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Chapter No. 1
SOLUTION OF QUADRATIC EQUATION
EXERCISE 1AQ.1 Solve the following quadratic equations by
factorisation:(a) x2 5x = 0Solution:
x2 5x = 0Take x as common
x(x 5) = 0Either
x = 0or
x 5 = 0x = 5
So, x = 0 or x = 5(b) 4x2 = 7x Solution:
4x2 = 7x 4x2 7x = 0
Take x as commonx(4x 1) = 0
Eitherx = 0
or4x 7 = 0
x = 47
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2 Emporium Career Maths Series
x = 143
So, x = 0, or x = 143
(c) 6t2 = t(6 4)Solution:
6t2 = t(t 4)6t2 t(t 4)
6t2 t2 + 4t = 05t2 + 4t = 0
t(5t + 4) = 0Either
t = 0or
5t + 4 = 0
t = 54
So, t = 0, or to = 54
(d) 5y2 = y(y + 3)Solution:
5y2 = y(y + 3)5y2 = y2 + 3y5y2 y2 3y = 0
y(4y 3) = 0Either
y = 0or
4y 3 = 0
y = 43
So, y = 0 or y = 43
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Emporium Career Maths Series 3
(e) a2 + 9a = 0Solution:
a2 + 9a = 0a(a+9) = 0
Eithera = 0
ora + 9 = 0
a = 9So, a = 0 or 9
(f) 3h2 = h(5 2h)Solution:
3h2 = h(5 2h)3h2 (5 2h) = 0 3h2 + 2h2 5h = 0
5h2 5h = 05h(h 1) = 0
Eitherh = 0
orh 1 = 0
h = 1So, h = 0 or 1
(g) x2 2x + 1 = 0Solution:
x2 2x + 1 = 0 x2 x x + 1 = 0
x(x 1) 1 (x 1) = 0(x 1)2 = 0
Eitherx 1 = 0
x = 1or
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4 Emporium Career Maths Series
x 1 = 0x = 1
So, x = 1, 1(h) 7a + a2 18 = 0 Solution:
7a + a2 18 = 0Re-arranging
a2 + 7a 18 = 0a2 + 9a 2a 18 = 0a(a+9) 2(a+9) = 0
(a+9) (a2) = 0Either
a + 9 = 0a = 9
ora 2 = 0
a = 2So, a = 9 or a = 2
(i) 2x2 + 5z 3 = 0Solution:
2z2 + 5z 3 = 02z2 + 6z z 3 = 0
2z(z+3) 1(z3) = 0(z+3) (2z1) = 0
Eitherz + 3 = 0
z = 3or
2z 1 = 0
z = 21
So, z = 3 or z = 21
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Emporium Career Maths Series 5
(j) c2 + 2c = 35 Solution:
c2 + 2c = 35c2 + 2c 35 = 0
c2 + 7c 5c 35 = 0c(c+7) 5(c7) = 0
(c+7) (c5) = 0Either
c + 7 = 0c = 7
orc 5 = 0
c = 5So, c = 7 or c = 5
(k) 8p 16 p2 = 0 8p 16 p2 = 0
Re-arrangingp2 8pr + 16 = 0
p2 4p 4p + 16 = 0p(p4) (p4) = 0
Eitherp 4 = 0
p = 4or
p 4 = 0p = 4
So, p = 4 or 4(l) 4 3b b2 = 0Solution:
4 3b b2 = 0b3 + 3b 4 = 0
b2 + 4b b 4 = 0b(b+4) 1(b+4) = 0
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6 Emporium Career Maths Series
(b+4) (b1) = 0Either
b + 4 = 0b = 4
orb 1 = 0
b = 1So, b = 4 or b = 1
(m) 12 a a2 = 012 a a2 = 0a2 + a 12 = 0
a2 + 4a 3a 12 = 0a(a+4) 3(a+4) = 0
(a+4)(a3) = 0Either
a + 4 = 0a = 4
ora 3 = 0
a = 3So, a = 4, or a = 3
(n) 10t2 t = 2Solution:
10t2 t = 210t2 t 2 = 0
10t2 5t + 4t 2 = 05t(2t1) +2(2t1) = 0
(2t1)(5t+2) = 0Either
2t 1 = 0
t = 21
or
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Emporium Career Maths Series 7
5t + 2 = 0
t = 52
So, t = 21
or = 52
(o) y2 22y + 96 = 0Solution:
y2 22y + 96 = 0y2 16y 6y + 96 = 0y(y16) 6(y16) = 0
(y16) (y6) = 0Either
y 16 = 0y = 16
ory 6 = 0
y = 6So, y = 16 or y = 6
(p) 12a2 16a 35 = 012a2 16a 35 = 0
12a2 30a + 14a 35 = 06a(2a5) +7(2a5) = 0
(2a 5) (6a + 7) = bEither
2a 5 = 0
a = 221
or6a + 7 = 0
6a = 7
a = 67
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8 Emporium Career Maths Series
a = 161
So, a = 221
or a = 161
(q) 15x2 + 4x 35 = 0Solution:
15x2 + 4x 35 = 015x2 + 25x 21x 35 = 0
5x(3x+5) 7(3x+5) = 0(3x+5) (5x7) = 0
Either3x + 5 = 0
x = 35
x = 132
or5x 7 = 0
x = 57
x = 152
So, x = 132
or a = 152
(r) 15x2 + 4x 35 = 0Solution:
28x2 85x + 63 = 028x2 36x 49x + 63 = 0
4x(7x9) 7(7x9) = 0(7x9) (4x7) = 0
Either7x 9 = 0, 7x = 9
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Emporium Career Maths Series 9
x = 79
x = 172
or4x 7 = 0, 4x = 7
x = 47
x = 143
So, x = 172
or x = 143
(s) 56x2 159x + 108 = 0Solution:
56x2 159x + 108 = 056x2 96x 63x + 108 = 08x(7x12) 9(7x12) =
0
(7x12) (8x9) = 0Either
7x 12 = 0 7x = 12
x = 7
12
x = 175
or8x 9 = 0, 8x = 9
x = 89
x = 181
So, x = 175
or x = 181
(t) 39x2 = 131x 44 = 0
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10 Emporium Career Maths Series
Solution:39x2 + 131x 44 = 0
39x2 + 143x 12x 44 = 013(3x11) 4(3x11) = 0
Either3x 11 = 0 3x = 11
x = 332
or13x 4 = 0 13x = 4
x = 134
So, x = 321
or x = 134
(u) 76x 96x2 15 = 0Solution:
76x 96x2 15 = 096x2 76x + 15 = 0
96x2 40x 36x + 15 = 08x(12x5) 3(12x5) = 0
(12x5) (8x3) = 0Either
12x 5 = 0
x = 125
or8x 3 = 0
x = 83
So, x = 125
or 83
Q. Form a quadratic equation in x with the giver roots for each
of the following:
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Emporium Career Maths Series 11
(a) 2, 3Solution:
Given roots arex = 2 and x = 3
x 2 = 0 x 3 = 0 the required equation is
(x 2) (x 3) = 0x2 2x 3x + 6 = 0
x2 5x + 6 = 0(b) 3, 4Solution:
Given roots arex = 3 and x = 4
x3 = 0 x+4 = 0 the required equation is
(x3) (x+4) = 0x2 3x + 4x 12 = 0
x2 + x 12 = 0(c) 5, 0Solution:
Given roots arex = 5 and x = 0
x + 5 = 0 x 6 = 0 the required equation is
(x + 5) (x 6) = 0x2 x 30 = 0
(d) 5, 21
Solution:
x = 5 and x = 21
x 5 = 0 2x 1 = 0 the required equation is
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12 Emporium Career Maths Series
(x 5) (2x 1) = 02x2 x 10x + 5 = 0
2x2 11x + 5 = 0
(e) 32 , 5
4
Solution:Given roots are
x = 32
and x = 54
3x 2 = 0 5x + 4 = 0 the required equation is
(3x 2) (5x + 4) = 015x2 + 12x 10x 8 = 0
15x2 + 2x 8 = 0
(f) 87 , 8
5
Solution:Given roots are
x = 87
and x = 65
8x = 7 6x = 58x + 7 = 0 6x 5 = 0
So our required equation is(8x + 7) (6x 5) = 0
48x2 40x + 42x 35 = 048x2 + 2x 35 = 0
(g) 2 21 , 1 4
3
Solution:Given factors
x = 221
and x = 1 43
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Emporium Career Maths Series 13
x = 25
, 2xn = 5 x = 47
x,n = 7
2x + 5 = 0 4x 7 = 0So our required equation is
(2x + 5) (4x + 7) = 08x2 + 14x + 20x + 35 = 0
8x2 + 34x + 35 = 0
(h) 1 21 , 3
2
Solution:Given factors
x = 121
and x = 32
, 3n = 2
x = 23
, 2n = 3 3x = 2
2x + 3 = 0 3x 2 = 0So our required equation is
(2x + 3) (3x 2) = 06x2 4x + 9x 6 = 0
6x2 + 5x 6 = 0
(i) 74 , 7
4
Solution:Given the factors
x = 74
and x = 74
7x + 4 = 0 7x + 4 = 0So our required equation is,
(7x + 4) (7x + 4) = 049x2 + 28x + 28x + 16 = 0
49x2 + 56x + 16 = 0
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14 Emporium Career Maths Series
EXERCISE No. 1BQ.1. Solve the following equations, giving your
answers correct
to 2 decimal places where necessary.(a) (x + 1)2 = 9
Solution:
(x + 1) = 9Taking the square root of each side.Either,
x + 1 = 3x = 2
orx + 1 = 3
x = 4(b) (2x + 1)2 = 16Solution:
(2x + 1) = 16Taking the square root of each side.Either,
2x + 1 = 4
x = 23
x = 1.5or
2x + 1 = 4
x = 25
x = 2.5(c) (3x + 2)2 = 49Solution:
(3x + 2) = 49Taking the square root of each side.Either.
3x + 2 7
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Emporium Career Maths Series 15
x = 35
x = 1.67or
3x + 2 = 7x = 3
(d)2
432
+x = 4529
Solution:Taking the square root of each side.Either,
2x + 43
= 57
2x = 57
43
2x = 201528
x = 4013
x = 0.33or
2x + 43
= 57
2x = 57
43
2x = 201528
x = 4043
x = 1.08(e) (5x 4)2 = 81Solution:
(5x 4) = 81
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16 Emporium Career Maths Series
Taking square root of each side.Either.
5x 4 = 9
x = 5
13
x = 2.60or
5x 4 = 9
x = 55
x = 1
(f) (3x +7)2 = 4925
Solution:Taking square root of each side.Either,
3x + 7 = 75
3x = 75
7
3x = 7495
x = 2144
x = 2.10or
3x + 7 = 75
3x = 75
7
3x = 7
595
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Emporium Career Maths Series 17
x = 2154
x = 2.57(g) (x-4)2 = 17
(x 4)2 = 7 Taking square rot of each side.Either,
x 5 = 4.12x = 8.12
orx 4 = 4.12
x = 0.12(h) (x + 3)2 = 11Solution:
(x + 3y) = 11Taking square root of each side.Either,
x + 3 = 3.32x = 0.32
orx + 3 = 3.32
x = 6.32(i) (2x 3)2 = 23Solution:
(2x 3) = 23Taking square root of each side.Either,
2x 3 = 4.80x = 3.90
or2x 3 = 4.80
x = 0.90(j) (3x +2)2 = 43Solution:
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18 Emporium Career Maths Series
(3x + 2) = 43Taking square root of each side.Either,
3x + 2 = 6.65x = 1.32
or3x + 2 = 6.56
x = 2.85(k) (5x 7)2 = 74Solution:
(5x 7) = 74Taking square root of each side.Either,
5x 7 = 8.6x = 3.12
or5x 7 = 8.6
x = 0.32(l) (3 + 7x)2 = 65Solution:
(3 + 7x)2 = 65Taking square root of each side.Either,
3 + 7x = 8.067x = 8.06 37x = 5.06x = 0.72
or3 + 7x = 8.06
7x = 8.06 37x = 11.06x = 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 19
(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.
x2 + 7x + 2
27
= 2
27x
+
2
27
must be added
(b) x2 3x Solution:
Given expression is x2 3x The coefficient of x is 3So square of
half of 3 will be added both side.
x2 3x + 2
23
= 2
23x
2
23
must be added
(c) x2 + 27 x
Solution:
Given expression is x2 + 27 x
The coefficient of x is 27
So square of half of 27 must be added.
x2 + 27 x +
2
27
=
+27x
2
27
=
1649
must be added.
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20 Emporium Career Maths Series
(d) x2 1.8x Solution:
Given expression is x2 1.8x
The coefficient of x is 1.8 = 59
So square of half of = 59 will be added both side.
x2 59
x +
10
9=
2
109x
10081
must be added.
(e) a2 + 2.4aSolution:
Given the expression is a2 + 2.4n
The coefficient of a is 2.4 = 5
12
So square of half of 5
12 will be added both side.
x2 + 5
12a +
2
56
= 2
56a
+
2536
must be added.
(f) c2 + 432 c
Solution:
Given the expression is c2 + 432
The coefficient of c is 3
14.
So square of half of 314 will be added both side.
c2 +3
14c +
2
37
= 2
37c
+
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Emporium Career Maths Series 21
2
37
=
949
must be added.
(g) y2 + 54 y
Solution:
Given the express in is y2 = 54
y
The coefficient of y is 54
.
So squared half of 54 will be added both side.
y2 + 5y
y + 2
52
= 2
52y
+
254
must be added.
(h) v2 321 v
Solution:
Given expression is v2 321
v
The coefficient of v is 27
p2 27
v + 2
47
= 22
47v
2
47
or
1649
must be added.
(i) b2 10kbSolution:
Given the expression is b2 10kb
The coefficient of b is 1-k. Half of this is 2
k10 = 5k.
b2 10kb + (5k)2 = (b 5k)2
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22 Emporium Career Maths Series
25k2 must be added.(j) g2 5kgSolution:
Given expression is g2 5kg
The coefficient of g is 5k. Half of this is 2
k5.
g2 5kg + 2
2k5
= 2
2k5y
4k25 2 must be added.
(k) h2 + 3mhSolution:
Given expression is h2 +3mh
The coefficient of h is 3 m. Half of this is 2m3
.
h2 + 3mh + 2
2m3
= 2
2m3h
+
4m9 2 must be added.
(l) k2 131 k
Solution:
Given expression is k2 131
k.
The coefficient of k is 34
. Half of this is 32
k2 34
+ 2
32
= 2
32k
94
must be added.
(m) d2 + 10xdSolution:
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Emporium Career Maths Series 23
Given expression is a2 + 10xdThe 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 5xkSolution:
Given expression is k2 5xkThis is a quadratic expression in
k.
The coefficient of k is 5x. Half of this is 2
x5.
k2 5xk +
2
x5=
2
2x5x
4x25 2 must be added.
(o) m2 5n2mSolution:
Given expression is m2 5n2mThis is a quadratic expression in
m.
The coefficient of m is 5n2. Half of this is 25
n2.
m2 5n2m + 22
2n5
= 22
2n5m
4n25 4 must be added.
EXERCISE NO.1CQ. 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 = 0Solution:
x2 + 2xz + 3 = 0
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24 Emporium Career Maths Series
x2 + 2x = 3x2 + 2x + (1)2 = 3 + (1)2
(x + 1)2 = 2Either,
x + 1 = + 2or
x + 1 = 2Hence, roots of x2 + 2x + 3 = 0 are complex.
(b) 5x2 4x 2 = 0Solution:
5x2 4x 2 = 0
x2 54
x 52
= 0
x2 54
x = 52
x2 54
x + 2
52
=
52
+ 2
52
2
52x
= 1514
Either,
x 52
= 2514
x 0.4 = 0.75x = 1.15
or
x 52
= 2514
x 0.4 = 0.75x = 0.35
(c) 2x2 + 7x + 2 = 0Solution:
2x2 + 7x + 2 = 0
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Emporium Career Maths Series 25
x2 + 27
x + 1 = 0
x2 + 27
x = 1
x2 + 27
x = 1
x2 + 27
x + 2
47
= 1 + 2
47
2
37x
+ = 1633
Either,
x + 47
= 1633
x + 1.75 = 1.44x = 0.31
or
x + 47
= 1633
x + 1.75 = 1.44x = 3.19
(d) 4x2 = 5x 21Solution:
4x2 = 5x 21
x2 = 45
x 4x
x2 45
x = 421
x2 45
x + 2
85
=
421
+ 2
85
2
85x
= 64311
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26 Emporium Career Maths Series
Either,
x 85
= + 64311
or
x 85
= 64311
Hence, roots of 4x2 = 5x 21 are complex.(c) 2x2 + 5x 3 =
0Solution:
2x2 + 5x 3 = 0
x2 + 25
x 23
= 0
x2 + 25
x = 23
x2 + 25
x + 2
45
=
23
+ 2
45
2
45x
+ = 1649
Either,
x + 45
= 47
x = 42
x = 21
or
x + 45
= 47
x = 412
x = 3(f) 3x2 + 8x + 2 = 0
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Emporium Career Maths Series 27
Solution:3x2 + 8x + 2 = 0
x2 + 38
x + 32
= 0
x2 + 38
x = 32
x2 + 38
x + 2
34
=
32
+ 2
34
2
34x
+ = 9
10
Either,
x + 34
= 9
10
x = 1.054 1.33x = 0.28
or
x + 34
= 9
10
x = 1.054 1.33x = 2:39
(g) 7x2 28x + 15 = 0Solution:
7x2 28x + 15 = 0
x2 4x + 7
15= 0
x2 4x = 715
x2 4x + (2)2 = 715
+ (2)2
(x 2)2 = 7
17
Either,
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28 Emporium Career Maths Series
x 2 = 7
13
x 2 = 1.36x 2 = 3.36
or
x 2 = 7
13
x 2 = 1.36x = 0.64
(h) 5x2 + 12x + 3 = 0Solution:
5x2 + 12x + 3 = 0
x2 + 5
12x +
53
= 0
x2 + 5
12x =
53
x2 + 5
12x +
2
56
=
53
+ 2
56
2
56x
+ = 2521
Either,
x + 56
= 2521
x + 1.10 = 0.918x = 0.918 1.10x = 2.02
(i) 2x2 + 3x 4 = 0Solution:
2x2 + 3x 4 = 0
x2 + 23
x 2 = 0
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Emporium Career Maths Series 29
x2 + 23
x = 2
x2 + 23
x + 2
43
= 2 + 2
43
2
43x
+ = 1641
Either,
x + 43
= 1641
x + 0.75 = 1.6x = 0.85
or
x + 43
= 1641
x + 0.75 = 1.6x = 2.35
(j) x2 12x + 36 = 0Solution:
x2 12x + 36 = 0(x)2 2(x) (6) + (6)2 = 0
(x6)2 = 0(x6) (x6) = 0
x = 6 or 6So, x may be 6 repeated.
(k) 5x2 + 30x 18 = 0 Solution:
5x2 + 30x 18 = 0
x2 + 6x 5
18= 0
x2 + 6x = 5
18
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30 Emporium Career Maths Series
x2 + 6x + (3)2 = 5
18 + (3)2
(x+3)2 = 563
Either,
x + 3 = 563
x + 3 = 3.55x = 0.55
or
x + 3 = 563
x + 3 = 3.55x = 6.55
(l) 3x2 4x + 7 = 0Solution:
3x2 4x + 7 = 0
x2 34
x + 34
= 0
x2 34
x = 47
(x)2 2(x)
32
+2
32
=
37
+ 2
32
2
32x
= 37
+ 94
2
32x
= 9
421 +
2
32x
= 917
Taking square root of each side.
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Emporium Career Maths Series 31
x 32
= 917
Hence, root of 3x2 4x + 7 = 0 are complex.(m) 3x2 + x 2 =
0Solution:
3x2 + x 2 = 0
x2 + 3x
32
= 0
x2 + 3x
= 32
x2 + 3x
+ 2
61
=
32
+ 2
61
2
61x
+ = 3625
Either,
x + 61
= 65
x = 64
x = 32
or
x + 61
= 65
x = 66
x = 1(n) x2 11x 26 = 0Solution:
x2 11x 26 = 0x2 11x = 26
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32 Emporium Career Maths Series
x2 11x + 2
211
= 26 + 2
211
2
211x
= 4
225
Either,
x 2
11=
215
x = 2
26
x = 13or
x 2
11=
215
x = 24
x = 2(o) 3x2 + 5x 2 = 0Solution:
3x2 + 5x 2 = 0
x2 + 35
x 32
= 0
x2 + 35
x = 32
x2 + 35
x + 2
65
=
32
+ 2
65
2
65x
+ = 3649
Either,
x + 65
= 67
x = 62
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Emporium Career Maths Series 33
x = 31
or
x + 65
= 67
x = 612
x = 2(p) 2x 3x2 4 = 0Solution:
2x 2x2 4 = 03x2 2x + 4 = 0
x2 32
x + 34
= 0
x2 32
x = 34
(x)2 2(x)
31
+2
31
=
34
+2
31
2
31x
= 34
+ 91
2
31x
= 9
112 +
2
31x
= 920
Taking square root of each side.
x 31
= 920
Hence, roots of 2x 4 = 0 are complex.(q) x2 7x 30 = 0
Solution:
x2 7x 30 = 0
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34 Emporium Career Maths Series
x2 7x = 20
x2 7x + 2
27
= 30 + 2
27
2
274
= 4
169
Either,
x 27
= 2
13
x = 2
20
x = 10or
x 27
= 213
x = 26
x = 3(r) (2x+3) (x2) x(x+1) = 0Solution:
(2x + 3) (x 2) x (x + 1) = 0(2x2 x 6) (x2 + x) = 0
x2 2x 6 = 0x2 2x = 6
x2 2x + (1)2 = 6 + (1)2 (x 1)2 = 7
Either,x 1 = 7
1 = 2.65x = 3.65
orx = 7
x 1 = 2.65
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Emporium Career Maths Series 35
x = 1.65(s) x2 6x 16 = 0Solution:
x2 6x 16 = 0x2 6x = 16
x2 6x + (3)2 = 16 + (3)2
(x 3)2 = 25Either,
x 3 = 5x = 8
orx 3 = 5
x = 2
(t) 2x2 x 321 = 0
Solution:
2x2 x 321
= 0
2x2 x 27
= 0
x2 2x
27
= 0
x2 2x
= 47
x2 2x
+ 2
41
=
47
+ 2
41
2
41x
= 1629
Either,
x 41
= 1629
x 0.25 = 1.35
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36 Emporium Career Maths Series
x = 1.60or
x 41
= 1629
x 0.25 = 1.35x = 1.10
(u) x2 16x 10 = 0Solution:
x2 16x 10 = 0x2 16x = 10
x2 16x + (8)2 = 10 + (8)2 (x 8)2 = 74
Either,x 8 = 74x 8 = 8.60
x = 16.60or
x 8 = 74x 8 = 8.60
x = 0.60(v) (2x + 1) (5x 4) (3x 2)2 = 0Solution:
(2x + 1) (5x 4) (3x 2)2 = 0(10x2 3x 4) (9x2 12x + 4) = 0
x2 + 9x 8 = 0x2 + 9x = 8
x2 + 9x + 2
29
= 8 + 2
29
2
29x
+ = 4
113
Either,
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Emporium Career Maths Series 37
x + 29
= 4
113
x + 4.5 = 5.32x = 0.82
or
x + 29
= 4
113
x + 4.5 = 5.32x = 9.82
(w) x2 2x 5 = 0Solution:
x2 2x 5 = 0x2 2x = 5
x2 2x + (1)2 = 5 +(1)2 (x 1)2 = 6
Either,x 1 = 6x 1 = 2.45
x =3.45or
x 1 = 6x 1 = 2.45
x = 1.45(x) 5x2 8x 30 = 0Solution:
5x2 8x 30 = 0
x2 5x
6 = 0
x2 5x
x = 6
x2 5x
+ 3
54
= 6+ 3
54
-
38 Emporium Career Maths Series
2
54x
= 25
166
Either,
x 54
= 25
166
x 0.8 = 2.58x = 3.38
or
x 54
= 25
166
x 0.8 = 2.58x = 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 = 02x2 x 1 = 0
x2 2x
21
= 0
x2 2x
= 21
x2 2x
+ 2
41
=
21
+ 2
41
2
41x
= 169
Either,
x 41
= 43
x = 44
x = 1
-
Emporium Career Maths Series 39
or
x 41
= 43
x = 42
x = 21
(z) 5x2 16x + 2 = 0Solution:
5x2 16x + 2 = 0
x2 5
16x +
52
= 0
x2 5
16x =
52
x2 5
16x +
2
58
=
52
+ 2
58
2
58x
= 2554
Either,
x 58
= 2554
x 1.6 = 1.47x = 3.07
or
x 58
= 2554
x 1.6 = 1.47x = 0.13
__________
EXERCISE 1AEXERCISE No. 1BEXERCISE NO.1C