Solution Manual ENG - Buku Teks Add Math KSSM TKT 4

Number line method:

Test point –4: Test point 0: Test point 2:

(–4)2 + 2(–4) – 3 0 (0)2 + 2(0) – 3 0 (2)2 + 2(2) – 3 0

+–+
x

–3 1
Since x2 + 2x – 3 , 0, the range of x is determined on the negative part of the number

line.
Thus, the solution for this quadratic inequality is –3 , x , 1.

Tabular method:

x , –3 –3 , x , 1 x.1

(x + 3) – + +

(x – 1) – – +

(x + 3)(x – 1) + – +

Since (x + 3)(x – 1) , 0, the range of x is determined on the negative part of the table.
Thus, the solution for this quadratic inequality is –3 , x , 1.

(f) (3x + 1)(5 – x) . 13
15x – 3x2 + 5 – x . 13
3x2 – 14x + 8 , 0 y
(3x – 2)(x – 4) , 0 8
2
x= 3   and   x = 4 0 –23
Test point 5:
Graph sketching method:
Since 3x2 – 14x + 8 , 0, the range of x is determined

on the curve of the graph below the x-axis. 4 x

Thus, the solution for this quadratic inequality
2
is 3 , x , 4.

Number line method:

Test point 0: Test point 3:

3(0)2 – 14(0) + 8 0 3(3)2 – 14(3) + 8 0 3(5)2 – 14(5) + 8 0

+–+
x
24
3
Since 3x2 – 14x + 8 , 0, the range of x is determined on the negative part of the

number line. 2
3
Thus, the solution for this quadratic inequality is , x , 4.

12

Tabular method: x , 2 2 , x , 4 x.4
3 3
(3x – 2) +
(x – 4) – + +
(3x – 2)(x – 4) +
– –

+ –

Since (3x – 2)(x – 4) , 0, the range of x is determined on the negative part of the table.
2
Thus, the solution for this quadratic inequality is 3 , x , 4.

2. 3x2 – 5x > 16 + x(2x + 1)
3x2 – 5x > 16 + 2x2 + x
x2 – 6x – 16 > 0
(x – 8)(x + 2) > 0
Thus, the range of x is x < –2 and x > 8.

Intensive Practice 2.1 (Page 44)

1. 3x(x – 5) = 2x – 1
3x2 – 15x = 2x – 1
3x2 – 17x + 1 = 0

Use the quadratic formula.

x = –b ± b2 – 4ac
2a

x = –(–17) ± (–17)2 – 4(3)(1)
2(3)

= 17 ± 277
6

x= 17 – 277 or x= 17 + 277
6 6
= 0.059 = 5.607

2. (a) 2(x – 5)2 = 4(x + 7)

2(x2 – 10x + 25) = 4x + 28

2x2 – 20x + 50 – 4x – 28 = 0

2x2 – 24x + 22 = 0

x2 – 12x + 11 = 0

(b) The sum of the roots = 12
The product of the roots = 11

3. 2x2 + 6x – 7 = 0
b c
α + β = – a , αβ = a

= – 6 = – 7
2 2

= –3

13

1 1 2α + 1 + 2β + 1 1 1 1
2α + 1 2β + 1 (2α + 1)(2β + 1) 2α + 1 2β + 1 2(α
( )( )(a)+ = , = 4αβ + + β) + 1

= 2(α + β ) + 2 1 ( )=4 – 7 1
4αβ + 2(α + β) + 2 + 2(–3) + 1

2(–3) + 2 = – 1
( )= 7 19
4 – 2 + 2(–3) + 1

= 4
19

Quadratic equation: x2 – 4 x – 1 = 0
19 19
19x2 – 4x – 1 = 0

5α 5β 5α2 + 5β 2 5α 5β 25αβ
β α αβ β α αβ
( )( )(b)+ = , =

= 5[(α + β)2 – 2αβ] = 25
αβ

( )= 7
5[(–3)2 – 2 – 2 ]

– 7
2
160
= – 7

Quadratic equation: x2 + 160 x + 25 = 0
7
7x2 + 160x + 175 = 0

(c) (α + 3β ) + (3α + β ) = 4α + 4β , (α + 3β )(3α + β ) = 3α2 + 10αβ + 3β2
= 4(α + β) = 3[(α + β )2 – 2αβ] + 10αβ

= 4(–3) = [3 (–3)2 – 2(– 7 )] + 10(– 7 )
= –12 2 2
= 13
Quadratic equation: x2 + 12x + 13 = 0

4. 3x2 + 19x + k = 0

When x = –7,
3(–7)2 + 19(–7) + k = 0

147 – 133 + k = 0

k = –14

5. rx2 + (r – 1)x + 2r + 3 = 0

a = r, b = r – 1, c = 2r + 3

(a) Assume α is the first root and –α is the second root.

α + (–α) = –(r – 1)
r
0 = –r + 1

r=1

14

(b) Assume a is the first root and 1 is the second root.
a
a1 1 2 = 2r + 3
a r

r = 2r + 3

r = –3

(c) Assume a is the first root and 2a is the second root.

a + 2a = (r – 1)
r
(r – 1)
3a = r

a = 1 – r …1
r
2r + 3
a(2a) = r

2a2 = 2r + 3 …2
r
Substitute 1 into 2.

21 1– r 22 = 2r + 3
3r r
2(1 – 2r + r2) 2r + 3
9r2 = r

2 – 4r + 2r2 = 18r2 + 27r

16r2 + 31r – 2 = 0

(16r – 1)(r + 2) = 0
1
r = 16 and r = –2

6. x2 – 8x + m = 0
Assume a is the first root and 3a is the second root.
–28 (–18)   , a3((332aa))22 ===
a + 3a = m
4a = m
a = m


m = 12

Substitute m = 12 into the equation

x2 – 8x + 12 = 0

(x – 2)(x – 6) = 0

x = 2 and x = 6

Thus, m = 12 and the roots are 2 and 6.

7. x2 + 2x = k(x – 1)

x2 + 2x – kx + k = 0

x2 + (2 – k)x + k = 0
Assume a is the first root and (a + 2) is the second root.
a + a + 2 = –2 + k
2a + 2 = –2 + k
k = 2a + 4 …1
a(a + 2) = k
a2 + 2a = k …2
15

Substitute 1 into 2.
a2 + 2a = 2a + 4
a2 = 4
a = 2 or –2
Since the roots are non-zero, the roots are 2 and 4.

Substitute a = 2 into 1.
k = 2(2) + 4
= 8

8. x2 + px + 27 = 0
Assume that a is the first root and 3a is the second root.
a + 3a = –p
4a = –p
p
a = – 4 …1

a(3a) = 27
3a2 = 27
a2 = 9 … 2

Substitute 1 into 2.
p
1– 4 22 = 9

p2 = 9
16
p2 = 144

p = –12 or 12

9. x2 + (k – 1)x + 9 = 0
3 + h + 1 = –k + 1
k = –h – 3 … 1
3(h + 1) = 9
3h + 3 = 9
3h = 6
h = 2 … 2

Substitute 2 into 1.
k = –2 – 3
= –5
Thus, the possible values for h and k are 2 and –5 respectively.

10. x2 – 8x + c = 0
a + a + 3d = 8
2a + 3d = 8
8 – 3d
a = 2 …1

a(a + 3d) = c
a2 + 3ad = c … 2

16

Substitute 1 into 2.

( ) ( )8 – 3d
2
2+3 8 – 3d d=c
2
64 – 48d + 9d2 24d – 9d2
4 + 2 =c

64 – 48d + 9d2 + 48d – 18d2 = 4c

64 – 9d2 = 4c
64 – 9d2
c= 4

11. (a) 2x2 x+1
2x2 – x – 1
0

(2x + 1)(x – 1) 0 – 1 or x 1.
Thus, the range of x is x 2

(b) (x – 3)2 5 – x

x2 – 6x + 9 5 – x
x2 – 5x + 4 0

(x – 1)(x – 4) 0
Thus, the range of x is 1 р x р 4.

(c) (1 – x)2 + 2x 17
1 – 2x + x2 + 2x 17
x2 – 16 0

(x + 4)(x – 4) 0
Thus, the range of x is –4 Ͻ x Ͻ 4.

12. (a) (x + 3)(x – 4) 0
x2 – x – 12 0
x2 – x 12

Comparing with x2 + mx Ͻ n

Thus, m = –1 and n = 12

(b) (x + 2)(x – 5) 0
x2 – 3x – 10 0
2x2 – 20 6x

Comparing with 2x2 + m Ͼ nx

Thus, m = –20 and n = 6

13. (x – 2)(x – a) 0 17
x2 – ax – 2x + 2a 0

x2 + (–a – 2)x + 2a 0
2x2 + 2(–a – 2)x + 4a 0
Comparing with 2x2 + bx + 12 Ͻ 0

4a = 12

a=3

b = 2(–a – 2)

= 2(–3 – 2)

= –10

Inquiry 3   (Page 45)

4. Equation Value of a Value of b Value of c Roots
y = x2 + 5x + 4 1 5 4 –1, –4
y = x2 – 6x + 9 1 –6 9 3, 3
y = 9x2 – 6x + 2 9 –6 2 No roots

Mind Challenge (Page 45)

When the discriminant b2 − 4ac < 0, the equation has an imaginary or complex roots.

Mind Challenge (Page 46)

To know if the graph touches only one point on the x-axis or intersects the x-axis on two
different points or does not intersect the x-axis.

Self Practice 2.4   (Page 46)

1. (a) x2 + 4x + 1 = 0
b2 – 4ac = 42 – 4(1)(1)

= 16 – 4
= 12 (. 0)

This equation has two different real roots.

(b) x2 = 8(x – 2)

x2 – 8x + 16 = 0

b2 – 4ac = (–8)2 – 4(1)(16)

= 64 – 64

= 0

This equation has two equal real roots.

(c) 5x2 + 4x + 6 = 0
b2 – 4ac = 42 – 4(5)(6)

= 16 – 120
= –104 (, 0)

This equation does not have real roots.

(d) –3x2 + 7x + 5 = 0
b2 – 4ac = 72 – 4(–3)(5)

= 49 + 60
= 109 (. 0)

This equation has two different real roots.

(e) –x2 + 10x – 25 = 0
b2 – 4ac = 102 – 4(–1)(–25)

= 100 – 100

= 0

This equation has two equal real roots.

18

(f) (2x – 1)(x + 3) = 0
2x2 + 5x – 3 = 0
b2 – 4ac = 52 – 4(2)(–3)

= 25 + 24
= 49 (. 0)

This equation has 2 different real roots.

Self Practice 2.5   (Page 48)

1. (a) 9x2 + p + 1 = 4px

9x2 – 4px + p + 1 = 0

For two equal real roots,

b2 – 4ac = 0

(–4p)2 – 4(9)(p + 1) = 0

16p2 – 36p – 36 = 0

4p2 – 9p – 9 = 0

(4p + 3)(p – 3) = 0 3
4
p = – or p = 3

(b) x2 + (2x + 3)x = p

x2 + 2x2 + 3x – p = 0

3x2 + 3x – p = 0

For two different real roots,
b2 – 4ac . 0
32 – 4(3)(–p) . 0
9 + 12p . 0
12p . –9
3
p . – 4

(c) x2 + 2px + (p – 1)(p – 3) = 0

x2 + 2px + p2 – 4p + 3 = 0

For having no real roots,
b2 – 4ac , 0

(2p)2 – 4(1)(p2 – 4p + 3) , 0
4p2 – 4p2 + 16p – 12 , 0
16p , 12
3
4
p ,

2. x2 + k = kx – 3
x2 – kx + k + 3 = 0

For two different real roots,
b2 – 4ac . 0
(–k)2 – 4(1)(k + 3) . 0
k2 – 4k – 12 . 0
(k + 2)(k – 6) . 0
k , –2 or k . 6

19

For two equal real roots,

b2 – 4ac = 0

(–k)2 – 4(1)(k + 3) = 0

k2 – 4k – 12 = 0

(k + 2)(k – 6) = 0

k = –2 or k = 6

3. (a) x2 + hx + k = 0

–2 + 6 = –h   , (–2)(6) = k

h = –4 k = –12

(b) x2 – 4x – 12 = c

x2 – 4x – 12 – c = 0

For having no real roots,
b2 – 4ac , 0

(–4)2 – 4(1)(–12 – c) , 0
16 + 48 + 4c , 0
64 + 4c , 0
4c , –64
c , –16


4. hx2 + 3hx + h + k = 0

For having two equal real roots,

b2 – 4ac = 0

(3h)2 – 4(h)(h + k) = 0

9h2 – 4h2 – 4hk = 0

4hk = 5h2
5
k = 4 h

5. ax2 – 5bx + 4a = 0

For having two equal real roots,

b2 – 4ac = 0

(–5b)2 – 4(a)(4a) = 0

25b2 – 16a2 = 0

25b2 = 16a2

1 a 22 = 25
b 16
a 5
b = 4

Thus, a : b = 5 : 4

Intensive Practice 2.2 (Page 48)

1. (a) x2 – 8x + 16 = 0
b2 – 4ac = (–8)2 – 4(1)(16)
= 64 – 64
= 0
The equation has two equal real roots.

20

(b) (x – 2)2 = 3
x2 – 4x + 4 – 3 = 0
x2 – 4x + 1 = 0
b2 – 4ac = (–4)2 – 4(1)(1)
= 16 – 4
= 12
The equation has different real roots.

(c) 2x2 + x + 4 = 0
b2 – 4ac = 12 – 4(2)(4)
= 1 – 32
= –31

The equation has no real roots.

2. (a) x2 + kx = 2x – 9

x2 + (k – 2)x + 9 = 0

b2 – 4ac = 0

(k – 2)2 – 4(1)(9) = 0

k2 – 4k + 4 – 36 = 0

k2 – 4k – 32 = 0

(k + 4)(k – 8) = 0

k = – 4 or k = 8

(b) kx2 + (2k + 1)x + k – 1 = 0

b2 – 4ac = 0

(2k + 1)2 – 4(k)(k – 1) = 0

4k2 + 4k + 1 – 4k2 + 4k = 0

8k + 1 = 0

8k = –1 1
8
k = –

3. (a) x(x + 1) = rx – 4

x2 + (1 – r)x + 4 = 0

b2 – 4ac 0

(1 – r)2 – 4(1)(4) 0

1 – 2r + r2 – 16 0

r2 – 2r – 15 0

(r + 3)(r – 5) 0

r –3 or r 5

(b) x2 + x = 2rx – r2

x2 + (1 – 2r)x + r2 = 0

b2 – 4ac 0

(1 – 2r)2 – 4(1)(r2) 0

1 – 4r + 4r2 – 4r2 0

1 – 4r 0

4r 1
1
r 4

21

4. (a) (1 – p)x2 + 5 = 2x

(1 – p)x2 – 2x + 5 = 0

b2 – 4ac 0

(–2)2 – 4(1 – p)(5) 0

4 – 20 + 20p 0

20p 16

p 4
5
(b) 4px2 + (4p + 1)x + p – 1 = 0

b2 – 4ac 0

(4p + 1)2 – 4(4p)(p – 1) 0

16p2 + 8p + 1 – 16p2 + 16p 0

24p –1 1
24
p –

5. (a) kx2 – 10x + 6k = 5

kx2 – 10x + 6k – 5 = 0

b2 – 4ac = 0

(–10)2 – 4(k)(6k – 5) = 0

100 – 24k2 + 20k = 0

6k2 – 5k – 25 = 0

(3k + 5)(2k – 5) = 0 5 5
3 2
k = – or k=

(b) Substitute k = – 5 into the equation,
3
5( )– 5
3 x2 – 10x + 6 – 3 –5=0

–5x2 – 30x – 30 – 15 = 0

5x2 + 30x + 45 = 0

x2 + 6x + 9 = 0

(x + 3)(x + 3) = 0

x = –3

Therefore, the root is x = –3.

6. x(x – 4) + 2n = m
x2 – 4x + 2n – m = 0

For 2 equal real roots,
b2 – 4ac = 0

(–4)2 – 4(1)(2n – m) = 0

16 – 8n + 4m = 0

4m = 8n – 16

m = 2n – 4

7. (a) b2 – 4c = 16……
b – c = –4
b = c – 4……

22

Substitute 2 into 1,

(c – 4)2 – 4c = 16

c2 – 8c + 16 – 4c = 16

c2 – 12c = 0

c(c – 12) = 0

Thus, c = 12
Substitute c = 12 into 2,

b = 12 – 4

= 8

Thus, b = 8 dan c = 12.

(b) x2 + 8x + 12 = 0
(x + 6)(x + 2) = 0
x = −6 or x = −2
Thus, the roots are –6 and –2.

8. (a) 2x2 – 5x + c = 0

For no real roots,
b2 – 4ac , 0
(–5)2 – 4(2)(c) , 0
25 – 8c , 0
8c . 25
c . 3.125



Thus, the possible values for c1 is 4 and c2 is 5.

(b) 2x2 – 5x + 1 (4 + 5) = 0
2
2x2 – 5x + 4.5 = 0

b2 – 4ac = (–5)2 – 4(2)(4.5)

= 25 – 36

= –11

Thus, the equation does not have two real roots.

Inquiry 4   (Page 49)

4. Changes in shape of position of the graph of function f(x) = ax2 + bx + c

Only the value • Change in value of a affects the shape and width of the graph however
of a changes the y-intercept remains unchanged.

• When a . 0, the shape of the graph is which passes through the
minimum point and when a , 0, the shape of the graph is which
passes through the maximum point.

• For the graphs a . 0, for example a = 1, when the value of a is larger
than 1, the width of the graph decreases. Conversely, when the value of a
is smaller than 1 and approaches 0, the width of the graph increases.

• For the graphs a , 0, for example, a = –1, when the value of a is smaller
than –1, the width of the graph decreases. Conversely, when the value of
a increases from –1 and approaches 0, the width of the graph increases.

23

Only the value • Change in value of b only affects the position of vertex with respect to the
of b changes y-axis, however the shape of the graph and the y-intercept are unchanged.

Only the value • When b = 0, the vertex is on the y-axis.
of c changes • For graphs a . 0, when b . 0, the vertex is on the left side of the y-axis

and when b , 0, the vertex is on the right side of the y-axis.
• For graphs a , 0, when b . 0, the vertex is on the right side of the

y-axis and when b , 0, the vertex is on the left side of the y-axis.

• Change in value of c only affects the position of graph of function
vertically upwards or downwards.

• The shape of the graph is unchanged.

Self Practice 2.6   (Page 51) (ii) When a cthhaenggreaspfhroinmcr–e1asteoa–nd41t,htehe
width of
4. (a) (i) When a changes from –1 to –3, the
width of the graph will decrease and y-intercept does not change
the y-intercept does not change.
y
y
6
6 y = –x2 + x + 6

–2 0 x 20 y = –x2 + x + 6
3
x
3

(b) When the value of b changes from 1 to –1, the vertex is on the left of the y-axis. All
points change except the y-intercept. The shape of the graph does not change.
y

6 y = –x2 + x + 6

–2 0 x
3

24

(c) When the value of c changes from 6 to –2, the graph moves 8 units downward. The
shape of the graph does not change.
y

6 y = –x2 + x + 6

–2 0 x
–2 3

Inquiry 5   (Page 51)

Discriminant Type of roots and position Position of graph of function
b2 – 4ac of graph f(x) = ax2 + bx + c

b2 – 4ac . 0 • 2 different real roots. a.0 a,0
• The graph intersects the
b2 – 4ac = 0 α βx α βx
x-axis at two different α=β x
points.

• 2 equal real roots.
• The graph touches the

x-axis at only one point.

α =β x

b2 – 4ac , 0 • No real roots. x
• The graph does not
x
intersect any point on the
x-axis.

Self Practice 2.7   (Page 54) x

1. (a) f(x) = –3x2 + 6x – 3
b2 – 4ac = 62 – 4(–3)(–3)
= 36 – 36
= 0
The quadratic function has two equal real roots. Since a , 0, the
graph f(x) is a parabola that passes through a maximum point and
touches the x-axis at one point.

25

(b) f(x) = x2 + 2x – 3 x
b2 – 4ac = 22 – 4(1)(–3) x
= 4 + 12
= 16
The quadratic function has two different real roots. Since a . 0,

the graph f(x) is a parabola that passes through a minimum point
and intersects the x-axis at two points.

(c) f(x) = 4x2 – 8x + 5
b2 – 4ac = (–8)2 – 4(4)(5)
= 64 – 80
= –16
The quadratic function has no real roots.
Since a . 0, the graph f(x) is a parabola that passes through a

minimum point and is above the x-axis.

2. (a) f(x) = x2 – 2hx + 2 + h

For two equal real roots,
b2 – 4ac = 0

(–2h)2 – 4(1)(2 + h) = 0
4h2 – 8 – 4h = 0
h2 – h – 2 = 0

(h + 1)(h – 2) = 0

h = –1 or h = 2

(b) f(x) = x2 – (h + 3)x + 3h + 1 h=5

For two equal real roots,
b2 – 4ac = 0

(–h – 3)2 – 4(1)(3h + 1) = 0
h2 + 6h + 9 – 12h – 4 = 0
h2 – 6h + 5 = 0

(h – 1) (h – 5) = 0

h = 1 or

3. (a) f(x) = 5x2 – (qx + 4)x – 2
= 5x2 – qx2 – 4x – 2
= (5 – q)x2 – 4x – 2

For two different real roots,
b2 – 4ac 0

(–4)2 – 4(5 – q)(–2) 0

16 + 40 – 8q 0

8q 56

q7

(b) f(x) = (q + 2)x2 + q(1 – 2x) – 5
= (q + 2)x2 – 2qx + q – 5

26

For two different real roots,
b2 – 4ac 0

(–2q)2 – 4(q + 2)(q – 5) 0

4q2 – 4(q2 – 3q – 10) 0

4q2 – 4q2 + 12q + 40 0

12q – 40

q – 10
3

4. (a) f(x) = rx2 + 4x – 6

For no real roots,

b2 – 4ac 0

42 – 4(r)(–6) 0

16 + 24r 0

24r –16

r – 2
3

(b) f(x) = rx2 + (2r + 4)x + r + 7

For no real roots,

b2 – 4ac 0

(2r + 4)2 – 4(r)(r + 7) 0

4r2 + 16r + 16 – 4r2 – 28r 0

12r 16

r 4
3

Inquiry 6   (Page 55)

3. Form of Quadratic x-intercept y-intercept Vertex Axis of
function function symmetry
(4, –4)
Vertex form f(x) = (x – 4)2 – 4 2 and 6 12 (4, –4) x=4
(4, –4)
General form f(x) = x2 – 8x + 12 2 and 6 12 x=4

Intercept form f(x) = (x – 2)(x – 6) 2 and 6 12 x=4

4. y x = 4

f(x) = (x – 4)2 – 4
12

f(x) = x2 – 8x + 12

f(x) = (x – 2)(x – 6)

02 6 x

(4, –4)

Mind Challenge   (Page 56)

Yes, I agree. Only the graph with vertex form or general form that has an x-intercept can be
expressed in intercept form.

27

Self Practice 2.8   (Page 57)

1. f(x) = 2(x – 3)2 – 8
= 2(x2 – 6x + 9) – 8
= 2x2 – 12x + 10
= 2(x2 – 6x + 5)
= 2(x – 1)(x – 5)
Comparing with f(x) = a(x – p)(x – q), therefore a = 2, p = 1 and q = 5

2. (a) f(x) = (x – 2)2 – 1
= x2 – 4x + 4 – 1
= x2 – 4x + 3
= (x – 1)(x – 3)
General form: f(x) = x2 – 4x + 3
Intercept form: f(x) = (x – 1)(x – 3)

(b) f(x) = 9 – (2x – 1)2
= 9 – (4x2 – 4x + 1)
= –4x2 + 4x + 8
= –4(x2 – x – 2)
= –4(x + 1)(x – 2)
General form: f(x) = –4x2 + 4x + 8
Intercept form: f(x) = –4(x + 1)(x – 2)

(c) f(x) = 2(x + 1)2 – 18
= 2(x2 + 2x + 1) – 18
= 2x2 + 4x – 16
= 2(x2 + 2x – 8)
= 2(x + 4)(x – 2)
General form: f(x) = 2x2 + 4x – 16
Intercept form: f(x) = 2(x + 4)(x – 2)

3. The vertex is (–4, –5).

General form

f(x) = – 1 (x + 4)2 – 5
2
1
=– 2 (x2 + 8x + 16) – 5

=– 1 x2 – 4x – 13
2

4. (a) From the graph, h = 2 and k = 16
At the point (0, 12),
12 = a(0 + 2)2 + 16
12 = 4a + 16
4a = –4
a = –1
Thus, a = –1, h = 2 and k = 16.

28

(b) f(x) = –(x + 2)2 + 16
= –(x2 + 4x + 4) + 16
= –x2 – 4x + 12
= –(x2 + 4x – 12)

= –(x + 6)(x – 2)
General form: f(x) = –x2 – 4x + 12

Intercept form: f(x) = –(x + 6)(x – 2)

5. (a) f(x) = x2 – x – 6

( ) ( )= x2 – x + –1 2 – –1 2 – 6
2 2
( )= 1 25
x– 2 2– 4

(b) f(x) = –x2 – 2x + 4

= – (x2 + 2x – 4)

[ ( ) ( ) ]= – x2 + 2x + 2 2– 2 2–4
2 2
= –[(x + 1)2 – 5]

= –(x + 1)2 + 5

(c) f(x) = –2x2 – x + 6

( )= –2 x2 + 1 x – 3
2
[ ( ) ( ) ]= –2 x2 +1 1 1
2 x + 4 2– 4 2–3

[( ) ]= –2x+ 1 2 – 49
4 16

( )= –2 x + 1 2+ 49
4 8

(d) f(x) = 3x2 – 2x – 9

( )=3 x2 – 2 x – 3
3
[ ( ) ( ) ]= 3 x2 –2 1 1
3 x + – 3 2– – 3 2–3

= 3[(x – )1 2 – 28 ]
9
3

( )= 3 x – 1 2– 28
3 3

(e) f(x) = (x + 2)(6 – x)

= 6x – x2 + 12 – 2x

= –(x2 – 4x – 12)

[ ( ) ( ) ]= – x2 – 4x + –4 2 – –4 2 – 12
2 2

= –[(x – 2)2 – 16]

= –(x – 2)2 + 16

29

(f) f(x) = 2 (x + 4)(x – 2)

= 2(x2 + 2x – 8)

[ ( ) ( ) ]= 2 x2 + 2x +2 2– 2 2–8
2 2
= 2[(x + 1)2 – 9]

= 2(x + 1)2 – 18

Inquiry 7 (Page 57)

5. Change in the shape and position of the graph of function

Only the value • The change in the value of a affects the shape and width of the graph.
of a changes • When a . 0, the shape of the graph is and it passes through a

Only the value minimum point and when a , 0, the shape of the graph is and it
of h changes passes through a maximum point.
• For graphs with a . 0, for example, a = 2, when the value of a gets
Only the value larger than 2, the width of the graph decreases. Conversely, when the
of k changes value of a gets smaller than 2 and approaches 0, the width of the graph
increases.
• For graphs with a , 0, for example, a = –2, when the value of a gets
smaller than –2, the graph shrinks. Conversely, when the value gets
larger than –2 and approaches 0, the graph becomes wider.
• The axis of symmetry and maximum or minimum value does not
change.

• The change in value of h only shows the horizontal movement of the
graph.

• When the value of h increases, the graph will move to the right and
when the value of h decreases, the graph will move to the left.

• The position of the axis of symmetry changes but the minimum and
maximum value will not change.

• The change in the value of k shows the vertical movement of the graph.
• When the value of k increases, the graph will move upwards and when

the value of k decreases, the graph will move downwards.
• The minimum and maximum value changes but the axis of symmetry

does not change.

Self Practice 2.9 (Page 59) y
4
1. (a) The maximum point is (2, 4) amd the equation for
the axis of symmetry is x = 2. f(x)= –3(x – 2)2 + 4

(b) (i) When the value of a changes from –3 to –10, x
the width of the graph decreases. The axis of
symmetry, x = 2 and the maximum value,
4 does not change.

02

30

(ii) When the value of h changes from 2 to 5, the y
graph with the same shape moves horizontally f(x)= –3(x – 2)2 + 4
3 units to the right. The equation of the axis
of symmetry becomes x = 5 and its maximum 4
value does not change, which is 4.
02 5 x
(iii) When the value of k changes from 4 to –2, the
graph with the same shape moves vertically y
6 units downwards. Its maximum value f(x)= –3(x – 2)2 + 4
becomes –2 and the axis of symmetry does
not change. 4

02 x
–2

2. (a) From the graph f(x) = (x – 3)2 + 2k and the maximum point (h, –6)

h = 3 2k = –6

 k = –3

Substitute the value of h and k into f(x), we obtain

f(x) = (x – 3)2 – 6

From point (0, p),

p = (0 – 3)2 – 6

= 9 – 6

= 3

Thus, h = 3, k = –3 and p = 3.

(b) When the graph moves 2 units to the right, the value of h increases by 2. Thus, the

equation of the axis of symmetry is x = 5.

(c) When the curve moves 5 units upwards, the value of k increases by 5. Thus, the

minimum value is –1.

3. (a) The graph moves 6 units to the right and the width of the graph increases. The equation of

the axis of symmetry becomes x = 6 and its minimum value does not change, which is 0.

(b) The graph moves 1 unit to the right and 5 units upwards and the width of the graph

decreases. The equation of the axis of symmetry becomes x = 1 and its minimum value

becomes 5.

(c) The graph moves 1 unit to the left and 4 units downwards and the width of the graph

increases. The equation of the axis of symmetry becomes x = –1 and its minimum value

becomes – 4. 31

Self Practice 2.10 (Page 61)

1. (a) f(x) = (x – 1)2 – 4

= x2 – 2x – 3
Since a . 0, f(x) has a minimum point.
y

b2 – 4ac = (–2)2 – 4(1)(–3)

= 4 + 12 f (x) = (x −1)2 − 4
= 16 (. 0)

The curve intersects the x-axis at 2 different points. −1 0 3 x
−3 (1, −4)
f(x) = x2 – 2x – 3 –2 22 –2
1 2 1 2 22
= x2 – 2x + – – 3

= (x – 1)2 – 4

The minimum point is (1, –4) and the axis of symmetry is x = 1.

When f(x) = 0

x2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x = 3 and x = –1

The intersection of the x-axis is at x = –1 and x = 3.

When x = 0,

f(0) = 02 – 2(0) – 3

= –3

The graph intersects the y-axis at (0, –3).

(b) f(x) = 2(x + 2)2 – 2
= 2(x2 + 4x + 4) – 2
= 2x2 + 8x + 6
Since a . 0, f(x) has a minimum value.

b2 – 4ac = 82 – 4(2)(6)

= 64 – 48 y
= 16 (. 0)

The curve intersects the x-axis at 2 different points. f (x) = 2 (x + 2)2 − 2

f(x) = 2x2 + 8x + 6 6

= 2(x2 + 4x + 3)

= 23x2 + 4x + 1 4 22 – 1 4 22 + 34 −3 −1 0 x
2 2 ( −2, −2)
= 2(x + 2)2 – 2

The minimum point is (–2, –2) and the axis of symmetry is x = –2

When f(x) = 0,

2x2 + 8x + 6 = 0

x2 + 4x + 3 = 0

(x + 3)(x + 1) = 0

x = –3 and x = –1

The graph intersects the x-axis at x = –3 and x = –1.

32

When x = 0,
f(0) = 2(0)2 + 8(0) + 6

= 6

The graph intersects the y-axis at (0, 6).

(c) f(x) = 9 – (x – 2)2

= 9 – (x2 – 4x + 4)

= –x2 + 4x + 5
Since a , 0, f(x) has a maximum point.

b2 – 4ac = (4)2 – 4(–1)(5)

= 16 + 20 y

= 36 (, 0) (2, 9)

The curve intersects the x-axis at 2 different points. 5
f(x) = –x2 + 4x + 5
f (x) = 9 − (x − 2)2
= –(x2 – 4x – 5) –4 x
–3x2 1 –4 22 1 2 22 54
= – 4x + 2 – – −1 0 5

= –[(x – 2)2 – 9]

= –(x – 2)2 + 9

The minimum point is (2, 9) and the axis of symmetry is x = 2.

When f(x) = 0,

–x2 + 4x + 5 = 0

x2 – 4x – 5 = 0

(x + 1)(x – 5) = 0

x = –1 and x = 5

The intersection of the x-axis is x = –1 and x = 5.

When x = 0,

f(0) = –(0)2 + 4(0) + 5

= 5

The graph intersects the x-axis at (0, 5).

(d) f(x) = –2(x – 1)(x – 3)
= –2(x2 – 4x + 3)
= –2x2 + 8x – 6
Since a , 0, f(x) has a maximum point.

b2 – 4ac = (8)2 – 4(–2)(–6)

= 64 – 48 y

= 16 (. 0) (2, 2) x
The curve intersects the x-axis at 2 different points. 01 3

f(x) = –2(x2 – 4x + 3) f (x) = −2 (x −1) (x − 3)
–4 –4 −6
= –23x2 – 4x + 1 2 22 – 1 2 22 + 34

= –2[(x – 2)2 – 1]

= –2(x – 2)2 + 2

The minimum point is (2, 2) and the axis of symmetry is x = 2.

33

When f(x) = 0

–2x2 + 8x – 6 = 0

x2 – 4x + 3 = 0

(x – 1)(x – 3) = 0

x = 1 and x = 3

The intersection of the x-axis is x = 1 and x = 3.

When x = 0,
f(0) = –2(0)2 + 8(0) – 6
= –6
The graph intersects the y-axis at (0, –6).

(e) f(x) = –(x + 3)(x + 5)
= –(x2 + 8x + 15)
= –x2 – 8x – 15
Since a , 0, f(x) has a maximum point.

b2 – 4ac = (–8)2 – 4(–1)(–15)
= 64 – 60
= 4 (. 0)
The curve intersects the x-axis at two different points.

f(x) = –(x2 + 8x + 15)

= –3x2 + 8x + 1 8 22 – 1 8 22 + 154
2 2
= –[(x + 4)2 – 1]

= –(x + 4)2 + 1

The maximum point is (– 4, 1) and the axis of symmetry is x = – 4. y

When f(x) = 0, ( − 4, 1)
−5 −3
–x2 – 8x – 15 = 0 0 x

x2 + 8x + 15 = 0

(x + 3)(x + 5) = 0 f (x) = − (x + 3) (x + 5)

x = –3 and x = –5 −15

The intersection of the x-axis is x = –3 and x = –5.

When x = 0,
f(0) = –(0)2 – 8(0) – 15

= –15

The graph intersects the y-axis at (0, –15).

(f) f(x) = 2(x + 1)(x – 3)
= 2(x2 – 2x – 3)
= 2x2 – 4x – 6
Since a . 0, f(x) has a minimum point.

b2 – 4ac = (–4)2 – 4(2)(–6)
= 16 + 48
= 64 (. 0)

34

The curve intersects the x-axis at two different points, y

f(x) = 2(x2 – 2x – 3) f (x) = 2 (x + 1) (x − 3)

= 23x2 – 2x + 1 –2 22 – 1 –2 22 – 34 −1 0 3 x
2 2
= 2[(x – 1)2 – 4]
−6
= 2(x – 1)2 – 8
The minimum point is (1, –8) and the axis of symmetry is x = 1. (1, − 8)

When f(x) = 0,

2x2 – 4x – 6 = 0

x2 – 2x – 3 = 0

(x + 1)(x – 3) = 0

x = –1 dan x = 3

The intersection of the x-axis is x = –1 and x = 3.

When x = 0,
f(0) = 2(0)2 – 4(0) – 6

= –6

The graph intersects the y-axis at (0, –6).

(g) f(x) = –x2 + 4x + 5
= –(x2 – 4x – 5)
Since a , 0, f(x) has a maximum point.

b2 – 4ac = (4)2 – 4(–1)(5)
= 16 + 20
= 36 (. 0)
The curve intersects the x-axis at two different points.

f(x) = –(x2 – 4x – 5)

= –3x2 – 4x + 1 –4 22 – 1 –4 22 – 54
2 2
= –[(x – 2)2 – 9]

= –(x – 2)2 + 9

The minimum point is (2, 9) and the axis of symmetry is x = 2.

When f(x) = 0, y
(2, 9)
–x2 + 4x + 5 = 0
f (x) = −x 2 + 4x + 5
x2 – 4x – 5 = 0 5

(x + 1)(x – 5) = 0

x = –1 and x = 5

The intersection with the x-axis is x = –1 and x = 5. 5x

When x = 0, −1 0
f(0) = –(0)2 + 4(0) + 5

= 5

The graph intersects the y-axis at (0, 5).

35

(h) f(x) = 2x2 + 3x – 2
Since a . 0, f(x) has a minimum point.

b2 – 4ac = (3)2 – 4(2)(–2)
= 9 + 16
= 25 (. 0)
The curve intersects the x-axis at two different points.

f(x) = 2x2 + 3x – 2 3 3
3 4 22 4 22
= 23x2 + 2 x + 1 – 1 – 14

= 231x + 3 22 – 25 4
4 16

= 21x + 3 22 – 25
4 8
1– 3 25 2 3
The minimum point is 4 , – 8 and the axis of symmetry is x = – 4 .

When f(x) = 0, y

2x2 + 3x – 2 = 0

(2x – 1)(x + 2) = 0 f (x) = 2x 2+3x − 2
1
x = 2 and x = –2 x

The intersection of the x-axis is x­ = 1 and x = –2. −2 _1
2 −22
( )–
When x = 0, 3_ , – 3 1_
4 8
f(0) = 2(0)2 + 3(0) – 2

= –2

The graph intersects the y-axis at (0, –2).

(i) f(x) = –x2 + 4x + 12
= –(x2 – 4x – 12)
Since a , 0, f(x) has a maximum point.

b2 – 4ac = (4)2 – 4(–1)(12)
= 16 + 48
= 64 (. 0)
The curve intersects the x-axis on two different points.

f(x) = –(x2 – 4x – 12) – 4
= –3x2 – 4x + 1 – 4 22 1 2 22 124
2 – –

= –[(x – 2)2 – 16]

= –(x – 2)2 + 16

The minimum point is (2, 16) and the axis of symmetry is x = 2. y
(2, 16)
When f(x) = 0,
–x2 + 4x + 12 = 0 12 f (x) = −x 2 + 4 x + 12

x2 – 4x – 12 = 0
(x – 6)(x + 2) = 0

x = 6 dan x = –2 −2 0 x
6
The intersection of the x-axis is x = –2 and x = 6.

36

When x = 0,
f(0) = –(0)2 + 4(0) + 12

= 12

The graph intersects the x-axis at (0, 12).

Self Practice 2.11   (Page 63)

1. Given h(t) = –5t2 + 8t + 4
(a) When t = 0, h(0) = –5(0)2 + 8(0) + 4

= 4

The height of the diving board from the surface of the water is 4 m.

(b) The x coordinate of the vertex = – b
2a
8
= – 2(–5)

= 0.8

The time taken for the diver to dive at maximum height is 0.8 seconds.

(c) When t = 0.8,
h(0.8) = –5(0.8)2 + 8(0.8) + 4

= 7.2

The maximum height that is achieved by the diver is 7.2 m.

(d) h(t) = 0

–5t2 + 8t + 4 = 0

5t2 – 8t – 4 = 0

(5t + 2)(t – 2) = 0 2
5
t = – or t = 2

The range for the time that the diver is in the air is 0 , t , 2 seconds.

2. Given h(x) = 15 – 0.06x2

(a) When x = 0,
h(0) = 15 – 0.06(0)2

= 15

The maximum height of the tunnel is 15 metres.

(b) When h(x) = 0,

15 – 0.06x2 = 0

0.06x2 = 15

x2 = 250

x = 15.81 meter

The width of the tunnel is 2(15.81) = 31.62 meter.

3. Width = 2(2)
= 4 meter
Depth = 1 meter

37

4. y= 1 x2 – x + 150
(a) 400 coordinates of
the vertex = – b
The 2a

=– (–1)

21 1 2
400
= 200

The length of the minimum point between each pole is 200 metres.

(b) f(200) = 1 (200)2 – 200 + 150
400

= 50

The height of the road above the water level is 50 metres.

Intensive Practice 2.3 (Page 63)

1. (a) f(x) = kx2 – 4x + k – 3

A quadratic function that has one intercept means that the function has an equal real root.

For two equal real roots,

b2 – 4ac = 0

(–4)2 – 4(k)(k – 3) = 0

16 – 4k2 + 12k = 0

k2 – 3k – 4 = 0

(k + 1)(k – 4) = 0

k = –1 or k = 4

(b) f(x) = 3x2 – 4x – 2(2k + 4)

A quadratic that intersects the x-axis at two different points means that the function has

two different real roots.

For two different and real roots,
b2 – 4ac . 0

(–4)2 – 4(3)(–4k – 8) . 0
16 + 48k + 96 . 0
48k . –112
7
k . – 3

2. f(x) = mx2 + 7x + 3
b2 – 4ac , 0
72 – 4(m)(3) , 0
49 – 12m , 0
12m . 49
m . 4.083



The smallest value of m is 5.

3. (a) f(x) = x2 + 6x + n 38
= x2 + 6x + 32 – 32 + n
= (x + 3)2 – 9 + n

(b) –9 + n = –5
n = 4

(c) f(x) = x2 + 6x + 4 y

b2 – 4ac = 62 – 4(1)(4)

= 36 – 16
= 20 (. 0)
The curve will intersect the x-axis at two different points. (x) = (x + 3)2 − 5 − 3 4 x

The minimum point is (–3, –5) and the axis of symmetry is x = –3. 0
f(0) = 02 + 6(0) + 4 −5

= 4

The y-intercept is 4.

4. rx + 4 = x2 – 4x + 5
x2 + (– 4 – r)x + 1 = 0
b2 – 4ac , 0
(– 4 – r)2 – 4(1)(1) , 0
16 + 8r + r2 – 4 , 0
r2 + 8r + 12 , 0
(r + 6)(r + 2) , 0
– 6 , r , –2

5. (a) The width of the graph decreases. The axis of symmetry and its minimum value does
not change.

(b) The graph with the same shape moves horizontally 3 units to the right.
(c) The graph with the same shape moves vertically 3 units upwards. Its minimum value

becomes 5 and the axis of symmetry does not change, which is x = 1.

6. (a) h(t) = 2(t – 3)2 h(t)
= 2t2 – 12t + 18 18 t = 3
= 2(t2 – 6t + 9)

b2 – 4ac = (–12)2 – 4(2)(18) t
= 144 – 144
= 0
The curve touches the t-axis at one point.

h(t) = 0 03

2(t2 – 6t + 9) = 0

2(t – 3)2 = 0

t=3

The curve touches the t-axis at t = 3.

h(0) = 2(0 – 3)2
= 18
The h-intercept of the curve is 18.

(b) r(t) = 2h(t)
= 2[2(t – 3)2]
= 4t2 – 24t + 36
= 4(t2 – 6t + 9)

39

b2 – 4ac = (–24)2 – 4(4)(36) r(t)

= 576 – 576 36
t=3
= 0

The curve touches the t-axis at one point.

r(t) = 0

4(t2 – 6t + 9) = 0

4(t – 3)2 = 0 03 t

t=3

The curve touches the t-axis at t = 3.

r(0) = 4(0 – 3)2

= 36

The r-intercept of the curve is 36.

(c) The graph of function h(t) with the value of a = 2 is wider than the graph of r(t) with
the value of a = 4. Thus, the bird that is represented by function r(t) moves at the
highest position, which is 36 m from the water level as compared to the bird that is
represented by the function h(t) with 18 m.

7. f(x) = 3 – 4k – (k + 3)x – x2
= –x2 – (k + 3)x + 3 – 4k

b2 – 4ac , 0
(–k – 3)2 – 4(–1)(3 – 4k) , 0
k2 + 6k + 9 + 12 – 16k , 0
k2 – 10k + 21 , 0
(k – 7)(k – 3) , 0

3 , k , 7



Thus, p = 3 and q = 7.

8. (a) The x coordinate of the vertex = – b
2a

4 = – b

21 1 2
8
b = –1
1
(b) y = 8 x2 – x +c

b2 – 4ac , 0
41 1 2c
(–1)2 – 8 , 0

1 c . 1
2
c . 2

(c) f(x) = 1 x2 – x + c
8
On the vertex (4, 2)
1
2 = 8 (4)2 – 4 + c

2 = –2 + c

c = 4

40

9. (a) The t coordinate of the vertex = – b
2a
32
=– 2(– 4)

=4

Therefore, the fireworks will explode at the time of 4 seconds.

(b) When t = 4,
  h(4) = –4(4)2 + 32(4)

= –64 + 128

= 64

Therefore, the fireworks will explode at a height of 64 m.

10. y = –(x – a)(x – b) (ii) b
(a) (i) a
a+ b
(iii) – ab (iv) 2 the
(b)
a+b is the x coordinate for the minimum point of graph and – ab is the
2
y-intercept for that graph.

11. f(x) = x2 – 4nx + 5n2 + 1 – 4n
1 – 4n 22 1 2 22
= x2 – 4nx + 2 – + 5n2 + 1

= (x – 2n)2 – 4n2 + 5n2 + 1

= (x – 2n)2 + n2 + 1

Given the minimum value for f(x) is m2 + 2n.

Therefore, n2 + 1 = m2 + 2n

m2 = n2 – 2n + 1

= (n – 1)2
\  m = n – 1 (as shown)

Reinforcement Practice   (Page 66)

1. 3x(x – 4) = (2 – x)(x + 5)
3x2 – 12x = 2x + 10 – x2 – 5x

4x2 – 9x – 10 = 0

x = –(–9) ± (–9)2 – 4(4)(–10)
2(4)

= 9 ± 241
8

x= 9 + 241 or x= 9 – 241
8 8
= 3.066 = –0.816

2. (a) (x – 4)2 = 3

x2 – 8x + 16 – 3 = 0

x2 – 8x + 13 = 0

41

(b) The sum of roots = – b
a
=8

The product of roots = c
a
= 13

(c) b2 – 4ac = (–8)2 – 4(1)(13)

= 64 – 52

= 12 ( 0)

The equation has two different real roots.

3. (a) x2 + kx = k – 8

x2 + kx – k + 8 = 0

For two equal real roots,

b2 – 4ac = 0

k2 – 4(1)(–k + 8) = 0

k2 + 4k – 32 = 0

(k + 8)(k – 4) = 0

k = –8 or k = 4

(b) For two different real roots,

b2 – 4ac 0

k2 + 4k – 32 0

(k + 8)(k – 4) 0

k –8 or k 4

(c) For real roots,

b2 – 4ac 0

k2 – 4k – 32 0

(k + 8)(k – 4) 0

k –8 or k 4

4. (a) When one root is –2,

3(–2)2 + p(–2) – 8 = 0

4 – 2p = 0

2p = 4

p 1 p=2
3 3
(b) – =

p = –1

5. 3hx2 – 7kx + 3h = 0

For two equal real roots,
b2 – 4ac = 0

(–7k)2 – 4(3h)(3h) = 0

49k2 – 36h2 = 0

h2 = 49
k2 36
h 7
k = 6

Thus, h : k = 7 : 6. 42

3hx2 – 7kx + 3h = 0…

h= 7 k…
6

Substitute ᕢ into ᕡ.

7( ) ( )3kx2– 7kx + 3 7 k =0
6 6
7 7
2 kx2 – 7kx + 2 k = 0

7kx2 – 14kx + 7k = 0…

Divide ᕣ by 7k.
x2 – 2x + 1 = 0

(x – 1)2 = 0

x=1

6. x2 – 7x + 10 0 0x7
(x – 2)(x – 5) 0 25
The range of x is x Ͻ 2 or x Ͼ 5.

x2 – 7x 0
x(x – 7) 0
The range of x is 0 р x р 7.

From the number line, the range of x2 x5 x
x for –10 x2 – 7x 0 is 0 7

0 x 2 or 5 x 7.

7. (a) The roots are 3 and 7

(b) p = –5 and – 1 q = 4
3 q = –12

(c) x = 5

(d) 3 x 7

8. (a) f(x) = x2 + bx + c
At (2, 0), 0 = 22 + b(2) + c
2b + c = – 4…

At (6, 0), 0 = 62 + b(6) + c
6b + c = –36…

– : 4b = –32
b = –8

Substitute b = –8 into ᕡ.
2(–8) + c = –4
c = 12

43

(b) f(x) = x2 – 8x + 12
When x = 4,
f(4) = 42 – 8(4) + 12
= – 4
The coordinates of the minimum point is (4, – 4).

(c) When f(x) is negative, the range of x is below the x-axis. Thus, 2 < x < 6.

(d) When the graph is reflected on the x-axis, the maximum value is 4.

9. Let the velocity of the boat be v.

Since the to-and-fro travel time is 6 hours,

24 + 24 = 6
v–3 v+3
24(v + 3) + 24(v – 3)
(v – 3)(v + 3) =6

48v = 6
v2 – 9
48v = 6(v2 − 9)

6v2 − 48v − 54 = 0

v2 − 8v − 9 = 0

(v + 1)(v − 9) = 0

v = −1 or v = 9

Thus, the velocity of the boat is 9 km/h.

10. Let the width be w. Thus,
w2 + (w + 6.8)2 = 1002

w2 + w2 + 13.6w + 46.24 = 10 000
2w2 + 13.6w – 9953.76 = 0

w = –13.6 ± 13.62 – 4(2)(–9953.76)
2(2)

= –13.6 ± 79815.04
4

w= –13.6 + 79815.04 or x= –13.6 – 79815.04
4 4
= 67.229 = –74.029

Thus, the width is 67.229 units.

11. (a) y= 1 x2 – 24x + 700
5
Let the floor be the x-axis and the wall of the house be the y-axis.

On the x-axis, y = 0

1 x2 – 24x + 700 = 0
5
x2 – 120x + 3 500 = 0

(x – 50)(x – 70) = 0

x = 50 or x = 70

The width of the opening of the drain = 70 – 50

= 20 units

44

(b) The x coordinate of the minimum point = 50 + 20
2
= 60
1
When x = 60, y = 2 (60)2 – 24(60) + 700

= –20

The minimum depth of the drain is 20 units.

12. (a) y = a(x – 3)2 + 2.5

At point (0, 2)

  2 = a(0 – 3)2 + 2.5
1
9a = – 2

 a = – 1
18
Thus, y = – 118(x – 3)2 + 2.5

(b) On the x-axis, y = 0

0 = – 1 (x – 3)2 + 2.5
18
1
18 (x – 3)2 = 2.5

(x – 3)2 = 45

x – 3 = ± 45

x = 3 + 45 or x = 3 – 45

= 9.708 = –3.708

The maximum length of a horizontal throw by Krishnan is 9.708 m.

45

CHAPTER 3 SYSTEM OF EQUATIONS

Inquiry 1 (Page 70)

4. Yes, the planes are intersect one another.
5. There are 3 shown axes, which are the x-axis, y-axis and z-axis. The linear equation in

three variables form a plane on each of those axes.

Inquiry 2 (Page 71)
4. There are 2 shown axes, which are the x-axis and y-axis. Each linear equation in two

variables form a straight line on each of those axes.

Self Practice 3.1 (Page 72)
1. Let x represents pants, y represents shirts and z represents shoes.

Thus, the linear equation in three variables is 3x + 2y + z = 750.
2. (a) 2m + 6n – 12p = 4

5m – n + p = 0
18m + 5p = 0
Yes, because all the three equations have three variables, m, n and p with the
power of the variables is 1. The equation has 0 for the value of n.
(b) 12e – f 2 – 6eg = 0
8e – 2f – 9g = –6
e – 17f = –6
No, because there are equations that has variables with power of 2.
(c) 7a – 6b – c = 0
10a + b + 4c = 3
61a + b – 2c = 0
Yes, because all the three equations have three variables, a, b and c with the power of
the variables is 1.
Inquiry 3 (Page 72)
3. Yes, there is an intersection point between three planes. The intersection point is
(1, −2, 3).

4. The intersection point (1, −2, 3) is the solution for those three linear equations.

Inquiry 4 (Page 72)
3. Those three planes do not only intersect on one point but also intersect on one straight

line.

1

Inquiry 5 (Page 73)
3. Those three planes do not have any intersection point.

Mind Challenge (Page 75)

3x – y – z = –120 …1
y – 2z = 30 …2
x + y + z = 180 …3
1 + 3: 4x = 60

x = 15

3 – 2: x + 3z = 150…4

Substitute x = 15 into 4.
15 + 3z = 150

z = 45

Substitute x = 15 and z = 45 into 3.
15 + y + 45 = 180
y = 120

x = 15, y =120 and z = 45, thus the solution that is obtained is the same as Example 3 that is
solved by the substitution method.

Self Practice 3.2 (Page 76)

1. (a) 7x + 5y − 3z = 16 … 1
3x − 5y + 2z = −8 …2
5x + 3y − 7z = 0 …3

1 + 2: 10x − z = 8 …4

2 × 3: 9x − 15y + 6z = −24 …5
3 × 5: 25x + 15y − 35z = 0 …6
5 + 6: 34x – 29z = −24 …7

4 × 29: 290x − 29z = 232 …8
8 − 7: 256x = 256
x=1

Substitute x = 1 into 4.
10(1) − z = 8
−z = −2
z=2

Substitute x = 1 and z = 2 into 1.
7(1) + 5y − 3(2) = 16

5y = 15

y=3

Thus, x = 1, y = 3 and z = 2 are the solutions for this system of linear equations.

2

(b) 4x − 2y + 3z = 1 …1
x + 3y − 4z = −7 …2
3x + y + 2z = 5 …3

2 × 4: 4x + 12y − 16z = −28 …4
4 − 1: 14y − 19z = −29 …5

2 × 3: 3x + 9y −12z = −21 …6
6 − 3: 8y − 14z = −26 …7

5 × 8: 112y − 152z = −232 …8
7 × 14: 112y − 196z = −364 …9
8 − 9:
44z = 132
z=3

Substitute z = 3 into 5.
14y − 19(3) = −29
14y – 57 = –29
14y = 28
y=2

Substitute y = 2 and z = 3 into 1.

4x − 2(2) + 3(3) = 1
4x + 5 = 1
4x = −4
x = −1

Thus, x = −1, y = 2, and z = 3 are the solutions for this system of linear equations.

2. (a) 2x + y + 3z = −2 …1
x − y − z = −3 …2
3 x − 2y + 3z = −12 …3

From 1, y = –2 – 2x – 3z…4

Substitute 4 into 2.

x – (–2 – 2x – 3z) – z = –3

x + 2 + 2x + 3z – z = –3

3x + 2z = –5 –5 – 3x
2
z= …5

Substitute (44) iknetod23alam (3).

3x – 2(–2 – 2x – 3z) + 3z = –12
1 2 3x + 4 + 4x + 6z + 3z = –12

7x + 9z = –16…6

Substitute 5 into 6. 6

( )7x + 9
–5 – 3x = –16
2

14x – 45 – 27x = –32

13x = –13

x = –1

3

Substitute x = –1 into 5.
–5 – 3(–1)
z= 2

= –1

Substitute x = –1 and z = –1 into 2.
–1 – y – (–1) = –3

y=3
Thus, x = −1, y = 3 and z = −1 are the solutions for this system of linear equations.
(b) 2 x + 3y + 2z = 16 …1

x + 4y − 2z = 12 …2
x + y + 4z = 20 …3

From 3, x = 20 – y – 4z …4
Substitute 4 into 1.

2(20 – y – 4z) + 3y + 2z = 16

40 – 2y – 8z + 3y + 2z = 16
y – 6z = –24
y = 6z – 24 …5

Substitute 4 into 2.
(20 – y – 4z) + 4y – 2z = 12
3y – 6z = –8 …6

Substitute 5 into 6.

3(6z – 24) – 6z = –8

18z – 72 – 6z = –8

12z = 64

z = 64
12

= 16
3

SGubanstitkuatne z = 16 keindtoala5m. 5.
3

( )y= 6 16 – 24
3
=8

SGubanstitkuatne z = 16 and y = 8 into 3.
3
( )x 16
+ 8 + 4 3 = 20

3x + 64 = 36

3x = –28

x = – 2_8_
3

Thus, x = – _2_8 , y = 8 and z = 1_6_ are the solutions for this system of linear equations.
3 3

4

Self Practice 3.3 (Page 77)

1. P + Q + R = 24 500 …1
0.04P + 0.055Q + 0.06R = 1 300 …2
P = 4Q …3

Substitute 3 into 1.

4Q + Q + R = 24 500
5Q + R = 24 500
R = 24 500 – 5Q…4

Substitute 3 into 2. …5

0.04(4Q) + 0.055Q + 0.06R = 1 300
0.215Q + 0.06R = 1 300

Substitute 4 into 5.

0.215Q + 0.06(24 500 – 5Q) = 1 300
0.215Q + 1 470 – 0.3Q = 1 300
−0.085Q = −170
Q = 2 000

Substitute Q = 2 000 into 3.
P = 4(2 000)

= 8 000

Substitute P = 8 000 and Q = 2 000 into 1.
8 000 + 2 000 + R = 24 500

R = 14 500

Thus, the amount of money in unit trust account P is RM8 000, Q is RM2 000 and
R is RM14 500.

2. Let x represents carnation, y represents rose and z represents daisy.
x + y + z = 200 …1

1.50x + 5.75y + 2.60z = 589.50 …2
y = z − 20 …3

Substitute 3 into 1.

x + (z − 20) + z = 200
x + 2z – 20 = 200
x = 220 – 2z…4

Substitute 3 into 2.

1.50 x + 5.75(z − 20) + 2.60z = 589.50
1.50x + 5.75z – 115 + 2.60z = 589.50

1.50x + 8.35z – 115 = 589.50
1.50x + 8.35z = 704.5 …5

Substitute 4 into 5. 5

1.50(220 – 2z) + 8.35z = 704.5
330 – 3z + 8.35z = 704.5
5.35z = 374.5
z = 70

Substitute z = 70 into 3.
y = z – 20
= 70 – 20
= 50

Substitute y = 50 and z = 70 into 1.
x + 50 + 70 = 200
x = 80

Thus, the number of carnations is 80, roses is 50 and daisies is 70.

3. Let x represents pens, y represents pencils and z represents notebooks.
5x + 3y + 9z = 102 …1
5x = 3y …2
x + y = z …3

[ ( )]DaFrripoamda 2, y =_5_x …4
3
5
5x + 5x + 9 x + 3 x = 102

( )10x+ 9 8 x= 102
3 34x = 102

x=3
Substitute x = 3 into 4.

y = _5_ x
3
=5

Substitute x = 3 and y = 5 into 3.

z =3 + 5
=8
Thus, the number of pens is 3, pencils is 5 and notebooks is 8.

Intensive Practice 3.1 (Page 78)

1. (a) Let x represent the largest angle, y represent the second largest angle and z represent the
smallest angle.
x + y + z = 180 …1
x − 20 = y + z …2 x −
10 = 3z 3…
From 3, x = 3z + 10 …4

Substitute 4 into 1. …5

3z + 10 + y + z = 180
y + 4z = 170

6

Substitute 4 into 2.
3z + 10 − 20 = y + z

−y + 2z = 10…6

5 + 6: 6z = 180
z = 30

Substitute z = 30 into 4.
x = 3(30) + 10

= 100.

Substitute z = 30 into 5.
y + 4(30) = 170
y = 50

Thus, the measurement for each angle in the triangle is 100°, 50° and 30°.

(b) Let x represents the first number, y represents the second number and z represents

the third number.
x + y + z = 19 …1
2x + y + z = 22 …2
x + 2y + z = 25 …3
2 − 1:
x=3
2 − 3: x − y = −3 …4

SGuabnsttiiktauntexx==33kkienetdodaalalam.m44..
3 − y = −3
y=6

Substitute x = 3 and y = 6 into 1.
3 + 6 + z = 19
z = 10

Thus, the values for those numbers are 3, 6 and 10.

2. (a) Elimination method:
x + y + z = 3 …1
x + z = 2 …2

2x + y + z = 5 …3

3 − 1: x = 2

Substitute x = 2 into 2.
2+z=2
z=0

Substitute x = 2 and z = 0 into 1.
2 + y + 0 = 3

y=1

Thus, x = 2, y = 1 and z = 0 are the solutions for this system of linear
equations.

7

…1
…2
…3

From 2, x = 2 − z …4

Substitute 4 into 1.

2 − z + y + z = 3
y=1

Substitute 4 into 3. …5

2(2 − z) + y + z = 5
4 − 2z + y + z = 5
−z + y = 1

Substitute y = 1 into 5.
−z + 1 = 1

z=0

Substitute z = 0 into 4.
x=2−0
=2

Thus, x = 2, y = 1 and z = 0 are the solutions for this system of linear equations.
(b) Elimination method:

2x + y − z = 7 …1
x − y + z = 2 …2
x + y − 3z = 2 …3

3 − 2: 2y − 4z = 0 …4
2 × 2: 2x − 2y + 2z = 4 …5
1 − 5: 3y − 3z = 3 …6
4 × 3: 6y − 12z = 0 …7
6 × 2: 6y − 6z = 6 …8

8 − 7: 6z = 6

z=1

Substitute z = 1 into 4.
2y − 4(1) = 0

2y = 4

y=2

Substitute y = 2 and z = 1 into 2.
x−2+1=2
x=3

Thus, x = 3, y = 2 and z = 1 are the solutions for this system of linear equations.

8

Substitution method:
2x + y − z = 7 …1
x − y + z = 2 …2
x + y − 3z = 2 …3

x = y − z + 2 …4

Substitute 4 into 1. …5

2(y − z + 2) +y − z = 7
2y − 2z + 4 +y − z = 7

3y − 3z = 3
y–z=1
y= 1+z

Substitute 4 into 3. …6

y − z + 2 + y − 3z = 2
2y − 4z = 0

Substitute 5 into 6.

2(1 + z) – 4z = 0
2 + 2z – 4z = 0
2z = 2
z=1

Substitute z = 1 into 5.
y=1+1

y=2

Substitute y = 2 and z = 1 into 4.
x=2–1+2
x=3

Thus, x = 3, y = 2 and z = 1 are the solutions for this system of linear equations.
(c) Elimination method:

x + y + z = 3 …1
2x + y − z = 6 …2
x + 2y + 3z = 2 …3

3 − 1: y + 2z = −1 …4

1 × 2: 2x + 2y + 2z = 6 …5
5 − 2: y + 3z = 0 …6

6 − 4: z=1

SGuabnsttiiktauntezz==11kientdoa4lam. 4.
y + 2(1) = −1
y = −3

Substitute y = −3 and z = 1 into 1.
x + (−3) + 1 = 3

x−3+1=3

x=5

Thus, x = 5, y = –3 and z = 1 are the solutions for this system of linear equations.
9

Substitution method:
x + y + z = 3 …1
2x + y − z = 6 …2

x + 2y + 3z = 2 …3

From 1, x = −y − z + 3…4

Substitute 4 into 2. …5

2(−y − z + 3) +y − z = 6
−2y − 2z + 6 + y − z = 6

−y − 3z = 0
y = –3z

Substitute 4 into 3. …6

−y − z + 3 + 2y + 3z = 2
y + 2z = −1

Substitute 5 into 6.

–3z + 2z = –1
z=1

Substitute z = 1 into 5.
y = –3(1)
y = −3

Substitute y = −3 and z = 1 into 1.
x + (−3) + 1 = 3

x−3+1=3

x=5

Thus x = 5, y = −3 and z = 1 are the solutions for this system of linear equations.
(d) Elimination method:

2x – y + z = 6 …1
3x + y – z = 2 …2
x + 2y – 4z = 8 …3

2 + 1: 5x = 8
8
x= 5

1 × 2: 4x – 2y + 2z = 12 …4
3 + 4: 5x – 2z = 20 …5

SGubanstitkuatne x = 8 keindtaolam 5.
5

( )5 8 – 2z = 20
5
8 – 2z = 20

2z = –12

z = –6

10

Substitute x = 8 and z = –6 into 1.
5

( )28 – y + (–6) = 6
5
16
5 –y–6= 6

y = – 454

Thus, x = 8 , y = – 44 and z = –6 are the solutions for this system of linear equations.
5 5

Substitution method: …1
2x – y + z = 6 …2
…3
3x + y – z = 2

x + 2y – 4z = 8

From 1, x = 6 +y – z …4
2

Substitute 4 into 2.

( )36+y – z +y–z=2
2
1 8 + 3y – 3z + 2y – 2z = 4

5y – 5z = –14
5y = –14 + 5z…5

SGubansttiitkuatne 4 ke idnatolam 3.
6+y–z
6 +2y – z
( ) + 2y – 4z = 8
+ 4y – 8z = 16

5y – 9z = 10…6

Substitute 5 into 6.

–14 + 5z – 9z = 10
4z = –24
z = –6

Substitute z = –6 into 5.
5y = –14 + 5(–6)
5y = –44
y = – 454

Substitute y = – 44 and z = –6 into 1.
5
( )2x – – 454 + (–6) = 6
16
2x = 5

x= 8
5

ThereTfhoures, x = 8 , y = – 44 and z = –6 are the solution for this system of linear equations.
5 5

11

(e) Elimination method:
x + y + 2z = 4 …1
x + y + 3z = 5 …2
2x + y + z = 2 …3

2 – 1: z=1

3 – 2: x – 2z = –3 …4

Substitute z = 1 into 4.
x – 2(1) = –3

x = –1

Substitute x = –1 and z = 1 into 1.
(–1) + y + 2(1) = 4
y=3

Thus, x = –1, y = 3 and z = 1 are the solutions for this system of linear equations.

Substitution method:
x + y + 2z = 4 …1
x + y + 3z = 5 …2
2x + y + z = 2 …3

From 1, x = 4 – y – 2z …4

Substitute 4 into 2.

4 – y – 2z + y + 3z = 5
z=1

Substitute 4 into 3.

2(4 – y – 2z) + y + z = 2
8 – 2y – 4z + y + z = 2
y + 3z = 6 …5

Substitute z = 1 into 5.
y + 3(1) = 6

y=3

Substitute y = 3 and z = 1 into 4.
x = 4 – 3 – 2(1)

= –1

Thus, x = –1, y = 3 are the solutions for this system of linear equations.
(f) Eliminition method:

x + 2y + z = 4 …1
x – y + z = 1 …2

2x + y + 2z = 2 …3

1 – 2: 3y = 3
y=1

12

Substitute y = 1 into 3. …4
2x + 1 + 2z = 2

2x + 2z = 1

3 + 2: 3x + 3z = 3
x+z=1
x = 1 – z …5

Substitute 5 into 4.

2(1 – z) + 2z = 1
2 – 2z + 2z = 1
0 ≠ –1

Thus, this equations has no solution.

Substitution method:
x + 2y + z = 4 …1
x – y + z = 1 …2
2x + y + 2z = 2 …3

x = 4 – 2y – z… 4

Substitute 4 into 2.

4 – 2y – z – y + z = 1
3y = 3
y=1

Substitute 4 and y = 1 into 3.
2(4 – 2(1) – z) + 1 + 2z = 2

2(2 – z) + 1 + 2z = 2
4 – 2z + 1 + 2z = 2
0 ≠ –3

Thus, this equations has no solution.

3. Let x represents butterscotch bread, y represents chocolate bread and z represents
coconut bread.
x + y + z = 2 150 …1
2x + 3y + 4z = 6 850 …2
x + 1.50y + 1.50z = 2 975 …3

3 − 1: 0.5y + 0.5z = 825 …4
1 × 2: 2x + 2y + 2z = 4 300 …5
2 − 5: y + 2z = 2 550 …6

4 × 2: y + z = 1 650 …7

6 − 7: z = 900

Substitute z = 900 into 7.
y + 900 = 1 650

y = 750

13

Substitute y = 750 and z = 900 into 1.
x + 750 + 900 = 2 150

x = 500

Thus, the number of butterscotch bread is 500 loaves, the amount of chocolate bread is
750 loaves and the amount of coconut bread is 900 loaves.

4. Let x represents small flower pots, y represents medium flower pots and z represents large
flower pots.
x = y + z …1
y = 2z …2
10x + 15y + 40z = 300 …3

Substitute 2 into 1.
x = 2z + z
= 3z …4

Substitute 2 and 4 into 3.

10(3z) + 15(2z) + 40z = 300
30z + 30z + 40z = 300
100z = 300
z=3

Substitute z = 3 into 2.
y = 2(3)

=6

Substitute z = 3 into 4.
x = 3(3)

=9

Thus, the minimum amount for small, medium and large flower pots is 9, 6 and 3
respectively.

5. Let x represents chickens, y represents rabbits and z represents ducks. The system
of equation formed is
20x + 50y + 30z = 1 500 …1

x + y + z = 50 …2
x = z …3

Substitute 3 into 1
20x + 50y + 30x = 1 500

50x + 50y = 1 500
x + y = 30 …4

Substitute 3 into 2

x + y + x = 50 …5
…6
5 – 4: 2x + y = 50

x = 20

Substitute x = 20 into 4
20 + y = 30
y = 10

Thus, the number of chickens is 20, rabbits is 10 and ducks is 20.

14

Inquiry 6 (Page 79)

3. Statement 1: Let x represents width and y represents length.
2x + 2y = 200
xy = 2 400

The number of equations is two with two variables.

Statement 2: Let x represents width and y represents length.
2x + 2y = 800

xy = 30 000
The number of equations is two with two variables.

Statement 3: Let x represents the first number and y represents the second number.
x–y=9

xy = 96
The number of equations is two with two variables.

Inquiry 7 (Page 80)

3. The intersection point for the liner equation x + 2y = 10 and non-linear equation y2 + 4x = 50
is the solution for both equations. The solution for both of these equations is known as
solution of simultaneous equations.

Mind Challenge (Page 80)
2x + y = 4 …1
y2 + 5 = 4x …2

From 1, y = 4 – 2x …3

Substitute 3 into 2.

(4 – 2x)2 + 5 = 4x

16 – 16x + 4x2 + 5 = 4x

4x2 – 20x + 21 = 0

(2x – 3)(2x – 7) = 0

2x – 3 = 0   , 2x – 7 = 0

x= 3 x= 7
2 2

Substitute x = 3 and x = 7 into 3.
2 2

y = 4 – 2( 3 ) , y = 4 – 2( 7 )
2 2
= 1 = –3

Thus, x = 3 , y = 1 and x = 7 , y = –3 are the solutions to these simultaneous equations.
2 2
The same solution is obtained in Example 7.

Self Practice 3.4 (Page 82)

1. (a) Substitution method:
2x − y = 7 …1

y2 − x(x + y) = 11 …2

y = 2x − 7 …3

15

Substitution 3 into 2. x – 2 = 0
x=2
(2x − 7)2 − x(x + 2x − 7) = 11
4x2 – 28x + 49 – x2 – 2x2 + 7x = 11

x2 − 21x + 38 = 0
(x − 19)(x − 2) = 0

x – 19 = 0   ,
x = 19

Substitution x = 19 into 3.
y = 2(19) − 7
= 31

Substitution x = 2 into 3.
y = 2(2) − 7
= −3

Thus, x = 19, y = 31 and x = 2, y = −3 are the solutions to these simultaneous equations.

Graph representation:

Thus, x = 19, y = 31 and x = 2, y = −3 are the solutions to these simultaneous equations.

(b) Eliminition method:
5y + x = 1 …1
x + 3y2 = −1 …2

2 – 1: 3y2 – 5y + 2 = 0

(3y – 2)(y – 1) = 0

3y – 2 = 0 y–1=0
2 y=1
y= 3

16