Form 4 Add Maths Chapter 2 Answers

CHAPTER 2 QUADRATIC FUNCTIONS

Inquiry 1   (Page 36)

4. The x-coordinate for point A and point B is the solution for the equation y = 3x2 + 11x − 4
when y = 0. This x-coordinate is called the roots for the quadratic equation 3x2 + 11x − 4 = 0.

Mind Challenge   (Page 38)

ax2 + bx + c = 0 Divide both sides of the
equation by a
x2 + b x = – c
a a
( ) ( )x2 +bb c b2
a x + 2a 2= – a + 2a Add both sides of the equation

coefficient of x 2
2
( )x+ b 2 = – 4ac + b2 ( )with
2a 4a2
± b2 – 4ac
x + b = 2a
2a

x = –b + b2 – 4ac or x = –b – b2 – 4ac
2a 2a

In general, the formula for solving a quadratic equation is

x = –b ± b2 – 4ac
2a

Self Practice 2.1 (Page 38)

1. (a) x2 + 4x – 9 = 0

x2 + 4x = 9

( ) ( )x2 + 4x +4 42
2 2=9+ 2

x2 + 4x + 22 = 9 + 22

(x + 2)2 = 13

x + 2 = ± 13

x = – 13 – 2 or x = 13 – 2
= 1.606
= −5.606
1
(b) x2 − 3x – 5 = 0

x2 − 3x = 5

( ) ( )x2 − 3x +–3 2 = 5 + –3 2
2 2
( )x − 3 29
2 2= 4

x− 3 =± 29
2 4

x=– 29 + 3 or x= 29 + 3
4 2 4 2

= −1.193 = 4.193

(c) –x2 – 6x + 9 = 0

x2 + 6x = 9

( ) ( )x2 + 6x +6 62
2 2=9+ 2

x2 + 6x + 32 = 9 + 32

(x + 3)2 = 18

x + 3 = ± 18

x = – 18 – 3 or x = 18 – 3

= −7.243 = 1.243

(d) 2x2 − 6x + 3 = 0

2x2 − 6x = −3

2(x2 – 3x) = −3
3
x2 – 3x = – 2

( ) ( )x2 − 3x +–3 2=– 3 + –3 2
2 2 2
( )x − 3 3
2 2= 4

x− 3 =± 3
2 4

x=– 3 + 3 or x= 3 + 3
4 2 4 2
= 0.634 = 2.366

(e) 4x2 − 8x + 1 = 0

2x2 − 8x = −1

4(x2 – 2x) = −1
1
x2 – 2x = – 4

( ) ( )x2 − 2x +–2 2 =– 1 + –2 2
2 4 2
3
(x − 1)2 = 4

x−1=± 3
4

x=– 3 +1 or x= 3 +1
4 4
= 0.134 = 1.866

(f) −2x2 + 7x + 6 = 0

2x2 − 7x = 6

( )2 x2 − 7 x =6
2
7
x2 − 2 x = 3

−( ) ( )x27x+ –7 2 = 3+ –7 2
2 4 4
2

( )x − 7 2= 97
4 16

x− 7 =± 97
4 16

x=– 97 + 7 or x= 97 + 7
16 4 16 4

= −0.712 = 4.212

2. (a) x2 – 4x – 7 = 0 with a = 1, b = −4 and c = −7.

x = –b ± b2 – 4ac
2a

x = –(–4) ± (– 4)2 – 4(1)(–7)
2(1)

= 4 ± 44
2

x= 4 – 44 or x= 4 + 44
2 2
= −1.317 = 5.317

(b) 2x2 + 2x – 1 = 0 with a = 2, b = 2 and c = −1.

x = –b ± b2 – 4ac
2a

x = –(2) ± (2)2 – 4(2)(–1)
2(2)

= –2 ± 12
4

x= –2 – 12 or x= –2 + 12
4 4
= −1.366 = 0.366

(c) 3x2 – 8x + 1 = 0 with a = 3, b = −8 and c = 1.

x = –b ± b2 – 4ac
2a

x = –(–8) ± (–8)2 – 4(3)(1)
2(3)

= 8 ± 52
6

x= 8 – 52 or x= 8 + 52
6 6
= 0.131 = 2.535

(d) 4x2 – 3x – 2 = 0 with a = 4, b = –3 and c = –2.

x = –b ± b2 – 4ac
2a

x = –(–3) ± (–3)2 – 4(4)(–2)
2(4)

= 3 ± 41 3
8

x= 3 – 41 or x= 3 + 41
8 8
= –0.425 = 1.175

(e) (x – 1)(x – 3) = 5

x2 – 4x + 3 = 5

x2 – 4x – 2 = 0 with a = 1, b = –4 and c = –2.

x = –b ± b2 – 4ac
2a

x = –(–4) ± (– 4)2 – 4(1)(–2)
2(1)

= 4 ± 24
2

x= 4 – 24 or x= 4 + 24
2 2
= –0.449 = 4.449

(f) (2x – 3)2 = 6

4x2 – 12x + 9 = 6

4x2 – 12x + 3 = 0 with a = 4, b = –12 and c = 3.

x = –b ± b2 – 4ac
2a

x = –(–12) ± (–12)2 – 4(4)(3)
2(4)

= 12 ± 96
8

x= 12 – 96 or x= 12 + 96
8 8
= 0.275 = 2.725

3. (a) Let x be the length and (x – 2) be the width. x cm
x2 + (x – 2)2 = 102
10 cm (x – 2) cm
x2 + x2 – 4x + 4 = 100
2x2 – 4x – 96 = 0
x2 – 2x – 48 = 0

(x – 8)(x + 6) = 0

x = 8 or x = –6 (Ignore)

Thus, the rectangle has a length of 8 cm and width of 6 cm.

(b) 2x + 2y = 26…
xy = 40…

From , y = 40 …
x

Substitute ᕣ into ᕡ.

( )2x + 240 = 26
x
2x2 + 80 = 26x

2x2 – 26x + 80 = 0

x2 – 13x + 40 = 0 4

(x – 5)(x – 8) = 0
x = 5 or x = 8

Substitute x = 5 into ᕣ.

y = 40
5
=8

Substitute x = 8 into ᕣ.

y = 40
8
=5

Thus, the measurement of the rectangle is 5 cm × 8 cm.

4. 1 × [(x + 3) + (3x + 2)] × (x – 1) = 17
2 (4x + 5) × (x – 1) = 34

4x2 + x – 5 = 34

4x2 + x – 39 = 0

(4x + 13)(x – 3) = 0 13
4
x = – (Ignore) or x=3

Thus, the value of x is 3.

Self Practice 2.2 (Page 41) (b) x2 – (α + β )x + αβ = 0
x2 – (–1 + 4)x + (–1)(4) = 0
1. (a) x2 – (α + β )x + αβ = 0 x2 – 3x – 4 = 0
x2 – (2 + 6)x + (2)(6) = 0
x2 – 8x + 12 = 0 (d) x2 – (α + β )x + αβ = 0
( ) ( )x2 –1 1
(c) x2 – (α + β )x + αβ = 0 5 –5 x+ 5 (–5) = 0
x2 – (–4 – 7)x + (–4)(–7) = 0
x2 + 11x + 28 = 0 x2 + 24 x – 1 = 0
5
2. α + β = –(p – 5), αβ = 2q 5x2 + 24x – 5 = 0
–(p – 5) = –3 + 6
–p + 5 = 3
p=2
2q = (–3)(6)
2q = –18
q = –9

3. 5x2 – 10x – 9 = 0
9
x2 – 2x – 5 =0

Thus, α + β = 2, αβ = – 9
5

5

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

=6 = – 9 + 2(2) + 4
5
31
= 5

Quadratic equation: x2 – 6x + 31 = 0
5
5x2 – 30x + 31 = 0

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

= 5(2) = 25(– 9 )
5
= 10 = –45

Quadratic equation: x2 – 10x – 45 = 0

(c) (α – 1) + (β – 1) = α + β – 2 , (α – 1)(β – 1) = αβ – α – β + 1
= αβ – (α + β) + 1
= 2–2

=0 = – 9 –2+1
5
14
= – 5

Quadratic equation: x2 – 0x – 14 =0
5
5x2 – 14 = 0

( )( )(d)α+ β = α +β , α β = αβ
3 3 3 3 3 9
9
= 2 = – 5
3
9
1
= – 5

Quadratic equation: x2 – 2 x – 1 = 0
3 5
15x2 – 10x – 3 = 0

4. 2x2 + 5x = 1

2x2 + 5x – 1 = 0

x2 + 5 x – 1 =0
2 2
5 1
Thus, α + β = – 2 , αβ = – 2

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

= (– 5 ) = 1
(– 2
1 ) (– 1 )
=5 2 = –2 2

Quadratic equation: x2 – 5x – 2 = 0

6

( ) ( )(b)α+1 + β + 1 = α + β + α+β , ( )( )α+ 1 β + 1 = αβ +1+1+ 1
β α αβ β α αβ

=– 5 +5 = – 1 +1+1–2
2 2
5 1
= 2 = – 2

Quadratic equation: x2 – 5 x – 1 =0
2 2
2x2 – 5x – 1 = 0

(c) α2 + β2 = (α + β)2 – 2αβ , (α2)(β2) = (αβ )2

= (– 5 )2 – 2(– 1 ) ( )= – 1 2
2 2 2
25 1
= 4 + 1 = 4

= 29
4
29 1
Quadratic equation: x2 – 4 x + 4 =0

4x2 – 29x + 1 = 0

α β α2 + β 2 α β
β α αβ β α
( ) ( ) ( )( )(d)+ = , =1

= (α + β)2 – 2αβ
αβ

= ( 29 )
4
(– 1 )
2

= – 29
2
29
Quadratic equation: x2 + 2 x + 1 = 0

2x2 + 29x + 2 = 0

5. 2x2 = 6x + 3

2x2 – 6x – 3 = 0

x2 – 3x – 3 =0
2
3
Thus, p + q = 3 , pq = – 2

p2q + pq2 = pq(p + q) , (p2q)(pq2) = p3q3

= – 3 (3) = (pq)3
2
9 ( )= 3
= – 2 – 2 3

= – 27
8
9 27
Quadratic equation x2 + 2 x – 8 = 0

8x2 + 36x – 27 = 0

7

Inquiry 2   (Page 41)

2. Graph sketching method:
•  The roots are x = –1 and x = 3
y

y>0 y>0 x
x>3
x < –1
–1 O 3
y<0

–1 < x < 3

Number line method:

•  The roots are x = –1 and x = 3

Test point –2: Test point 0: Test point 4:
(4 + 1)(4 – 3) > 0
(–2 + 1)(–2 – 3) > 0 (0 + 1)(0 – 3) < 0

+– +

x < –1 –1 –1 < x < 3 3 x > 3

Tabular method: x –1 Range of values of x –1 x 3
– x3 +
(x + 1) – + –
(x – 3) + + –
(x + 1) (x – 3) +

3. When (x + 1)(x – 3) . 0, the range of values of x is x , –1 or x . 3 and when
(x + 1)(x – 3) , 0, the range of x is 1 , x , 3.

Self Practice 2.3   (Page 44)

1. (a) x2 , 4
x2 – 4 , 0
(x – 2)(x + 2) , 0
y
x = –2 and x = 2 –2 0 2

Graph sketching method: –4
Since x2 – 4 , 0, the range of x is determined on the
x

graph below the x-axis.

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

8

Number line method:

Test point –3: Test point 1: Test point 3:

(–3)2 – 4 0 (1)2 – 4 0 (3)2 – 4 0

+–+
x

–2 2
Since x2 – 4 , 0, the range of x is determined on the negative part of the number line.

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

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

Since x2 – 4 , 0, the range of x is determined on the negative part of the table.
Thus, the solution for this quadratic inequality is –2 , x , 2.

(b) (2 – x)(8 – x) , 0 y
x – 2   and  x = 8 02 8x

Graph sketching method:
Since (2 – x)(8 – x) , 0, the range of x is determined

on the curve of the graph below the x-axis.
Thus, the solution for this quadratic inequality is

2 , x , 8.

Number line method:

Test point 1: Test point 3: Test point 9:

(2 – 1)(8 – 1) 0 (2 – 3)(8 – 3) 0 (2 – 9)(8 – 9) 0

+–+
x

28
Since (2 – x)(8 – x) , 0, the range of x is determined on the negative part of the

number line.
Thus, the solution for this quadratic inequality is 2 , x , 8.

9

Tabular method: x,2 2,x,2 x.8
+ – –
(2 – x) + + –
(8 – x) + – +
(2 – x)(8 – x)

Since (2 – x)(8 – x) , 0, the range of x is determined on the negative part of the table.
Thus, the solution for this quadratic inequality is 2 , x , 8.

(c) x2 < 4x + 12
x2 – 4x – 12 < 0
(x + 2)(x – 6) < 0
y

x = –2   and  x = 6

Graph sketching method: –2 0 6 x
Since x2 – 4x – 12 < 0, the range of x is determined
–12
on the curve of the graph below the x-axis.
Thus, the solution for this quadratic inequality is

–2 < x < 6.

Number line method:

Test point –3: Test point 1: Test point 7:

(–3)2 – 4(–3) – 12 0 (1)2 – 4(1) – 12 0 (7)2 – 4(7) – 12 0

+–+
x

–2 6
Since x2 – 4x – 12 < 0, the range of x is determined on the negative part of the number

line.
Thus, the solution for this quadratic inequality is –2 < x < 6.

Tabular method: x < –2 –2 < x < 6 x>6
– + +
(x + 2) + – +
(x – 6) + – +
(x + 2)(x – 6)

Since (x + 2)(x – 6) < 0, the range of x is determined on the negative part of the table.
Thus, the solution for this quadratic inequality is –2 < x < 6.

10

(d) x(x – 2) > 3 y
  x2 – 2x – 3 > 0
(x + 1)(x – 3) > 0
x = –1   and  x = 3

Graph sketching method: –1 0 3x
Since x2 – 2x – 3 > 0, the range of x is determined –3

on the curve of the graph above the x-axis.
Thus, the solution for this quadratic inequality is x < –1

and x > 3.

Number line method: Test point 2: Test point 4:
Test point –2:

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

+–+

–1 3
Since x2 – 2x – 3 > 0, the range of x is determined on the positive part of the number line.
Therefore, the solution for this quadratic inequality is x < –1 and x > 3.

Tabular method: x < –1 –1 < x < 3 x>3
– + +
(x + 1) – – +
(x – 3) + – +
(x + 1)(x – 3)

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

(e) (x + 2)2 , 2x + 7 y
x2 + 4x + 4 , 2x + 7
x2 + 2x – 3 , 0 –3 0 1 x
(x + 3)(x – 1) , 0 –3
x = –3   and  x = 1

Graph sketching method:
Since x2 + 2x – 3 , 0, the range of x is determined

on the curve of the graph below the x-axis.
Thus, the solution for this quadratic inequality is

–3 , x , 1.

11

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