Chapter No.

SOLUTION OF
QUADRATIC EQUATION
EXERCISE 1A
Q.1 Solve the following quadratic equations by factorisation:
(a) x2 5x = 0
Solution:
x2 5x = 0
Take x as common
x(x 5) = 0
Either
x=0
or
x5=0
x=5
So, x = 0 or x = 5
(b) 4x2 = 7x
Solution:
4x2 = 7x
4x2 7x = 0
Take x as common
x(4x 1) = 0
Either
x=0
or
4x 7 = 0
7
x=
4

Emporium Career Maths Series

3
4
3
So, x = 0, or x = 1
4
2
(c) 6t = t(6 4)
Solution:
x=1

6t2 = t(t 4)
6t2 t(t 4)
6t2 t2 + 4t = 0
5t2 + 4t = 0
t(5t + 4) = 0
Either
t=0
or
5t + 4 = 0
4
t=
5
So, t = 0, or to

4
5

(d) 5y2 = y(y + 3)

Solution:
5y2
5y2
5y2
y(4y 3)

=
=

y(y + 3)
y2 + 3y
y2 3y = 0
0

Either
y=0
or
4y 3 = 0
3
y=
4
So, y =

0 or y =

3
4

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(e) a2 + 9a = 0
Solution:
a2 + 9a = 0
a(a+9) = 0
Either
a=0
or
a+9=0
a=9
So, a = 0 or 9
(f) 3h = h(5 2h)
Solution:
3h2 = h(5 2h)
3h2 (5 2h) = 0
3h2 + 2h2 5h = 0
5h2 5h = 0
5h(h 1) = 0
Either
h=0
or
h1=0
h=1
So, h = 0 or 1
2
(g) x 2x + 1 = 0
Solution:
x2 2x + 1 = 0
x2 x x + 1 = 0
x(x 1) 1 (x 1) = 0
(x 1)2 = 0
Either
x1=0
x=1
or
2

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x1=0
x=1
So, x = 1, 1
(h) 7a + a2 18 = 0
Solution:
7a + a2 18 = 0
Re-arranging
a2 + 7a 18 = 0
a2 + 9a 2a 18 = 0
a(a+9) 2(a+9) = 0
(a+9) (a2) = 0
Either
a+9=0
a = 9
or
a2=0
a=2
So, a = 9 or a = 2
(i)
2x2 + 5z 3 = 0
Solution:
2z2 + 5z 3 = 0
2z2 + 6z z 3 = 0
2z(z+3) 1(z3) = 0
(z+3) (2z1) = 0
Either
z+3=0
z=3
or
2z 1 = 0
1
z=
2
1
So, z = 3 or z
=
2

Emporium Career Maths Series

(j) c2 + 2c = 35
Solution:
c2 + 2c = 35
c2 + 2c 35 = 0
c2 + 7c 5c 35 = 0
c(c+7) 5(c7) = 0
(c+7) (c5) = 0
Either
c+7=0
c=7
or
c5=0
c=5
So, c = 7 or c = 5
(k) 8p 16 p2 = 0
8p 16 p2 = 0
Re-arranging
p2 8pr + 16 = 0
p2 4p 4p + 16 = 0
p(p4) (p4) = 0
Either
p4=0
p=4
or
p4=0
p=4
So, p = 4 or 4
(l)
4 3b b2 = 0
Solution:
4 3b b2 = 0
b3 + 3b 4 = 0
b2 + 4b b 4 = 0
b(b+4) 1(b+4) = 0

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(b+4) (b1) = 0
Either
b+4=0
b=4
or
b1=0
b=1
So, b = 4 or b = 1
(m) 12 a a2 = 0
12 a a2 = 0
a2 + a 12 = 0
a2 + 4a 3a 12 = 0
a(a+4) 3(a+4) = 0
(a+4)(a3) = 0
Either
a+4=0
a=4
or
a3=0
a=3
So, a = 4, or a = 3
2
(n) 10t t = 2
Solution:
10t2 t = 2
10t2 t 2 = 0
10t2 5t + 4t 2 = 0
5t(2t1) +2(2t1) = 0
(2t1)(5t+2) = 0
Either
2t 1 = 0
1
t=
2
or

Emporium Career Maths Series

5t + 2 = 0
2
t=
5
So, t =

1
or
2

2
5

(o) y2 22y + 96 = 0
Solution:
y2 22y + 96 = 0
y2 16y 6y + 96 = 0
y(y16) 6(y16) = 0
(y16) (y6) = 0
Either
y 16 = 0
y = 16
or
y6=0
y=6
So, y = 16 or y = 6
2
(p) 12a 16a 35 = 0
12a2 16a 35 = 0
12a2 30a + 14a 35 = 0
6a(2a5) +7(2a5) = 0
(2a 5) (6a + 7) = b
Either
2a 5 = 0
1
a=2
2
or
6a + 7 = 0
6a = 7
7
a=
6

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a = 1

1
6

1
1
or a = 1
2
6
2
(q) 15x + 4x 35 = 0
Solution:
15x2 + 4x 35 = 0
15x2 + 25x 21x 35 = 0
5x(3x+5) 7(3x+5) = 0
(3x+5) (5x7) = 0
Either
3x + 5 = 0
5
x=
3
2
x = 1
3
or
5x 7 = 0
7
x=
5
2
x=1
5
2
2
So, x = 1 or a = 1
3
5
2
(r) 15x + 4x 35 = 0
Solution:
28x2 85x + 63 = 0
28x2 36x 49x + 63 = 0
4x(7x9) 7(7x9) = 0
(7x9) (4x7) = 0
Either
7x 9 = 0, 7x = 9
So, a = 2

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9
7
2
x=1
7
x=

or
4x 7 = 0, 4x = 7
7
x=
4
3
x=1
4
2
3
So, x = 1 or x = 1
7
4
2
(s) 56x 159x + 108 = 0
Solution:
56x2 159x + 108 = 0
56x2 96x 63x + 108 = 0
8x(7x12) 9(7x12) = 0
(7x12) (8x9) = 0
Either
7x 12 = 0 7x = 12
12
x=
7
5
x=1
7
or
8x 9 = 0, 8x = 9
9
x=
8
1
x=1
8
5
1
So, x = 1 or x = 1
7
8
2
(t) 39x = 131x 44 = 0

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Solution:
39x2 + 131x 44 = 0
39x2 + 143x 12x 44 = 0
13(3x11) 4(3x11) = 0
Either
3x 11 = 0 3x = 11
2
x=3
3
or
13x 4 = 0 13x = 4
4
x=
13
1
So, x = 3 or x = 4 13
2
(u) 76x 96x2 15 = 0
Solution:
76x 96x2 15 = 0
96x2 76x + 15 = 0
96x2 40x 36x + 15 = 0
8x(12x5) 3(12x5) = 0
(12x5) (8x3) = 0
Either
12x 5 = 0
5
x=
12
or
8x 3 = 0
3
x=
8
5
3
So, x =
or
12
8
Q. Form a quadratic equation in x with the giver roots for
each of the following:

Emporium Career Maths Series

(a) 2, 3
Solution:
Given roots are
x = 2 and x = 3
x2=0
x3=0

the required equation is

(x 2) (x 3) = 0
2
x 2x 3x + 6 = 0
x2 5x + 6 = 0
(b) 3, 4
Solution:
Given roots are
x = 3 and x = 4
x3 = 0
x+4 = 0

the required equation is

(x3) (x+4) = 0
2
x 3x + 4x 12 = 0
x2 + x 12 = 0
(c) 5, 0
Solution:
Given roots are
x = 5 and x = 0
x+5=0
x6=0

the required equation is

(x + 5) (x 6) = 0
x2 x 30 = 0
1
(d) 5,
2
Solution:
1
x = 5 and
x=
2
x5=0
2x 1 = 0

the required equation is

11

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Emporium Career Maths Series

(x 5) (2x 1) = 0
2x2 x 10x + 5 = 0
2x2 11x + 5 = 0
2 4
,
3 5
Solution:
Given roots are
2
4
x=
and x =
3
5
3x 2 = 0
5x + 4 = 0

the required equation is

(3x 2) (5x + 4) = 0
15x2 + 12x 10x 8 = 0
15x2 + 2x 8 = 0
7 5
(f)
,
8 8
Solution:
Given roots are
7
5
x=
and
x=
8
6
8x = 7
6x = 5
8x + 7 = 0
6x 5 = 0
So our required equation is
(8x + 7) (6x 5) = 0
2
48x 40x + 42x 35 = 0
48x2 + 2x 35 = 0
1 3
(g) 2 , 1
2 4
Solution:
Given factors
1
3
x = 2
and
x=1
2
4
(e)

Emporium Career Maths Series

5
7
, 2xn = 5
x=
x,n = 7
2
4
2x + 5 = 0
4x 7 = 0
So our required equation is
(2x + 5) (4x + 7) = 0
2
8x + 14x + 20x + 35 = 0
8x2 + 34x + 35 = 0
1 2
(h) 1 ,
2 3
Solution:
Given factors
1
2
x = 1
and x = , 3n = 2
2
3
3
x=
, 2n = 3
3x = 2
2
2x + 3 = 0
3x 2 = 0
So our required equation is
(2x + 3) (3x 2) = 0
6x2 4x + 9x 6 = 0
6x2 + 5x 6 = 0
4 4
(i)
,
7 7
Solution:
Given the factors
4
4
x=
and
x=
7
7
7x + 4 = 0
7x + 4 = 0
So our required equation is,
(7x + 4) (7x + 4) = 0
49x2 + 28x + 28x + 16 = 0
49x2 + 56x + 16 = 0
x=

13

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EXERCISE No. 1B
Q.1. Solve the following equations, giving your answers correct
to 2 decimal places where necessary.
(a) (x + 1)2 = 9
Solution:
(x + 1) = 9
Taking the square root of each side.
Either,
x+1 =3
x =2
or
x+1 =3
x =4
2
(b) (2x + 1) = 16
Solution:
(2x + 1) = 16
Taking the square root of each side.
Either,
2x + 1 = 4
3
x =
2
x = 1.5
or
2x + 1 = 4
5
x =
2
x = 2.5
(c) (3x + 2)2 = 49
Solution:
(3x + 2) = 49
Taking the square root of each side.
Either.
3x + 2 7

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5
3
x = 1.67
x =
or
3x + 2 = 7
x =3
2
29
3

(d) 2x + =
45
4

Solution:
Taking the square root of each side.
Either,
3
7
2x +
=
4
5
7 3
2x =

5 4
28 15
2x =
20
13
x =
40
x = 0.33
or
3
7
2x +
=
4
5
7
3
2x =

5
4
28 15
2x =
20
43
x =
40

x = 1.08
(e) (5x 4) = 81
Solution:
2

(5x 4) = 81

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Emporium Career Maths Series

Taking square root of each side.

Either.
5x 4 = 9
13
x =
5
x = 2.60
or
5x 4 = 9
5
x =
5
x =1
25
(f) (3x +7)2 =
49
Solution:
Taking square root of each side.
Either,
5
3x + 7 =
7
5
3x =
7
7
5 49
3x =
7
44
x =
21
x = 2.10
or
5
3x + 7 =
7
5
3x =
7
7
5 59
3x =
7

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Emporium Career Maths Series

54
21
x = 2.57
x =
(g)

(x-4)2 = 17

(x 4)2
Taking square rot of each side.
Either,
x5
x
or
x4
x
2
(h) (x + 3) = 11
Solution:
(x + 3y)
Taking square root of each side.
Either,
x+3
x
or
x+3
x
2
(i)
(2x 3) = 23
Solution:
(2x 3)
Taking square root of each side.
Either,
2x 3
x
or
2x 3
x
2
(j) (3x +2) = 43
Solution:

=7

= 4.12
= 8.12
= 4.12
= 0.12

= 11

= 3.32
= 0.32
= 3.32
= 6.32

= 23

= 4.80
= 3.90
= 4.80
= 0.90

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Emporium Career Maths Series

(3x + 2) = 43
Taking square root of each side.
Either,
3x + 2 = 6.65
x = 1.32
or
3x + 2 = 6.56
x = 2.85
2
(k) (5x 7) = 74
Solution:
(5x 7) = 74
Taking square root of each side.
Either,
5x 7 = 8.6
x = 3.12
or
5x 7 = 8.6
x = 0.32
2
(l)
(3 + 7x) = 65
Solution:
(3 + 7x)2 = 65
Taking square root of each side.
Either,
3 + 7x = 8.06
7x = 8.06
3
7x = 5.06
x = 0.72
or
3 + 7x = 8.06
7x = 8.06 3
7x = 11.06
x = 1.38
Q.2 What number must be added to each of the following
expressions to obtain a perfect square?

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Emporium Career Maths Series

(a) x2 + 7x
Solution:
Given expression is x2 + 7x
The coefficient of x is 7.
So square of half of will be added both side.
7
x + 7x +
2

= x +
2

7
must be added
2
2
(b) x 3x
Solution:
Given expression is x2 3x
The coefficient of x is 3
So square of half of 3 will be added both side.

3
x2 3x +

= x
2

must be added
2
7
(c) x2 + x
2
Solution:
7
Given expression is x2 + x
2
7
The coefficient of x is
2
So square of half of 7 2 must be added.

2
7
7
x + x+
2
2
2

2
49
7
must be added.
=
16
2

= x +
2

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Emporium Career Maths Series

(d) x2 1.8x
Solution:
Given expression is x2 1.8x
The coefficient of x is 1.8 =

9
5

So square of half of = 9 5 will be added both side.

x2

9
9
x+

5
10

= x

10

81
must be added.
100
(e) a2 + 2.4a
Solution:
Given the expression is a2 + 2.4n
12
The coefficient of a is 2.4 =
5
12
So square of half of
will be added both side.
5

x2 +

2
12
6
a+
5
5

= a +
5

36
must be added.
25
2
(f) c2 + 4 c
3
Solution:

Given the expression is c2 + 4

2
3

14
.
3
So square of half of 14 3 will be added both side.
The coefficient of c is

c2 +

2
14
7
c+
3
3

= c +
3

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Emporium Career Maths Series

2
49
7
must be added.
=
9
3
4
(g) y2 + y
5
Solution:
4
Given the express in is y2 = y
5
4
The coefficient of y is .
5
So squared half of 4 5 will be added both side.

y2 +

2
y
2
y+
5
5

= y +
5

4
must be added.
25
1
(h) v2 3 v
2
Solution:

Given expression is v2 3
The coefficient of v is
p2

1
v
2

7
2
2
7
7
v+

2
4

= v2
4

2
49
7

must be added.

or
16
4
(i)
b2 10kb
Solution:
Given the expression is b2 10kb

10k
= 5k.
2
= (b 5k)2

The coefficient of b is 1-k. Half of this is

b2 10kb + (5k)2

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Emporium Career Maths Series

25k2 must be added.

(j) g2 5kg
Solution:
Given expression is g2 5kg
The coefficient of g is 5k. Half of this is
5k
g2 5kg +

5k
.
2

5k

= y
2

25k 2
must be added.
4
(k) h2 + 3mh
Solution:
Given expression is h2 +3mh

The coefficient of h is 3 m. Half of this is

3m
h2 + 3mh +

3m
.
2

3m

= h +

9m 2
must be added.
4
1
(l)
k2 1 k
3
Solution:
1
Given expression is k2 1 k.
3
4
2
The coefficient of k is
. Half of this is
3
3

4 2 2
k
+

3 3
2

4
must be added.
9
(m) d2 + 10xd
Solution:

= k
3

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Emporium Career Maths Series

Given expression is a2 + 10xd

The coefficient of d is 10x. Half of this is 5x.
D2 + 10xd + (5x)2 = (d + 5x)2

(5x)2 = 25x2 must be added.

(n) h2 5xk
Solution:
Given expression is k2 5xk
This is a quadratic expression in k.
5x
The coefficient of k is 5x. Half of this is
.
2
5x
k 5xk +

2
2

5x

= x
2

25x 2
must be added.
4
(o) m2 5n2m
Solution:
Given expression is m2 5n2m
This is a quadratic expression in m.

The coefficient of m is 5n2. Half of this is

5n 2

m 5n m +
2

5 2
n.
2

5n 2
= m

25n 4
must be added.
4

EXERCISE NO.1C
Q.

Solve the following equations by factorisation where

possible or by completing the square. If the answers involve
decimal places, give them correct to 2 decimal places. If an
equation has no real roots, indicate that this is so.
(a) x2 + 2x + 3 = 0
Solution:
x2 + 2xz + 3 = 0

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Emporium Career Maths Series

x2 + 2x = 3
x2 + 2x + (1)2 = 3 + (1)2
(x + 1)2 = 2
Either,
2

x+1 =+
or

x+1 = 2
Hence, roots of x + 2x + 3 = 0 are complex.
(b) 5x2 4x 2 = 0
Solution:
5x2 4x 2 = 0
4
2
x2 x
=0
5
5
4
2
x2 x =
5
5
2

x2

2
4
2
x+
5
5

x
5

2 2 2
+
5 5

14
15

Either,
x

2
5

14
25

x 0.4 = 0.75
x = 1.15
or
x

2
5

14
25

x 0.4 = 0.75
x = 0.35
(c) 2x2 + 7x + 2 = 0
Solution:
2x2 + 7x + 2 = 0

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Emporium Career Maths Series

7
x+1 =0
2
7
x2 + x = 1
2
7
x2 + x = 1
2

x2 +

2
7
7
x + x+
2
4
2

x +
3

7
= 1 +
4

33
16

Either,
7
33
=
4
16
x + 1.75 = 1.44
x = 0.31
x+

or
x+

7
4

33
16

x + 1.75 = 1.44
x = 3.19
(d) 4x2 = 5x 21
Solution:
4x2 = 5x 21
5
x
x2 = x
4
4
5
21
x2 x =
4
4
2
5
5
x x+
4
8
2

x
8

21 5 2
=
+
4
8

311
64

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Emporium Career Maths Series

Either,
x

5
8

=+

311
64

5
8

311
64

or

Hence, roots of 4x2 = 5x 21 are complex.

(c) 2x2 + 5x 3 = 0
Solution:
2x2 + 5x 3 = 0
5
3
x2 + x
=0
2
2
5
3
x2 + x =
2
2
x2 +

5

4

x +
4

5
x+
2

3 5 2
+
2 4

49
16

Either,
x+

5
4

7
4
2
x =
4
1
x =
2

x+

5
4

or

(f)

3x2 + 8x + 2 = 0

7
4
12
x =
4
x =3
=

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Emporium Career Maths Series

Solution:
3x2 + 8x + 2 = 0
8
2
x2 + x +
=0
3
3
8
2
x2 + x =
3
3
x2 +

4

3

x +
3

8
x+
3

2
2
4
+
3
3

10
9

Either,
x+

4
3

10
9

x = 1.054 1.33
x = 0.28
or
x+

4
3

10
9

x = 1.054 1.33
x = 2:39
(g) 7x2 28x + 15 = 0
Solution:
7x2 28x + 15 = 0
15
x2 4x +
=0
7
15
x2 4x =
7
15
x2 4x + (2)2 =
+ (2)2
7
17
(x 2)2 =
7
Either,

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Emporium Career Maths Series

13
7

x2 =

x 2 = 1.36
x 2 = 3.36
or
13
7

x2 =

x 2 = 1.36
x = 0.64
(h) 5x2 + 12x + 3 = 0
Solution:
5x2 + 12x + 3 = 0
12
3
x2 +
x+
=0
5
5
12
3
x2 +
x =
5
5
x2 +

2
12
6
x+
5
5

x +
5

3 6 2
+
5
5

21
25

Either,
x+

6
5

21
25

x + 1.10 = 0.918
x = 0.918 1.10
x = 2.02
(i)
2x2 + 3x 4 = 0
Solution:
2x2 + 3x 4 = 0
3
x2 + x 2 = 0
2

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Emporium Career Maths Series

x2 +
x2 +

3
x =2
2

3

4

x +
4

3
x+
2

3
=2+
4
=

41
16

Either,
3
41
=
4
16
x + 0.75 = 1.6
x = 0.85
x+

or
x+

3
4

41
16

x + 0.75 = 1.6
x = 2.35
(j) x2 12x + 36 = 0
Solution:
x2 12x + 36
(x)2 2(x) (6) + (6)2
(x6)2
(x6) (x6)
x
So, x may be 6 repeated.
(k) 5x2 + 30x 18 = 0
Solution:
5x2 + 30x 18
18
x2 + 6x
5

=0
=0
=0
=0
=6

=0
=0

x2 + 6x =

18
5

or 6

30

Emporium Career Maths Series

18
+ (3)2
5
63
=
5

x2 + 6x + (3)2 =
(x+3)2
Either,

63
5
x + 3 = 3.55
x = 0.55
x+3 =

or
x+3 =

63
5

x + 3 = 3.55
x = 6.55
(l)
3x2 4x + 7 = 0
Solution:
3x2 4x + 7 = 0
4
4
x2 x +
=0
3
3
4
7
x2 x =
3
4
2 2 2
(x)2 2(x) +
3 3
2

x
3

x
3

7 2 2
+
3 3

7
4
+
3
9

21 + 4
9

17
9

x
3

Taking square root of each side.

31

Emporium Career Maths Series

2
3

17
9

Hence, root of 3x2 4x + 7 = 0 are complex.

(m) 3x2 + x 2 = 0
Solution:
3x2 + x 2 = 0
x
2
x2 +

=0
3
3
x
2
x2 +
=
3
3
x2 +

2
x
1
+
3
6

x +
6

2 1 2
+
3 6

25
36

Either,
x+

1
6

5
6
4
x =
6
2
x =
3

x+

1
6

or
5
6
6
x =
6
x = 1
=

(n) x2 11x 26 = 0
Solution:
x2 11x 26 = 0
x2 11x = 26

32

Emporium Career Maths Series

11
x2 11x +
2

11

x
2

11
= 26 +
2

225
4

Either,
x

11
2

15
2
26
x =
2
x = 13

11
2

or
15
2
4
x =
2
x =2
=

(o) 3x2 + 5x 2 = 0
Solution:
3x2 + 5x 2 = 0
5
2
x2 + x
=0
3
3
5
2
x2 + x =
3
3
x2 +

5

6

x +
6

5
x+
3

2 5 2
+
3 6

49
36

Either,
x+

5
6

7
6
2
x =
6
=

33

Emporium Career Maths Series

x =

1
3

or
x+

7
6
12
x =
6
x =2

5
6

(p) 2x 3x2 4 = 0
Solution:
2x 2x2 4
3x2 2x + 4
2
4
x2 x +
3
3
2
x2 x
3
1 1
(x)2 2(x) +
3 3

=0
=0
=0
=

4
3

4 1 2
+
3 3

4
1
+
3
9

12 + 1
9

20
9

x
3

x
3

x
3

Taking square root of each side.

x

1
3

Hence, roots of 2x 4 = 0 are complex.

(q) x2 7x 30 = 0
Solution:
x2 7x 30 = 0

20
9

34

Emporium Career Maths Series

x2 7x = 20
7
x2 7x +
2

4
2

7
= 30 +
2
=

169
4

Either,
x

7
2

13
2
20
x =
2
x = 10

7
2

or
13
2
6
x =
2
x =3

(r) (2x+3) (x2) x(x+1) = 0

Solution:
(2x + 3) (x 2) x (x + 1)
(2x2 x 6) (x2 + x)
x2 2x 6
x2 2x
x2 2x + (1)2
(x 1)2
Either,

=0
=0
=0
=6
= 6 + (1)2
=7

x1 =

1 = 2.65
x = 3.65
or
x =

x 1 = 2.65

35

Emporium Career Maths Series

x = 1.65
(s) x 6x 16 = 0
Solution:
2

x2 6x 16
x2 6x
x2 6x + (3)2
(x 3)2

=0
= 16
= 16 + (3)2
= 25

Either,
x3 =5
x =8
or
x3 =5
x =2
(t)

2x2 x 3

1
=0
2

Solution:
1
2
7
2x2 x
2
x
7
x2

2
2
x
x2
2

2x2 x 3

x2

2
x
1
+
2
4

x
4

=0
=0
=0
=

7
4

7 1 2
+
4 4

29
16

Either,
x

1
4

29
16

x 0.25 = 1.35

36

Emporium Career Maths Series

x = 1.60
or
x

1
4

29
16

x 0.25 = 1.35
x = 1.10
(u) x2 16x 10 = 0
Solution:
x2 16x 10
x2 16x
x2 16x + (8)2
(x 8)2

=0
= 10
= 10 + (8)2
= 74

Either,
x8 =

74

x 8 = 8.60
x = 16.60
or
x8 =

74

x 8 = 8.60
x = 0.60
2
(v) (2x + 1) (5x 4) (3x 2) = 0
Solution:
(2x + 1) (5x 4) (3x 2)2 = 0
(10x2 3x 4) (9x2 12x + 4) = 0
x2 + 9x 8 = 0
x2 + 9x = 8

Either,

9
x2 + 9x +
2

x +
2

9
=8+
2
=

113
4

37

Emporium Career Maths Series

9
113
=
2
4
x + 4.5 = 5.32
x = 0.82
x+

or
x+

9
2

113
4

x + 4.5 = 5.32
x = 9.82
(w) x2 2x 5 = 0
Solution:
x2 2x 5
x2 2x
x2 2x + (1)2
(x 1)2

=0
=5
= 5 +(1)2
=6

Either,
x1 =

x 1 = 2.45
x =3.45
or
x1 =

x 1 = 2.45
x = 1.45
(x) 5x2 8x 30 = 0
Solution:
5x2 8x 30 = 0
x
x2
6 =0
5
x
x2 x = 6
5
3
x
4
x
+
5
5
2

4
= 6+
5

38

Emporium Career Maths Series

x
5

166
25

Either,
x

4
5

166
25

x 0.8 = 2.58
x = 3.38
or
x

4
5

166
25

x 0.8 = 2.58
x = 1.78
(y) 4x(3x 1) 2 = (2x 1) (5x + 1)
Solution:
4x(3x 1) 2 = (2x 1) (5x + 1)
12x2 4x 2 = 10x2 3x 1
12x2 4x 2 + 3x + 1 = 0
2x2 x 1 = 0
x
1
x2

=0
2
2
x
1
x2
=
2
2
x2

2
x
1
+
2
4

x
4

1 1 2
+
2 4

9
16

Either,
x

1
4

3
4
4
x =
4
x =1
=

39

Emporium Career Maths Series

or
x

3
4
2
x =
4
1
x =
2

1
4

(z) 5x2 16x + 2 = 0

Solution:
5x2 16x + 2 = 0
16
2
x2
x+
=0
5
5
16
2
x2
x =
5
5
2
16
8
x
x+
5
5
2

x
5

2
2
8
=
+
5
5

54
25

Either,
x

8
5

54
25

x 1.6 = 1.47
x = 3.07
or
x

8
5

54
25

x 1.6 = 1.47
x = 0.13
__________