Exercise No 1.3 (Solutions Exercise) Math For Studens.

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Equations Reducible to Quadratic Form

Exercise 1.3
Type (i) The equation of the types

ax 4 bx 2 c 0

Replacing x 2 y and x4 y2 in equation Solve the following equation.

i) 2 x 4 11x 2 5 0
Reducing quadratic form
ax 4 bx 2 c 0 , we get a quadratic equation in
x2 y , x4 y2
y. 2 y 2 11 y 5 0
a2 , b 11 , c5
Example 1: By quadratic formula
b b 2 4ac
Solve the equation x
2a
By putting value
x 4 13 x 2 36 0
(11) (11) 2 4(2)(5)
y
Solve: 2(2)
11 121 40
Let: x 2 y, then x4 y2 y
4
11 81 11 9
Equation (i) become y y
4 4
y 2 13 y 36 0 , Which can be factorized as, 11 9 11 9
y , y
4 4
y 2 9 y 4 y 36 0 20 2
y , y
4 4
y ( y 9) 4( y 9) 0
1
y5 , y
( y 9)( y 4) 0 2
Putting x 2 y
Either: 1
x2 5 , x2
2
y 9 0 , y40 Taking square root
1
y9 , y4 x2 5 , x2
2
1
Put y x 2 x 5 , x
2
x2 9 , x2 4
1
S .S 5, Ans
Taking under root on both sides 2

x2 9 , x2 4

x 3 , x 2

S .S 3, 2 Ans
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Matric Math Solution Ex 1.3 MATH FOR STUDENTS
ii) 2 x4 9 x2 4 iv) x
2
3
54 15 x
1
3

Solve: Solve:
2 x4 9 x2 4 0 2 1
Reducing quadratic form x 3
15 x 3
54 0
x2 y , x4 y2 According to type (i) ax bx 2 c 0 4

2 y2 9 y 4 0 Let
a2 , b 9 , c4 x
1
3
y
By quadratic formula 2
b b 2 4ac x 3
y2
x
2a y 2 15 y 54 0
Do yourself same as above question a 1 b 15 c 54
1 1
iii) 5x 7 x 2 2 4 By quadratic formula
According to type (i) ax 4 bx 2 c 0 b b 2 4ac
Let y
1 2a
x 4
y
(15) (15)2 4(1)(54)
1
x 2y 2 y
2(1)
5 y2 7 y 2 0
15 225 216
a 5, b 7, c2 y
2
By quadratic formula
15 9 15 3
b b 2 4ac y , y
y 2 2
2a
15 3 15 3
(7) (7) 2 4(5)(2) y , y
y 2 2
2(5) 18 12
y , y
7 49 40 2 2
y
10 y 9 , y6
7 9 73
y y 1
10 10 Put y x 3

73 73
y y
1 1
, x 3 9 , x 3 6
10 10
10 4 Taking cube root on both sides
y y
(6)
, 1 3 1 3
10 10 x 3
(9)3 , x 3 3

2
y 1 , y
5 x 729 , x 216

Put y x
1
4 S .S 729 , 216 Ans
1 1 2 v) 3x 2 5 8 x 1
x 4
1 , x 4

5 Solve:
Taking power root on both side
3 x 2 8 x 1 5 0
x (1)
4
1 4
2
1 4
According to Type (i) ax 4 bx 2 c 0
4 4
, x 4

5 Let:
16
x 1 , x x 1 y x 2 y 2
625
3y2 8 y 5 0
16 a3 b 8 c5
S .S 1 , Ans
625 According to quadratic formula

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Matric Math Solution Ex 1.3 MATH FOR STUDENTS
b b 2 4ac 42
y y
2a 2
Put the values 42 42
y , y
(8) (8) 2 4(3)(5) 2 2
y 6 2
2(3) y , y
2 2
8 64 60 y 3 y 1
y ,
6
Put 2 x 2 1 y
8 4 8 2
y y 2 x2 1 3 , 2 x2 1 1
6 6
8 2 82 2 x2 3 1 , 2 x2 1 1
y y
6 6 2 x2 2 , 2x2 0
10 6 2
y y x2 , x2 0
6 6 2
5 x2 1 , x2 0
y y 1
3 Taking square root on both sides
Put x 1 y x2 1 , x2 0

x 1
5
, x 1 1 x 1 , x0
3 S .S 1 , 0 Ans
1 5 1
, 1 x x 3
vii) 4 4
x
x 3 x
x 3
3
x , x 1 Solve:
5 Let
3 x x3
S .S , 1 Ans y y
5 x3 x
1
vi) 2x 2
1
2x 1
3
2
4 y 4 4
y
Solve: 4 y2 4
y 4 4
2x 2
1
3
2x 1
2
4 ........ (i y y
Let y2 4 4 y y2 4 y 4 0
2x 2
1 y a 1
By quadratic formula
b 4 c4

Put the value in equation (i)

b b 2 4ac
3 y2 3 y
y 4 4 2a
y y
(4) (4) 2 4(1)(4)
y2 3 4 y y2 4 y 3 0 y
2(1)
a 1 b 4 c3
By quadratic formula 4 16 16 4 0
y y
b b 2 4ac 2 2
y 40 40
2a y y
2 2
(4) (4) 2 4(1)(3)
y 4 4
2(1) y y
2 2
4 16 12 4 4 y2 y2
y y
2 2

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Matric Math Solution Ex 1.3 MATH FOR STUDENTS
x xa xa 7
Put y xi)
x3 x a x a 12
x x Same as question (viii) (Do yourself)
2 , 2
x3 x3
x 2x 6 , x 2x 6 Type (iii) Reciprocal Equation
x6 0 , x6 0
x6 x6 1 1
, a x2 2 b x c 0
x x
S .S 6 , 6 Ans
ax bx cx bx a 0
4 3 2

4x 1 4x 1 1
viii) 2 Example:
4x 1 4x 1 6 2 x 4 5 x3 14 x 2 5 x 2 0
Solve:
Dividing by x 2
4x 1 4x 1 1
y 2 x 4 5 x 3 14 x 2 5 x 2
4x 1 4x 1 y 2 2 2 2 0
x2 x x x x
1 13 y 2 1 13
y 5
2 x 2 5 x 14 2 0
2
y 6 y 6 x x
6 y 6 13 y
2
6 y 13 y 6 0
2
2 5
2 x 2 2 5 x 14 0
a6 b 13 c6 x x
By quadratic formula 1 1
2 x 2 2 5 x 14 0 ......... (i
b b 2 4ac x x
y
2a Let
(13) (13) 2 4(6)(6) y x
1
y x
2(6)
1 1
13 169 144 13 25 x2 2 2 y 2 x2 2 y 2 2
y y x x
12 12 Putting value in equation (i)
13 5 2( y 2 2) 5 y 14 0
y
12
13 5 13 5 2 y 2 4 5 y 14 0
y , y 2 y 2 5 y 18 0
12 12
18 8 a2 b 5 c 18
y , y By quadratic formula
12 12
3 2 b b 2 4ac
y , y y
2 3 2a
4x 1 (5) (5) 2 4(2)(13)
Put y y
4x 1 2(2)
4x 1 3 4x 1 2 5 25 144 5 169
, y y
4x 1 2 4x 1 3 4 4
8 x 2 12 x 3 , 12 x 3 8 x 2 5 13
12 x 8 x 2 3 , 12 x 8 x 2 3 y
4
4x 5 , 4 x 5 5 13 5 13
y , y
5 5 4 4
x , x
4 4 18 8
y , y
4 4
5 5 9
S .S , Ans y , y 2
4 4 2

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Matric Math Solution Ex 1.3 MATH FOR STUDENTS
1 1 2
Put y x x2 2
2x 2 0
x x x
x
1 9
x
1
2
1 1
, x2 2 2 x 2 0
x 2 x x x
x2 1 9 x2 1 Let
, 2
x 2 x 1
y x
2x 2 9x
2
, x 2 1 2 x x
2x 9x 2 0
2
, x2 2 x 1 0 Taking square on the both sides
2
Taking 1
2x 9x 2 0
2 ( y)2 x
x
a2 b 9 c2
1
Using quadratic formula y2 x2 2 2
x
b b 2 4ac
y 1
2a y 2 2 x2 2
x
(9) (9) 2 4(2)(2)
y By putting the values
2(2) y2 2 2 y 2 0
9 81 16 9 65 y2 2 y 0 y ( y 2) 0
y y
4 4 y0 , y20
Now taking 1 1
x 0 , x 20
x2 2 x 1 0 x x
a 1 b2 c 1 x 1
2
x 1
2

b b 4ac2 0 , 2
y x x
2a x2 1 0 , x2 1 2 x
2 (2) 4(1)(1)
2
x2 1 , x2 2x 1 0
y
2(1) x2 1 , x2 2x 1 0
2 4 4 2 0 x 1 x2 2x 1 0
y y ,
2 2 Now taking
2 0 2 0 x2 2x 1 0
y , y
2 2 a 1 b 2 c 1
2 2
y , y
2 2 b b 2 4ac
y 1 , y 1 x
2a
9 65 (2) (2) 2 4(1)(1)
S .S 1 , Ans x
4 2(1)
2 44 2 8
x) x 2x 2x 2x 1 0
4 3 2 x x
2 2
Solve:
x 4 2 x3 2 x 2 2 x 1 0 x
22 2
x

2 1 2
Dividing by x 2 2 2
x 4 2 x3 2 x 2 2 x 1 x 1 2
2 2 2 0
x2 x2 x x x
2 1
x2 2x 2 2 0
S .S 1 , 1 2 Ans
x x
Arranging the equation

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Matric Math Solution Ex 1.3 MATH FOR STUDENTS
xi) 2 x 4 x3 6 x 2 x 2 0 xvi) 2 x 64.2 x 20 0
Do yourself. Same as question (x) Solve:
xii) 4.22 x 1 9.2 x 1 0 64
2 x x 20 0
Solve: 2
4.22 x 1 9.2 x 1 0 Let:
4 22 x 21 9.2 x 1 0 y 2x
4 2 x 21 9.2 x 1 0 y 2 64
2
64
y 20 0 20
8 2
x 2 y y
9.2 1 0
x

y 64 20 y
2
y 2 20 y 64 0
Let
a 1 b 20 c 64
2x y
b b 2 4ac
8 y2 9 y 1 0 y
2a
a 8 b 9 c 1
By quadratic formula (20) (20) 2 4(1)(64)
y
b b 2 4ac 2(1)
y
2a 20 400 256 20 144
y y
(9) (9) 2 4(8)(1) 2 2
y 20 12
2(8) y
2
9 81 32 9 49 20 12 20 12
y y y , y
16 16 2 2
97
y y
32
, y
8
16 2 2
97 97
y , y y 16 , y4
16 16
16 2 Put y 2x
y , y
16 16 2 x 16 , 2x 4
1 (2) x (2) 4 , (2) x (2) 2
y 1 , y
8 x4 , x2
Put 2x y S .S 2 , 4 Ans
1 xv) ( x 1)( x 3)( x 5)( x 7) 192
2x 1 , 2x
8 Solve:
1 By arranging
2 2 2x
x 0
, ( x 1)( x 5)( x 3)( x 7) 192
2
3

( x 2 5 x x 5)( x 2 7 x 3x 21) 192

2 1 2x 2
x 0 3
,
( x 2 4 x 5)( x 2 4 x 21) 192
x0 , x 3 Let:
S .S 0 , 3 Ans y x2 4 x
xiii) 32 x 2 12.3x 3 ( y 5)( y 21) 192
Solve: y 2 21y 5 y 105 192
32 x 2 12.3x 3 0
y 2 26 y 105 192 0
32 x.32 12.3x 3 0
y 2 26 y 87 0
3
x 2
.9 12.3 3 0
x
a 1 b 26 c 87
Let b b 2 4ac
y3 x y
2a
Do yourself same as (xii)

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Matric Math Solution Ex 1.3 MATH FOR STUDENTS
(26) (26) 2 4(1)(87) xvi) ( x 1)( x 2)( x 8)( x 5) 360 0
y
2(1) Solve:
26 676 348 26 1024 ( x 1)( x 2)( x 8)( x 5) 360 0
y y
2 2 ( x 2 2 x x 2)( x 2 5 x 8 x 40) 360 0
26 32
y ( x 2 3x 2)( x 2 3x 40) 360 0
2
26 32 26 32 Let:
y y
2 2 y x 2 3x
58 6 ( y 2)( y 40) 360 0
y y
2 2
y 2 40 y 2 y 80 360 0
y 29 y 3
Put y x2 4 x y 2 38 y 280 0
x 2 4 x 3 x2 4 x 3 0 a 1 b 38 c 280
a 1 b 4 c3 b b 2 4ac
y
(4) (4) 2 4(1)(3) 2a
y
2(1)
(38) (38) 2 4(1)(280)
4 16 12 4 4 y
y y 2(1)
2 2
42 38 1444 1120
y y
2 2
42 42 38 324 38 18
y , y y y
2 2 2 2
6 2
y , y 38 18 38 18
2 2 y , y
2 2
y3 , y 1
56 20
As y 29 y , y
2 2
x 4 x 29
2
x 2 4 x 29 0 y 28 , y 10
a 1 b 4 c 29
(4) (4) 2 4(1)(3) Put y x 2 3x
y
2(1) x 2 3x 28 , x 2 3x 10
(4) (4) 2 4(1)(29) x 2 3x 28 0 , x 2 3x 10 0
y
2(1) Now take x 2 3x 28 0
4 16 116 4 132 a 1 b 3 c 28
y y
2 2
(3) (3) 2 4(1)(28)
4 16 116 4 33 4 y
y y 2(1)
2 2

y
4 2 33
y

2 2 33 y
3 9 112
y
3 121
2(1) 2
2 2
y 2 33 3 11
y
2

S .S 1, 3, 2 33 Ans y
3 11
2
, y
3 11
2
y7 , y 4

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Matric Math Solution Ex 1.3 MATH FOR STUDENTS
Now take
x 2 3x 10 0
a 1 b 3 c 10

3 9 40 3 49
y y
2 2

3 7
2
3 7
y y
2 2

3 7 37
y , y
2 2

10 4
y , y
2 2

y5 , y 2

S .S 5 , 2, 5, 2 Ans

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