chapter 2 - quadratic expessions and equations

10

CCTS Thinking systematically

CCTS Identifying

Example

Example

2.1 Quadratic Expressions

State whether the following expressions are quadratic expressions.

(a) 1

x2 – 3x + 5 (b) 6 – 5x (c) 3p2 – 7 ––

2

(d) –7q2 + 5q (e)

1 + 3x – 5

–– x

(a) Quadratic expression (b) Not a quadratic expression (c) Quadratic expression

(as a = 0)

(d) Quadratic expression (e) Not a quadratic expression (as a = –1)

1. Identify the following expressions as quadratic or linear.

Expand

(a) x(5x + 2) (b) –3x(4 – 2x)

(a) (b)

= 5x × x + 2 × x = 5x2 + 2x

= –3x × 4 – 3x(–2x)

= –12x + 6x2

Quadratic expression(a) 5x2 – 8x + 2

(e) 6x – x2 (f) 40q2 + 3q – 1

(c) 1

x(6x – 9) + 6 –– 3

(c) 1

x × 6x – 1 x × 9 + 6 –– ––

3 3

= 2x2 – 3x + 6

2. Expand the following.

(a) x (24 – 8x) –– 4

= 6x – 2x2

(c) –3x(6 – x)

= –18x + 3x2

(d) –px(x – p)

= –px2 + p2x

(e) (3x – 4)(–3x)

= –9x2 + 12x(f) 13x(11 – 13x)

= 143x – 169x2

(g) –5x(–16 – 3x)

= 80x + 15x2

(h) hx(kx – 1)

= hkx2 – hx

(b) 3x(

x –

) + 2

=

1 x2 – x + 2 –– 2

(b) 1

x + 7 4

(d) n2 + n – 70

x(5x + 2)

Solution:

Solution:

(c) 7 – 2x + x2

Linear expression

Quadratic expression Quadratic expression

Quadratic expression Quadratic expression

1— 2

1— 5

2— 7

1— 6

1— 3

CHAPTER 2 Quadratic Expessions and Equations

11

CCTS Interpreting

Example

Example

3. Expand the following to form quadratic expressions.

(a) (x – 2)(x – 3)

= x2 – 3x – 2x + 6

= x2 – 5x + 6

(b) (2x – 1)(3 + x)

= 6x + 2x2 – 3 – x = 2x2 + 5x – 3

(c) (2 – 5x)(2x – 4)

= 4x – 10x2 – 8 + 20x = –10x2 + 24x – 8

(d) (5x – 1)(2x + 4)

= 10x2 + 20x – 2x – 4

= 10x2 + 18x – 4

(e) (5x + 2)2

= (5x + 2)(5x + 2)

= 25x2 + 10x + 10x + 4

= 25x2 + 20x + 4

(f) (–2x + 1)(3x + 4)

= –6x2 + 3x – 8x + 4

= –6x2 – 5x + 4

(g) (–2x – 3)(x – 4)

= –2x2 + 8x – 3x + 12

= –2x2 + 5x + 12

(h) (x – 9)(2 – 5x)

= 2x – 5x2 – 18 + 45x = –5x2 + 47x – 18

(i) (–2x + 3)2

= (–2x + 3)(–2x + 3)

= 4x2 – 6x – 6x + 9

= 4x2 – 12x + 9

(j) (1 x – 3)2 ––

3

= ( 1

x – 3)( 1 x – 3) –– ––

3 3

=

1 x2 – x – x + 9 ––

9

=

1 x2 – 2x + 9 ––

9

(k) (5x – 1)(x + 2)

= 5x2 + 10x – x – 2

= 5x2 + 9x – 2

(l) (2x + 1)(– x – 5)

= –2x2 – 10x – x – 5

= –2x2 – 11x – 5

Expand the following.

(a) (x – 3)(2x + 5)

(a) (x – 3)(2x + 5)

The diagram above shows a fi eld and its path in the

shape of a rectangle. Find the total area in term s of x.

6 m

path

x m

3 m

x m

field

= 2x2 + 5x – 6x – 15

= 2x2 – x – 15

(b) (–2 – 3x)2

(b) (–2 –3x)2

= (–2 – 3x)(–2 – 3x)

= 4 + 6x + 6x + 9x2

= 9x2 + 12x + 4 +5x – 6x = –11x

Common Error

Solution:

Area = length × width

= (6 + x) × (3 + x)

= 18 + 6x + 3x + x2

= x2 + 9x + 18

Solution:

12

CCTS Recognising

Example

2.2 Factorisation of Quadratic Expressions

1. Factorise the following quadratic expressions.

Factorise the following expressions completely.

(a) 2x2 – 4 (b) 84x – 12x2 (c) 10p2 – p

Solution:

(a) 2x2 – 4 (b) 84x – 12x2 (c) 10p2 – p

= 2x2 – 2 × 2 = 12x × 7 – 12x × x = 10 × p × p – p × 1

= 2(x2 – 2) = 12x(7 – x) = p(10p – 1)

12 cm

2x cm

x cm

8 cm

.......

...........

...

...

..........

3x + 2

2x + 3

(a) 5p2 – 4p = p(5p – 4)

(b) 10 – 2x2

= 2(5 – x2)

(c) 28x + x2

= x(28 + x)

(d) –12 – 4n2

= – 4(3 + n2)

(e) 144x2 – 24x = 24x(6x – 1)

(f) 25x2 + 15

= 5(5x2 + 3)

(g) 5x2 – x = x(5x – 1)

(h) 6x + 108x2

= 6x(1 + 18x)

(i) 121x2 + 33

= 11(11x2 + 3)

(j) 12x2 – 4x = 4x(3x – 1)

(k) 81x2 + 18x = 9x(9x + 2)

(l) 4x – 8x2

= 4x(1 – 2x)

The diagram above shows a rubber plantation.

There are (3x + 2) columns of rubber trees and

(2x + 3) rows of rubber trees. Find the total number

of rubber trees in the plantation.

Total number of rubber trees

= (3x + 2) × (2x + 3)

= 6x2 + 9x + 4x + 6

= 6x2 + 13x + 6

5.

4.

Find the area, in cm2, of the shaded region in the

diagram above.

Length = (8 – x) cm

Width = (12 – 2x) cm

Area of the shaded region

= (8 – x)(12 – 2x)

= 96 – 16x – 12x + 2x2

= 96 – 28x + 2x2

13

Example

CCTS Recognising

CCTS Recognising

Example

Factorise the following quadratic expressions.

(a) x2 – 1 (b) p2 – 49 (c) 50x2 – 8

Solution:

(a) x2 – 12 = (x + 1)(x – 1) (b) p2 – 49 = p2 – 72 (c) 50x2 – 8 = 2(25x2 – 4)

= (p + 7)(p – 7) = 2[(5x)2 – 22]

= 2(5x + 2)(5x – 2)



(a) x2 + 3x + 2 (b) x2 + x – 2 (c) x2 – 6x + 8

(a) x2 + 3x + 2

= x2 + x + 2x + 2

= x(x + 1) + 2(x + 1)

= (x + 1)(x + 2)

(b) x2 + x – 2

= x2 – x + 2x – 2

= x(x – 1) + 2(x – 1)

= (x – 1)(x + 2)

(c) x2 – 6x + 8

= x2 – 2x – 4x + 8

= x(x – 2) – 4(x – 2)

= (x – 2)(x – 4)

Tips

(b) x2 + x – 2

(c) x2 – 6x + 8

(a) x2 + 3x + 2

1 × 2

1x + 2x = 3x –1 × 2

–x + 2x = x (–2) × (– 4)

–2x – 4x = – 6x

(a) 1 – p2

= (1 – p)(1 + p)

(b) x2 – 36

= x2 – 62

= (x + 6)(x – 6)

(c) 25 – 121x2

= 52 – (11x)2

= (5 + 11x)(5 – 11x)

(d) 12x2 – 3

= 3(4x2 – 1)

= 3[(2x)2 – 12]

= 3(2x + 1)(2x – 1)

(f) 36 – 4x2

= 4[9 – x2]

= 4[32 – x2]

= 4(3 + x)(3 – x)

(g) 100 – 4x2

= 4[25 – x2]

= 4[52 – x2]

= 4(5 + x)(5 – x)

(h) 144x2 – 1

= (12x)2 – 12

= (12x + 1)(12x – 1)

(k) 121x2 – 81

= (11x)2 – 92

= (11x + 9)(11x – 9)

(l) 1 – 25p2

= 12 – (5p)2

= (1 + 5p)(1 – 5p)

(e) 25 – 9x2

= 52 – (3x)2

= (5 + 3x)(5 – 3x)

(i) 18x2 – 50

= 2(9x2 – 25)

= 2[(3x)2 – 52]

= 2(3x + 5)(3x – 5)

(j) 49 – 25x2

= 72 – (5x)2

= (7 + 5x)(7 – 5x)

(m) 49x2 – 16

= (7x)2 – 42

= (7x + 4)(7x – 4)

(n) 100 – 64x2

= 4[25 – 16x2]

= 4[(5)2 – (4x)2]

= 4(5 + 4x)(5 – 4x)

(o) 400 – x2

= 202 – x2

= (20 + x)(20 – x)

(p) 169 – 64p2

= 132 – (8p)2

= (13 + 8p)(13 – 8p)

Solution:

14

Example


(a) x2 + 6x + 8

= x2 + 2x + 4x + 8

= x(x + 2) + 4(x + 2)

= (x + 2)(x + 4)

(b) x2 + 3x – 10

= x2 + 5x – 2x – 10

= x(x + 5) – 2(x + 5)

= (x + 5)(x – 2)

(c) x2 – 2x – 8

= x2 – 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x – 4)(x + 2)

(d) x2 – 4x – 21

= x2 – 7x + 3x – 21

= x(x – 7) + 3(x – 7)

= (x – 7)(x + 3)

(e) x2 – 7x + 10

= x2 – 2x – 5x + 10

= x(x – 2) – 5(x – 2)

= (x – 2)(x – 5)

(f) x2 – 5x – 36

= x2 – 9x + 4x – 36

= x(x – 9) + 4(x – 9)

= (x – 9)(x + 4)

(g) x2 + x – 12

= x2 + 4x – 3x – 12

= x(x + 4) – 3(x + 4)

= (x + 4)(x – 3)

(h) x2 – 2x – 15

= x2 – 5x + 3x – 15

= x(x – 5) + 3(x – 5)

= (x – 5)(x + 3)

(i) x2 + 8x + 15

= x2 + 3x + 5x + 15

= x(x + 3) + 5(x + 3)

= (x + 3)(x + 5)


(a) 2x2 + x – 6 (b) 10x2 + x – 2 (c) 42 – 17x – 15x2

Solution:


(a) 6x2 + 7x – 3

6x2 + 7x – 3

= (3x – 1)(2x + 3)

(b) 10x2 – 13x – 3

10x2 – 13x – 3

= (5x + 1)(2x – 3)

(j) x2 + x – 56

= x2 – 7x + 8x – 56

= x(x – 7) + 8(x – 7)

= (x – 7)(x + 8)

(k) x2 + 9x – 22

= x2 + 11x – 2x – 22

= x(x + 11) – 2(x + 11)

= (x + 11)(x – 2)

(l) x2 – 13x – 30

= x2 – 15x + 2x – 30

= x(x – 15) + 2(x – 15)

= (x – 15)(x + 2)

(c) 10x2 – 19x – 15

10x2 – 19x – 15 = (5x + 3)(2x – 5)

3x –1 –2x2x 3 9x6x2 –3 7x

5x 1 2x2x –3 –15x10x2 –3 –13x

5x 3 6x2x –5 –25x10x2 –15 –19x

2x × x = 2x2

–3 × 2 = –6

–3x + 4x = x

∴ 2x2 + x – 6

= (2x – 3)(x + 2)

5x –2 –4x 2x 1 5x10x2 –2 x

5x × 2x = 10x2

–2 × 1 = –2

–4x + 5x = x

∴ 10x2 + x – 2

= (5x – 2)(2x + 1)

–5x 6 18x 3x 7 –35x–15x2 42 –17x

–5x × 3x = –15x2

6 × 7 = 42

18x – 35x = –17x

∴ 42 – 17x – 15x2

= (–5x + 6)(3x + 7)

2x –3 –3x x 2 4x2x2 –6 x

15

CCTS Recognising & representing

Example


(a) 10x2 – 5x – 15

= 5(2x2 – x – 3)

= 5(2x – 3)(x + 1)

(b) –12x2 + 21x + 6

= –3(4x2 – 7x – 2)

= –3(4x + 1)(x – 2)

(c) 20 – 28x + 8x2

= 4(5 – 7x + 2x2)

= 4(5 – 2x)(1 – x)

(d) 18 + 3x – 3x2

= 3(6 + x – x2)

= 3(3 – x)(2 + x)

(e) –6x2 – 48x – 90

= –6(x2 + 8x + 15)

= –6(x + 5)(x + 3)


(a) 3x2 – 12x – 36 (b) 24x2 + 28x – 20

Solution:

(a) 3x2 – 12x – 36

= 3(x2 – 4x – 12)

= 3(x + 2)(x – 6)

(b) 24x2 + 28x – 20

= 4(6x2 + 7x – 5)

= 4(3x + 5)(2x – 1)

(d) –2x2 + 7x – 3

–2x2 + 7x – 3

= (–2x + 1)(x – 3)

or (1 – 2x)(x – 3)

(e) 2x2 + 5x – 7

2x2 + 5x – 7

= (2x + 7)(x – 1)

(f) 10x2 + 13x – 3

10x2 + 13x – 3 = (5x – 1)(2x + 3)

(g) 9x2 + 26x – 3

9x2 + 26x – 3

= (9x – 1)(x + 3)

(h) 21x2 – 2x – 8

21x2 – 2x – 8

= (7x + 4)(3x – 2)

(i) –12x2 + 11x – 2

–12x2 + 11x – 2 = (–3x + 2)(4x – 1)

or (2 – 3x)(4x – 1)

(f) 2x2 – 2x – 40

= 2(x2 – x – 20)

= 2(x – 5)(x + 4)

–2x 1 x x –3 6x– 2x2 –3 7x

2x 7 7x x –1 –2x2x2 –7 5x

5x –1 –2x 2x 3 15x10x2 –3 13x

9x –1 –x x 3 27x9x2 –3 26x

7x 4 12x 3x –2 –14x21x2 –8 –2x

–3x 2 8x 4x –1 3x–12x2 –2 11x

TipsTake out the common factor 4, then

factorise. This way is easier than

factorising it without taking out the

factor 4.

16

CCTS Representing

CCTS Identifying

CCTS Identifying

Example

Example

2.3 Quadratic Equations

Identify the following equations and state the number of unknowns.

(a) 5x2 + 3x = 0 (b) 7x + 5 = 0 (c) 3xy + y = 0

Solution:

(a) It is a quadratic equation

in one unknown. (i.e. x)

(b) It is not a quadratic

equation. It has one

unknown. (i.e. x)

(c) It is not a quadratic

equation. It has

two unknowns.

(i.e. x and y)

1. Identify the quadratic equations with one unknown from the following equations.

(c) x2 + 2 =

x – 1 ––––– ––––– 3 2

(c) x2 + 2 =

x – 1 ––––– –––– 3 2

2x2 + 4 = 3x – 3

2x2 – 3x + 4 + 3 = 0

2x2 – 3x + 7 = 0

(c) x2 + 3

= x ––––– 4

It is a quadratic equation in one unknown.

(a) 15 + 3x = 6x2

It is a quadratic equation in one unknown.

(b) 7 = 1 + 4y –– x

It is not a quadratic equation.

(d) 16 = pq2 + 2p

It is not a quadratic equation.

2. Write the following quadratic equations in general form.

(a) 7x2 = 4x + 3

7x2 – 4x – 3 = 0(b)

x + 2 = x2

5

x + 2 = 5x2

–5x2 + x + 2 = 0

5x2 – x – 2 = 0

(c) 9 = 1

x2 + 3x 2

18 = x2 + 6x 0 = x2 + 6x – 18

x2 + 6x – 18 = 0

(d) 1 (x2 + 1) =

x – 3 –– –––– 8 3

x2 + 1 =

x – 3 ––––– ––––– 8 3

3x2 + 3 = 8x – 24

3x2 – 8x + 3 + 24 = 0

3x2 – 8x + 27 = 0

(f) (x + 3)(x – 2) = 5

x2 + 3x – 2x – 6 – 5 = 0

x2 + x – 11 = 0

(e) (x – 1)(x + 2) = (2x – 1)2

(x – 1)(x + 2) = (2x – 1)(2x – 1)

x2 – x + 2x – 2 = 4x2 – 2x – 2x + 1

x2 + x – 2 = 4x2 – 4x + 1

0 = 4x2 – x2 – 4x – x + 1 + 2

3x2 – 5x + 3 = 0

Write the following quadratic equations in general form.

(a) 2 – x = 5x2 (b) 5x – 2

= x2

4

Solution:

(a) 2 – x = 5x2

–5x2 – x + 2 = 0

5x2 + x – 2 = 0

(b) 5x – 2

= x2

4

5x – 2 = 4x2

–4x2 + 5x – 2 = 0

4x2 – 5x + 2 = 0

17

Example

Area of trapezium

= [(2x + 3) + (x – 2)] y= (3x + 1) y= (3x + 1) × (5 – x)

= –3x2 + 14x + 5

5.

The diagram above shows a kite with p = q,

q = 3x – 3 and r = 2x – 1. Given that the area of

the kite is 36 cm2, show that 2x2 – 3x – 8 = 0.

Area of kite

= 2 × area of triangle

= 2 × 1 × [(3x – 3) + 1

(3x – 3)] × (2x – 1) 2 3

= (3x – 3 + x – 1) × (2x – 1)

= (4x – 4)(2x – 1)

= 8x2 – 12x + 4

8x2 – 12x + 4 = 36

4(2x2 – 3x + 1) = 36

2x2 – 3x + 1 = 9

2x2 – 3x – 8 = 0

4.

The diagram above shows the cross-section of

a roof. Find a quadratic equation involving x as

a variable.

82 = (x + 1)2 + (x + 1)2

64 = x2 + 2x + 1 + x2 + 2x + 1

64 = 2x2 + 4x + 2

0 = 2x2 + 4x – 62

or 2x2 + 4x – 62 = 0

Solution:

Total number of students = (x + 3)(2x – 3)

420 = (x + 3)(2x – 3)

420 = 2x2 + 6x – 3x – 9

2x2 + 3x – 9 – 420 = 0

2x2 + 3x – 429 = 0

1––

3

There are 420 form one students in SMK Berapit. They are divided into (x + 3) classes and each class has

(2x – 3) number of students. Write a quadratic equation for this situation.

6.

The diagram above shows four congruent

rectangles arranged in the position shown. Find

the area of the shaded region, A cm2.

If A = 50, show that 4x2 – 12x – 41 = 0.

Length of side of region A = 3x – 2 – (x + 1)

= 3x – 2 – x – 1

= 2x – 3

A is a square.

∴Area of A = (2x – 3)2

50 = (2x – 3)(2x – 3)

50 = 4x2 – 12x + 9

0 = 4x2 – 12x – 41

4x2 – 12x – 41 = 0

Given that area = 70

⇒ 70 = –3x2 + 14x + 5

3x2 – 14x – 5 + 70 = 0

3x2 – 14x + 65 = 0

3.

The diagram above shows a trapezium with an

area of 70 cm2. Given that y = 5 – x, show that

3x2 – 14x + 65 = 0.

(x + 1) m (x + 1) m

8 m

A

(x + 1) cm

(3x 2) cm

(2x + 3) cm

2y cm

(x – 2) cm

p cmr cm

q cm

p

r

q

18

CCTS Interpreting

Example

2.4 Roots of Quadratic Equations

1. Given the following values of x and its corresponding equations, determine whether the given value is a root

of the equation.

(a) x = 2, 2x2 – 3x – 2 = 0

When x = 2,

2x2 – 3x – 2

= 2(2)2 – 3(2) – 2

= 8 – 6 – 2

= 0

� x = 2 is a root of the quadratic equation.

(b) x = 3, 3x2 – 16x + 5 = 0

When x = 3,

3x2 – 16x + 5

= 3(3)2 – 16(3) + 5

= 27 – 48 + 5

= –16 ≠ 0

� x = 3 is not a root of the quadratic equation.

(c) x = –2, 1 x2 –

1 x + 2 = 0 –– ––

4 2

When x = –2,

1 x2 –

1 x + 2 –– ––

4 2

=

1 (–2)2 –

1 (–2) + 2 –– ––

4 2

=

1 × 4 + 1 + 2 –– 4

= 1 + 3

= 4 ≠ 0

� x = –2 is not a root of the quadratic equation.

(d) x = 1

, 2 – x

= x2

2 6

2 – x = x2

6

2 – x = 6x2

6x2 + x – 2 = 0

When x = 1

, 2

6x2 + x – 2

= 6 × 1 × 1 +

1 – 2

2 2 2

=

3 +

1 – 2

2 2

= 0

� x = 1

is a root of the quadratic equation. 2

Determine whether 2 or 4 is a root of 2x2 – 5x – 12 = 0.

Solution:

(a) When x = 2, (b) When x = 4,

2x2 – 5x – 12

= 2(2)2 – 5(2) – 12

= 8 – 10 – 12

= –14 ≠ 0

Therefore, x = 2 is not a root

of the equation.

2x2 – 5x – 12

= 2(4)2 – 5(4) – 12

= 32 – 20 – 12

= 0

Therefore, x = 4 is a root of the

equation.

TipsIf the fi nal value is zero, then the

value of x is a root of the equation.

Tips: Change it to

general form.

19

CCTS Recognising & interpreting

Example

(a) 2x2 – 3x – 2 = 0

(a) When x = 1,

2x2 – 3x – 2

= 2(1)2 – 3(1) – 2

= 2 – 3 – 2

= –3 ≠ 0

When x = 2,

2(2)2 – 3(2) – 2

= 8 – 6 – 2

= 0

∴ x = 2 is a root of the quadratic equation.

When x = 0,

2x2 – 3x – 2

= 2(0)2 – 3(0) – 2

= –2

When x = –1

2(–1)2 – 3(–1) – 2

= 2 + 3 – 2

= 3

Try x =

– 1

, –– 2

2(– )2 – 3(– ) – 2

= + – 2

= 0

∴ x = – is a root of the quadratic equation.

(b) 2x2 + 3x – 9 = 0

(b) When x = 1,

2(1)2 + 3(1) – 9

= 2 + 3 – 9

= –4

When x = 2,

2(2)2 + 3(2) – 9

= 8 + 6 – 9

= 5

Try x = ,

2( )2 + 3( ) – 9

= + – 9

= 0

∴ x = is a root of the quadratic equation.

When x = –2,

2(–2)2 + 3(–2) – 9

= 8 – 6 – 9

= –7

When x = –3,

2(–3)2 + 3(–3) – 9

= 18 – 9 – 9

= 0

∴ x = –3 is a root of the quadratic equation.

(e) x = 1

, (2x – 1)(x + 1) = 0 –– 2

When x = 1 , ––

2

(2x – 1)(x + 1)

= (2( ) – 1)( + 1)

= (0)( )

= 0

� x = is a root of the quadratic equation.

Determine the solutions (roots) of the quadratic equations below by trial and error.

1––

2

1––

2

1––

2

1–– 2

1–– 2

1–– 2

3–– 2

1–– 2

1–– 2

3–– 2

3–– 2

9–– 2

9–– 2

3–– 2

3–– 2

Note:Changes from –ve to

+ve. Therefore, there

is a root between

x = 1 and x = 2.

Note:When the value of

x changes from

0 to –1, the answer

changes from –2

(–ve) to 3 (+ve).

This means that

there is a root

between

x = 0 and x = –1.

Solution:

(f) x = 1

, 3x2 – 4x = –1 ––

3

3x2 – 4x + 1 = 0

When x =

1 –– 3

3x2 – 4x + 1

= 3( )2 – 4( ) + 1

= – + 1

= 0

� x = is a root of the quadratic equation.

1––

3

1––

3

1––

31––

3

4––

3

20

⇒ (2x – 9)(x + 1) = 0

2x – 9 = 0 or x + 1 = 0

2x = 9

∴ x = or x = –1

Example

Change it to the

general form.

2. Find the solutions of the following quadratic equations by trial and error.

(a) x2 + 4x – 5 = 0

When x = 1,

x2 + 4x – 5

= 12 + 4(1) – 5

= 0


When x = –4,

(–4)2 + 4(– 4) – 5

= 16 –16 – 5

= –5

When x = –5,

(–5)2 + 4(–5) – 5

= 25 – 20 – 5

= 0


(b) x2 + x – 6 = 0

When x = 2,

x2 + x – 6

= 22 + 2 – 6

= 4 + 2 – 6

= 0


When x = –2,

(–2)2 + (–2) – 6

= 4 – 2 – 6

= –4

When x = –3,

(–3)2 + (–3) – 6

= 9 – 3 – 6

= 0


(b) 9 + 7x = 2x2

(b) 9 + 7x = 2x2

2x2 – 7x – 9 = 0

3. Solve the following quadratic equations by factorisation.

Solve the following quadratic equations by factorisation.

(a) 3x2 – 5x = 2

Solution:

(a) 3x2 – 5x = 2

3x2 – 5x – 2 = 0

(c) 2x2 + 35

= x 19

2x2 + 35 = 19x 2x2 – 19x + 35 = 0

(2x – 5)(x – 7) = 0

2x – 5 = 0 or x – 7 = 0

2x = 5

x = or x = 7

(a) 3x2 – 21x = –30

3x2 – 21x + 30 = 0

(3x – 6)(x – 5) = 0

3x – 6 = 0 or x – 5 = 0

3x = 6

x = 2 or x = 5

(b) x2

– 10

= –x –– –– 3 3

x2 – 10 = –3x x2 + 3x – 10 = 0

(x – 2)(x + 5) = 0

x – 2 = 0 or x + 5 = 0

x = 2 or x = –55–– 2

⇒ (3x + 1)(x – 2) = 0

3x + 1 = 0 or x – 2 = 0

3x = –1

∴ x = – or x = 2

9–– 2

2x –9 –9x x 1 2x2x2 –9 –7x

3x 1 x x –2 –6x3x2 –2 –5x

1–– 3

21

CCTS Solving problems

Example

(h) 10 = 3x(9x – 13)

10 = 27x2 – 39x 0 = 27x2 – 39x – 10

0 = (9x + 2)(3x – 5)

9x + 2 = 0 or 3x – 5 = 0

9x = –2 or 3x = 5

x = –

or

x =

A man and his son are painting a house. When both of them paint together, they need 4 days to complete. If

the father paints the house alone, he can complete the job 6 days earlier than the son. Find the number of days

needed for the son to paint the house alone.

1–– x

(d) –8x2 + 14x = 3

–8x2 + 14x – 3 = 0

(4x – 1)(–2x + 3) = 0

4x – 1 = 0 or –2x + 3 = 0

4x = 1 or –2x = –3

x = or x =

x =

(e) 1 + 12x = 6x2 – 7x + 16

0 = 6x2 – 7x – 12x + 16 – 1

0 = 6x2 – 19x + 15

0 = (2x – 3)(3x – 5)

2x – 3 = 0 or 3x – 5 = 0

2x = 3 3x = 5

x = 3

or x =

5 –– –– 2 3

(f) 6x2 + 7 = 23x 6x2 – 23x + 7 = 0

(3x – 1)(2x – 7) = 0

3x – 1 = 0 or 2x – 7 = 0

3x = 1 or 2x = 7

x = or x =

1–– 4

–3–– –23–– 2

1––

3

7––

2

(g) 10 – 8x = 21x(x + 1)

10 – 8x = 21x2 + 21x 0 = 21x2 + 21x + 8x – 10

0 = 21x2 + 29x – 10

0 = (7x – 2)(3x + 5)

7x – 2 = 0 or 3x + 5 = 0

7x = 2 or 3x = –5

x = or x = –

(i) 4 + x = 15 x2

2

8 + 2x = 15x2

15x2 – 2x – 8 = 0

(3x + 2)(5x – 4) = 0

3x + 2 = 0 or 5x – 4 = 0

3x = –2 or 5x = 4

x = – or x =

2––

9

5––

3 2––

3

4––

5

8x – 24 = x2 – 6x 0 = x2 – 14x + 24

0 = (x – 12)(x – 2)

x – 12 = 0 or x – 2 = 0

x = 12 or x = 2 (not applicable)

The son can complete the job in 12 days.

2––

7

5––

3

Solution:

Let x be the number of days needed if the son paints

alone.

In one day, he will complete amount of work.

Let (x – 6) be the number of days needed if the father

paints alone.

In one day, he will complete ( ) amount of work.

If both paint together, in one day they will complete

of the amount of work.

1 x – 6

1 +

1 =

x x – 6

x – 6 + x =

x(x – 6)

2x – 6 =

x(x – 6)

1––4

1––4

1––4

1–– 4

22

4. The length of a rectangle is fi ve times its breadth.

When the breadth of the rectangle is increased by

10 m, while the length of the rectangle is decreased

by 25 m, the area of the rectangle remains the same.

Find the dimensions of the original rectangle.

Let x and 5x be the dimensions of the original

rectangle. Then (x + 10) and (5x – 25) will be its

new dimensions.

Area = 5x2 = (x +10)(5x – 25)

5x2 = 5x2 + 25x – 250

0 = 25x – 250

25x = 250

x = 10

Thus, the dimensions of the original rectangle are

10 m and 50 m.

5. John has a rectangular fi sh pond with an area of

1 575 m2. If he reduces the length of one side by 30

m and the other by 20 m, the new pond will become

a square. Find the length of the new pond.

Let x be the length of the new pond.

Then (x + 20) and (x + 30) will be the length of the

sides of the original pond.

⇒ 1 575 = (x + 20)(x + 30)

1 575 = x2 + 50x + 600

0 = x2 + 50x – 975

0 = (x – 15)(x + 65)

x – 15 = 0 or x + 65 = 0

x = 15 or x = – 65 (not acceptable)

The side of the new pond is 15 m.

⇒ 180 + 3 = 180

x + 3 x

180

x + 3

7. Faizal thinks of two positive integers. The difference

between the two integers is 8 and the sum of the

square of each integer is 370. Find the value of both

integers.

Let x and (x + 8) represent the two positive integers.

x2 + (x + 8)2 = 370

x2 + x2 + 16x + 64 = 370

2x2 + 16x – 306 = 0

x2 + 8x – 153 = 0

(x – 9)(x + 17) = 0

x – 9 = 0 or x + 17 = 0

x = 9 or x = –17 (not acceptable)

The second integer = 9 + 8 = 17

Therefore, the two integers are 9 and 17.

6. There are 180 pencils to be distributed equally

among a group of students. When 3 new students

join the group, each of the students will receive

3 pencils less. Find the initial number of students

in the group.

Let x be the initial number of students in the group.

Initially each student will receive 180 pencils. x

After 3 new students join the group, each will

receive pencils.

multiply by x(x + 3),

180x + 3x(x + 3) = 180(x + 3)

180x + 3x2 + 9x = 180x + 540

3x2 + 9x – 540 = 0

x2 + 3x – 180 = 0

(x – 12)(x + 15) = 0

x – 12 = 0 or x + 15 = 0

x = 12 or x = –15 (not acceptable)

Initially, the group has 12 students.

23

SPM20012006

1. (3x – y)2 – x(2x – y) =

A 7x2 – 5xy + y2

B 7x2 – 7xy + y2

C 11x2 – 5xy + y2

D 11x2 – 7xy + y2

2. – 2x(1 – x) + 5x2 =

A 7x2 – 2x

B 9x2

C 2x – 7x2

D 7x2 + 2x

3. Expand 4(3 – 2x)(x – 1).

A –8x2 + 5x – 3

B –8x2 – 5x + 3

C 8x2 – 20x + 12

D –8x2 + 20x – 12

4. Expand px(4x – 3).

A 4px2 – 3 C 4px2 – 3px B 4px2 + 3px D 4p2x – 3px

5.

Form a quadratic expression for the area of the

triangle shown above.

A 1 (4x – 1) C 2x(4x – 3) ––

2

B x(4x + 3) D 1 x(3 + 4x) ––

2

6. Factorise

x2 –

x completely.

A x (x – 2) C 3x(x – 2) ––

3

B x(x – 2) D 1

x(x2 – 2) –– 3

7. Factorise 16x2 – 1.

A 4x(4x + 1)

B 4x(4x – 1)

C (4x + 1)(4x – 1)

D (4x + 1)2

8. Factorise 3x2 – 13x – 10.

A (3x + 2)(x + 5)

B (3x + 2)(x – 5)

C (3x – 2)(x – 5)

D (3x – 2)(x + 5)

9. Factorise 3(x – 1) – x(x – 1).

A (x + 1)(3 – x) C (3x – 3)(3 – x)

B (x – 1)(3 – x) D (x – 1)(3 + x)

10. Factorise 9x2 – m2.

A (3x + m)2 C (3x – m)(3x + m)

B (3x – m)2 D (x – m)(3x + 3m)

11. Factorise 15x2 + 10x – 25 completely.

A 5(x – 1)(3x + 5)

B (5x – 5)(3x + 5)

C 5(1 – x)(3x + 5)

D (5x – 1)(3x + 5)

12. Which of the following is not a quadratic

equation?

A 4x – 1 = 4 C mx – 3x = 2 ––––– ––

5 x

B 5x – 4x2 = 0 D 7 =

5 ––– ––––– x2 x + 1

13. Which of the following quadratic equations is

not in the general form?

A 1 x2 + 2x – 5 = 0 C 16x2 +

2 x – 1

= 0 2 3 5

B 3x – x2 = 2 D 6x2 + 7x – 8 = 0

14.

The diagram above shows the length and breadth

of a rectangle. If the length is increased by 1 m

and its breadth is decreased by 2 m, fi nd the area,

in m2, of the new rectangle.

A (2x – 1)(x + 2)

B (2x – 1)(x + 3)

C 2x2 + 3x – 2

D 2x2 – 3x – 2

4x 3

2x

2x m

x m

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

SPM20012005

2— 3

1— 3

Paper 1

SPM Practice 2

24

15. Determine which of the following is a root of

(5x – 1)(x – 3) = 0.

A 1 C 3

B 2 D 4

16. Determine which of the following is a root of

3x2 + 7x – 6 = 0.

A –3 C 0

B –1 D 2

17. One of the roots of 8x2 – 6x – 9 = 0 is

A –1 1 C

3 –– –– 2 4

B 1 1 D

4 –– –– 2 3

18. The solutions of 6x2 + 11x – 2 = 0 are

A 6 and 1 C

1 and –2 –– ––

2 6

B – 1

and 2 D 6 and –2 –– 6

19. The roots of (2x – 3)(3 – 5x) = 0 are

1 5 A – — and — C 3 and –5 3 3

1 5 3 3 B — and – — D — and — 3 3 2 5

20. The roots of x2 + 5 x – 2 = 0 are

A 1

and –6 C 3 and –

1 –– –– 3 6

B 2

and 4 D 2 and –5 ––

3

✓

✓✓

✓

2. Expand

(a) (x – 1)(5 – 4x)

(b)

1 (x + 2)(6x + 9) ––

3

(a) (x – 1)(5 – 4x) = 5x – 4x2 – 5 + 4x

= – 4x2 + 9x – 5

(b) 1

(x + 2)(6x + 9) = 1 (x + 2)3(2x + 3) ––

3 3

= (x + 2)(2x + 3)

= 2x2 + 4x + 3x + 6

= 2x2 + 7x + 6

4. Expand (–10x + 3)2.

(–10x + 3)(–10x + 3) = 100x2 – 30x – 30x + 9 = 100x2 – 60x + 9

6. Factorise 64p2 – 25.

64p2 – 25 = (8p)2 – 52

= (8p + 5)(8p – 5)

1. Solve the quadratic equation 6x2 – 35 = 11x by

factorisation

6x2 – 11x – 35 = 0

(2x – 7)(3x + 5) = 0

∴ 2x – 7 = 0 or 3x + 5 = 0

7 5 x = or x = – 2 2

3. Given that 3x2 + 2x – 8 = (x + 2)(ax + p). Find the

values of a and p.

3x2 + 2x – 8 = (3x – 4)(x + 2)

⇒ a = 3 and p = –4

5. Factorise 2x2 – 9x – 143.

2x2 – 9x – 143 = (x – 11)(2x + 13)

✓

✓

2—3

Paper 2

SPM20012007

25

SPM20012006

SPM20012005

7. Solve

(a) x2 – 7x = 0

3 – 11x (b) ——— = 5x2

4

(a) x2 – 7x = 0

x(x – 7) = 0

x = 0 or x – 7 = 0

x = 0 or x = 7

3 – 11x (b) ——— = 5x2

4

20x2 + 11x – 3 = 0

(5x – 1)(4x + 3) = 0

5x – 1 = 0 or 4x + 3 = 0

5x = 1 or 4x = –3

1 3 x = –– or x = – —

5 4

9.

The diagram above shows a ladder leaning against

a wall. When the ladder slides down x m, form

a quadratic equation for the ladder at the new

position.

(Note: When A moves x m, B moves x m.)

172 = (15 – x)2 + (8 + x)2

289 = 225 – 30x + x2 + 64 + 16x + x2

289 = 2x2 – 14x + 289

2x2 – 14x = 0

8. Find the roots of the quadratic equations:

x – 30 (a) x2 – 8x = –––––– (b) 6x2 + x – 1 = 0 2

x – 30(a) x2 – 8x = ––––– 2

2x2 – 16x = x – 30

2x2 – 17x + 30 = 0

(2x – 5)(x – 6) = 0

x = or x = 6

(b) 6x2 + x – 1 = 0

(2x + 1)(3x – 1) = 0

∴ 2x + 1 = 0 or 3x – 1 = 0

1 1 x = – — or x = — 2 3

5––

2

{

{

17 m15 m

8 m

x

x

B

A 10.

When a metal plate is heated, it expands to a new

dimensions as shown in the diagram. Find the

increase in the area of the plate, in m2.

Increase in area = (6 + 2x)(5 + 2x) – 5 × 6

= 30 + 22x + 4x2 – 30

= 4x2 + 22x

Metal

plate

5 m

6 m

x

x

x

x

chapter 2 - quadratic expessions and equations

Documents