0
` x = 16 or x = -14 (rejected)
19. (a) Draw the graph of y = 3x - x2 + 1 from
x = -1 to x = 4.
y = 3x - x2 + 1
x -1 0 1 2 3 4
y -3 1 3 3 1 -3
x
y
–1 1 2 3 4–1
1
2
3
4
5
6
–2
–3
0
y = 3x – x2+ 1
y = 2x + 4
(b) Add the graph of y = 2x + 4 to the figure
in (a).
y = 2x + 4
x -1 -0.5 0 1
y 2 3 4 6
From the figure, the two graphs have no
points of intersection.
` The simultaneous equations have no
real solutions.
01 NSS TM 5A (E)-2P-KJ.indd 56 2010/7/20 4:37:30 PM
Chapter 1: More about Equations 57
20. (a) Draw the graph of y = x2 + 5x + 10 from
x = -10 to x = 5.
x -10 -8 -5 0 3 5
y 60 34 10 10 34 60
x
y
–2–4–6–8–10 2 4–10
10
20
30
40
50
60
–20
–30
–40
0
y = x2+ 5x + 10
y = –15 – 5x
(b) Add the graph of y = -15 - 5x to the
figure in (a).
y = -15 - 5x
x -10 -5 0 5
y 35 10 -15 -40
From the figure, the solution of the
simultaneous equations is x = -5, y = 10.
21. x y
yx5 3
2 2
25
+ =
=-*
Substitute (ii) into (i):
x2 + (5 - 3x)
2 =
25
x2 + 25 - 30x + 9x
2 -
25
= 0
4x2 - 12x + 9 = 0
(2x - 3)2 = 0
` x = 23
(repeated)
Substitute x = 23
into (ii):
y = 5 - 323a k
= 21
` The solution of the simultaneous equations
is x = 23
, y = 21
.
22. ( )( )x y
x y
8 4
2
5
6
- - =
- =
-*
From (ii): x = 2y + 6 .................................... (iii)
Substitute (iii) into (i):
(2y + 6 - 8)(y - 4) = -5
2y2 - 2y - 8y + 8 + 5 = 0
2y2 - 10y + 13 = 0
The discriminant of the equation is:
D = (-10)2 - 4(2)(13)
= -4
1 0
` There are no real solutions for y.
i.e. The simultaneous equations have no real
solutions.
.................................................... (i)
..................................................... (ii)
...................................... (i)
.................................................... (ii)
01 NSS TM 5A (E)-2P-KJ.indd 57 2010/7/20 4:37:33 PM
58 Solutions
23. x y2 11
2 2
y y x
3
3
2
+ =
=+ -*
From (ii): 2(y + 3) = 3(y - x)
y = 3x + 6 ....................... (iii)
Substitute (iii) into (i):
2x2 + (3x + 6)
2 = 11
2x2 + 9x
2 + 36x + 36 = 11
11x2 + 36x + 25 = 0
(11x + 25)(x + 1) = 0
` x = -1125
or x = -1
Substitute these values of x into (iii):
When x = -1125
, y = 31125
-a k + 6 = -119
.
When x = -1, y = 3(-1) + 6 = 3.
` The solutions of the simultaneous equations
are x = -1125
, y = -119
and x = -1, y = 3.
24. Rewrite the given equation as:
x y
y x
2 3
4 3
2 2- =
=-*
From (ii): y = 4x + 3 .................................... (iii)
Substitute (iii) into (i):
2x2 - (4x + 3)
2 = 3
2x2 - 16x
2 - 24x - 9 = 3
7x2 + 12x + 6 = 0
The discriminant of the equation is:
D = (12)2 - 4(7)(6)
= -24
1 0
` There are no real solutions for x.
i.e. The simultaneous equations have no real
solutions.
................................................. (i)
............................................... (ii)
................................................... (i)
.................................................... (ii)
25. Rewrite the given equation as:
9 6 9xy x
x y
y
3 2 2
222
=
=
+
+ -*
From (i): y = 1 - x23 ................................... (iii)
Substitute (iii) into (ii):
9x x123
-a k + 6 x12
23
-a k - 9x2 = 2
9x - 227
x2 + 6 - 18x +
227
x2 - 9x
2 - 2 = 0
9x2 + 9x - 4 = 0
(3x + 4)(3x - 1) = 0
` x = -34
or x = 31
Substitute these values of x into (iii):
When x = -34
, y = 1 - 34
23
-a k = 3.
When x = 31
, y = 1 - 23
31a k =
21
.
` The solutions of the simultaneous equations
are x = -34
, y = 3 and x = 31
, y = 21
.
26. y mx
xy
1
1
=
=
-*
Substitute (i) into (ii):
x(mx - 1) = 1
mx2 - x - 1 = 0
The discriminant of the equation is:
D = (-1)2 - 4(m)(-1)
= 1 + 4m
Since the simultaneous equations have no real
solutions, there are no real solutions for x.
i.e. D = 1 + 4m 1 0
` m 1 -41
................................................... (i)
..................................... (ii)
..................................................... (i)
............................................................ (ii)
01 NSS TM 5A (E)-2P-KJ.indd 58 2010/7/20 4:37:39 PM
Chapter 1: More about Equations 59
27. 2
2( )
x
x1 + - 4 = 0
2
2( )
x
x1 + = 4
(1 + x)2 = 4x
2
1 + 2x + x2 = 4x
2
3x2 - 2x - 1 = 0
(x - 1)(3x + 1) = 0
` x = 1 or x = 31
-
28. 2( )x 1
3
- -
x 11-
= 41
2( )
( )
x
x
1
3 1
-
- - =
41
2( )x
x
1
4
-
- =
41
(x - 1)2 = 4(4 - x)
x2 - 2x + 1 = 16 - 4x
x2 + 2x - 15 = 0
(x - 3)(x + 5) = 0
` x = 3 or x = 5-
29. xx
23
-+
- xx1 -
= 417
( )
( ) ( ) ( )
x x
x x x x
2
3 1 2
-
+ - - - =
417
2
2 2
x x
x x x x
2
3 3 2
-
+ - + + =
417
2
2
x x
x
2
2 2
-
+ =
417
17(x2 - 2x) = 4(2x
2 + 2)
9x2 - 34x - 8 = 0
(9x + 2)(x - 4) = 0
` x = 92
- or x = 4
30. 3x x2 11
2 11
++ +
a k = 10
x2 11+
:( )
x
x
2 1
1 3 2 1
+
+ + = 10
2( )x
x
2 1
6 4
+
+ = 10
10(2x + 1)2 = 6x + 4
40x2 + 40x + 10 = 6x + 4
20x2 + 17x + 3 = 0
(5x + 3)(4x + 1) = 0
` x = 53
- or x = 41
-
31. Let u = 3x, then the original equation becomes
u243
= 36 - u
243 = 36u - u2
u2 - 36u + 243 = 0
(u - 27)(u - 9) = 0
` u = 27 or u = 9
When u = 27, 3x = 27
` x = 3
When u = 9, 3x = 9
` x = 2
32. Let u = 4x, then u:4
2 = 4
x:4
2 = 4
x + 2.
The original equation becomes
u:42 + u
1 = 10
16u2 - 10u + 1 = 0
(8u - 1)(2u - 1) = 0
` u = 81
or u = 21
When u = 81
, 4x =
81
22x
= 2-3
2x = -3
` x = 23
-
When u = 21
, 4x =
21
22x
= 2-1
2x = -1
` x = 21
-
33. log (x + 1) - log (3x2 - 5) = -1
log 2x
x
3 5
1
-
+ = log
101
` 2x
x
3 5
1
-
+ =
101
3x2 - 5 = 10(x + 1)
3x2 - 10x - 15 = 0
01 NSS TM 5A (E)-2P-KJ.indd 59 2010/7/20 4:37:51 PM
60 Solutions
x = 2
( ) ( )( )
( )
10 10 4 3 15
2 3
! - - -
= 6
10 280!
= 5 70
3+
or 5 70
3-
(rejected)
= .4 46, cor. to 2 d.p.
34. log (102x
- 56) - x = 0
log (102x
- 56) = x
` 102x
- 56 = 10x
(10x)2 - 10
x - 56 = 0
(10x - 8)(10
x + 7) = 0
` 10x = 8 or 10
x = -7 (rejected)
x = log 8
= .0 90, cor. to 2 d.p.
35. tan i + tan
1i
= -2
tan2i + 1 = -2 tan i
tan2i + 2tan i + 1 = 0
(tan i + 1)2 = 0
tan i = -1 (repeated)
` i = 135c, 315c
36. 5 cos i + 2 sin2i + 1 = 0
5 cos i + 2(1 - cos2i) + 1 = 0
5 cos i + 2 - 2 cos2i + 1 = 0
2 cos2i - 5 cos i - 3 = 0
(cos i - 3)(2 cos i + 1) = 0
` cos i - 3 = 0 or 2 cos i + 1 = 0
cos i = 3 (rejected) or cos i = -21
` i = 120c, 240c
37. tan i = sincos
32
i
i+
cossin
i
i =
sincos
32
i
i+
3 sin2i = cos i(2 + cos i)
3(1 - cos2i) = 2 cos i + cos
2i
4 cos2i + 2 cos i - 3 = 0
cos i = 2
( )( )
( )
2 2 4 4 3
2 4
!- - -
= 2 52
8!-
= 1 13
4- +
or 1 13
4- -
(rejected)
` i = 49.35c, 310.65c, cor. to 2 d.p.
38. (a) 3 sin2x + 4 sin x cos x - 4 cos
2x = 0
2
2
cos
sin
x
x3 + 2
cos
sin cos
x
x x4 - 2
2
cos
cos
x
x4 = 0
3 tan2x + 4 tan x - 4 = 0
(b) 3 sin2x + 4 sin x cos x - 4 cos
2x = 0
3 tan2x + 4 tan x - 4 = 0 (From (a))
(tan x + 2)(3 tan x - 2) = 0
` tan x + 2 = 0 or 3 tan x - 2 = 0
tan x = -2 or tan x = 32
` x = 116.57c, 296.57c or
x = 33.69c, 213.69c, cor. to 2 d.p.
39. (a) The original number is 10x + y and the
new number is 10y + x.
From the question,
( )x y
xy
y x
35
10 10 18
=
+ - =+*
i.e. xy
x y
35
2
=
- =*
(b) 35xy
x y 2
=
- =*
From (ii): y = x - 2 .............................. (iii)
Substitute (iii) into (i):
x(x - 2) = 35
x2 - 2x - 35 = 0
(x - 7)(x + 5) = 0
` x = 7 or x = -5 (rejected)
Substitute x = 7 into (iii):
y = 7 - 2 = 5
` The original two-digit number is 75.
.................................................. (i)
.............................................. (ii)
01 NSS TM 5A (E)-2P-KJ.indd 60 2010/7/20 4:37:56 PM
Chapter 1: More about Equations 61
40. (a) From the question,
( )
x y
x yx y 3 1
500
3 00
2 2+ =
+ + =-*
i.e. x y
x y
500
02 5
2 2+ =
+ =*
(b) x y
x y
500
2 50
2 2+ =
+ =*
From (ii): y = 50 - 2x ......................... (iii)
Substitute (iii) into (i):
x2 + (50 - 2x)
2 = 500
x2 + 2 500 - 200x + 4x
2 = 500
x2 - 40x + 400 = 0
(x - 20)2 = 0
` x = 20 (repeated)
Substitute x = 20 into (iii):
y = 50 - 2(20) = 10
` The length of the longest side of the
foundation
= (x + y) m
= (20 + 10) m
= 30 m
41. Let the number of members of the club be x.
From the question,
x 42 000
- - x
2 000 = 25
( )
( )
x x
x x
000 2 000 4
4
2 - -
- = 25
2x x4
8 000
- = 25
25(x2 - 4x) = 8 000
25x2 - 100x - 8 000 = 0
x2 - 4x - 320 = 0
(x - 20)(x + 16) = 0
` x = 20 or x = -16 (rejected)
` The number of members of the club is 20.
........................................ (i)
.......................................... (ii)
42. Let the time taken for the cyclist to finish the
journey by the longer route be x hours.
From the question,
x24
- .x 0 5
20+
= 4
( . )
( . )
x x
x x
0 5
24 0 5 20
+
+ - = 4
2.x x
x
0 5
4 12
+
+ = 4
4x + 12 = 4x2 + 2x
4x2 + 2x - 4x - 12 = 0
2x2 - x - 6 = 0
(x - 2)(2x + 3) = 0
` x = 2 or x = -23
(rejected)
` It takes 2 hours for the cyclist to finish the
journey by the longer route.
43. According to the question, we need to find the
value of x when P = 0.
` 11 000(2-0.02x
) - 8 000(2-0.04x
) - 3 000 = 0
11(2-0.02x
) - 8(2-0.04x
) - 3 = 0
Let u = 2-0.02x
, then u2 = (2
-0.02x)2 = 2
-0.04x.
The original equation becomes
11u - 8u2 - 3 = 0
8u2 - 11u + 3 = 0
(u - 1)(8u - 3) = 0
` u = 1 or u = 83
When u = 1, 2-0.02x
= 1
x = 0 (rejected)
When u = 83
, 2-0.02x
= 83
-0.02x = log
log
283
x = . log
log
0 02 283
-
= 70.75, cor. to 2 d.p.
` The machine can produce paint at most for
70.75 months.
01 NSS TM 5A (E)-2P-KJ.indd 61 2010/7/20 4:38:02 PM
62 Solutions
44. From the question, H = 5.
` 5 = 12 tan i - 59
(1 + tan2i)
25 = 60 tan i - 9 - 9 tan2i
9 tan2i - 60 tan i + 34 = 0
tan i = 2
( ) ( )( )
( )
60 60 4 9 34
2 9
! - -
= 18
60 2 376!
= 10 66
3+
or 10 66
3-
` i = 32.02c, 80.60c, cor. to 2 d.p.
` The maximum value of i is 80.60c.
45. Let the speed of the cruise ship be x km/h.
From the question,
x 3840+
+ x 3840-
= 37.5
( ) ( )
( ) ( )
x x
x x
3 3
840 3 840 3
+ -
- + + = 37.5
2x
x
9
1680
- = 37.5
37.5(x2 - 9) = 1 680x
37.5x2 - 337.5 - 1 680x = 0
5x2 - 224x - 45 = 0
(x - 45)(5x + 1) = 0
` x = 45 or x = -51
(rejected)
` The time taken for the cruise ship to arrive
at town B
= 45 3
840+
hours
= 17.5 hours
` The cruise ship arrived at town B at
11:30 p.m.
46. (a) Let the time taken for tap A alone to fill up
the tank be x minutes.
From the question,
x30
+ x 25
30-
= 1
( )
( )
x x
x x
25
30 25 30
-
- + = 1
2x x
x
25
60 750
-
- = 1
60x - 750 = x2 - 25x
x2 - 85x + 750 = 0
(x - 75)(x - 10) = 0
x = 75 or x = 10 (rejected)
` The time taken for tap A alone to fill
up the tank is 75 minutes.
(b) The time taken for tap B alone to fill up
the tank
= (75 - 25) minutes
= 50 minutes
Integrated Questions (cross-chapters)
47. Substituting x = -3 into the equation
2x2 + ax - 12b = 0, we have
2(-3)2 + a(-3) - 12b = 0
i.e. 6 - a - 4b = 0 .................. (i)
Since the equation x2 - ax + b = 0 has two
equal real roots,
the discriminant D = 0
i.e. (-a)2 - 4(1)(b) = 0
a2 - 4b = 0 .............................. (ii)
(ii) - (i):
a2 + a - 6 = 0
(a - 2)(a + 3) = 0
` a = 2 or a = -3
From (i): b = 41
(6 - a) ................................ (iii)
Substitute these values of a into (iii):
When a = 2, b = 41
(6 - 2) = 1.
When a = -3, b = 41
[6 - (-3)] = 49
.
` a = 2, b = 1 or a = -3, b = 49
.
01 NSS TM 5A (E)-2P-KJ.indd 62 2010/7/20 4:38:06 PM
Chapter 1: More about Equations 63
48. (a) From the figure, the y-intercept is 18.
` c = 18
(b) y = ax2 + bx + c
= a x x2
ab
+` j + c
= a x x2
2 2
ab
ab
ab
2 2++ -a ak k; E + c
= a x2
ab
2+a k + c -
2
ab4
a The vertex of the parabola is (3 , 0).
` c - 2
ba4
= 0
i.e. 18 - 2
ab4
= 0
2
ba4
= 18
b2 = 72a
(c) a The vertex of the parabola is (3 , 0).
` -a
b2
= 3
` b = -6a .................................... (i)
Substitute (i) into the equation of (b):
(-6a)2 = 72a
36a2 = 72a
a2 - 2a = 0
a(a - 2) = 0
` a = 0 (rejected) or a = 2
Substitute a = 2 into (i):
b = -6(2) = 12-
49. From the question, 21
ab = 1241 .................... (i)
The equation of the straight line can be written
as y = - ab
x + b
Given that the line passes through (2 , 3), we
have
3 = - ab
(2) + b
i.e. 3a = -2b + ab ................................... (ii)
From (i): a = b2
49 ........................................... (iii)
Substitute (iii) into (ii):
3b2
49a k = -2b + b2
49a kb
b2
147 = -2b +
249
4b2 - 49b + 147 = 0
(b - 7)(4b - 21) = 0
` b = 7 or b = 421
Substitute these values of b into (iii):
When b = 7, a = ( )2 749
= 27
.
When b = 421
, a = 2
49
421a k
= 3
14.
When a = 27
and b = 7,
the equation of the line AB is
y = -7
27
x + 7
i.e. 2x + y - 7 = 0
When a = 3
14 and b =
421
,
the equation of the line AB is
y = -
314421
x + 421
i.e. 9x + 8y - 42 = 0
01 NSS TM 5A (E)-2P-KJ.indd 63 2010/7/20 4:38:14 PM
64 Solutions
Lively Maths Problem
50. Let the number of chickens he bought be x.
From the question,
18x1 000
+` j(x - 5) +
10 5x1 000
#-` j = 1 000 + 220
(1 000 + 18x)(x - 5) +
5(1 000 - 10x) = 1 220x
1 000x + 18x2 - 5 000 -
90x + 5 000 - 50x = 1 220x
18x2 - 360x = 0
x2 - 20x = 0
x(x - 20) = 0
` x = 0 (rejected) or x = 20
` He bought 20 chickens.
Revision Test (P.47)
1. D From the figure, the points of intersection of
the graphs of y = x2 - 4 and y = -x + 2
are (-3 , 5) and (2 , 0).
` The solutions of the simultaneous
equations are (-3 , 5) and (2 , 0).
2. A x x
y
y
x
1
5 3
2= + +
=
-
-*
Substitute (i) into (ii):
-x2 + x + 1 = 5 - 3x
-x2 + x + 3x + 1 - 5 = 0
x2 - 4x + 4 = 0
(x - 2)2 = 0
` x = 2 (repeated)
..................................... (i)
............................................ (ii)
3. D Rewrite the given equation as:
y x
x y
x3
4
1
3 2 1
2- + =
- =-*
(i) # 2 + (ii):
-2x2 + 9x - 4 = 3
2x2 - 9x + 7 = 0
(x - 1)(2x - 7) = 0
` x = 1 or x = 27
From (ii): y = 12
(3x - 5) ..................... (iii)
Substitute these values of x into (iii):
When x = 1, y = 12
[3(1) - 5] = -1.
When x = 27
, y = 3 521
27
-a k: D = 411
.
` y = 1- or 114
4. B x41-
+ x2
1+
= 43
( ) ( )x x
x x4 22 4
- ++ + -
= 43
( ) ( )x x4 2
6- +
= 43
2x x8 2
2
+ - =
41
8 + 2x - x2 = 8
x2 - 2x = 0
x(x - 2) = 0
` x = 0 or x = 2
5. A Let u = 2x, then u
2 = (2
x)2 = 2
2x.
The original equation becomes
u2 + 2u - 8 = 0
(u + 4)(u - 2) = 0
` u = -4 or u = 2
When u = -4, 2x = -4 (rejected)
When u = 2, 2x = 2
` x = 1
..................................... (i)
.................................... (ii)
01 NSS TM 5A (E)-2P-KJ.indd 64 2010/7/20 4:38:20 PM
Chapter 1: More about Equations 65
6. C log3 (2x + 1) + log3 (x - 2) = 1
log3 (2x + 1)(x - 2) = log3 3
` (2x + 1)(x - 2) = 3
2x2 - 3x - 5 = 0
(x + 1)(2x - 5) = 0
` x = -1 (rejected) or x = 25
7. A 4 cos2i - 9 cos (180c + i) + 2 = 0
4 cos2i + 9 cos i + 2 = 0
(cos i + 2)(4 cos i + 1) = 0
` cos i + 2 = 0 or 4 cos i + 1 = 0
cos i = -2 (rejected) or cos i = -41
a 0c G i 1 180c
` i = 104.48c, cor. to 2 d.p.
i.e. The equation has 1 root.
8. B The speed of the car is (x - 5) km/h after
decreasing and the time taken will be
x 5900-
hours.
From the question, we have
x 5900-
- x900
= 2.
HKCEE Questions
9. D 10. B
11. E 12. D
13. D 14. B
15. D 16. C
01 NSS TM 5A (E)-2P-KJ.indd 65 2010/7/20 4:38:22 PM