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Equations Reducible to Quadratic EquationsExercise 4.2
Solve the following equations:
1. 4 26 8 0x x− + =Solution:
4 26 8 0x x− + =Let 2y x= and 2 4y x=The above equation becomes: 2 6 8 0y y− + =
2
2
6 8 0
4 2 8 0
( 4) 2( 4) 0
( 2)( 4) 0
y y
y y y
y y y
y y
− + =− − + =
− − − =− − =2 0y − = and 4 0y − =2y = 4y =
As, 2y x= 2 2x = 2 4x =
2x = ± 2x = ±solution set = { 2, 2,2, 2}− −
2. 2 110 3x x− −− =Solution:
2 1
2 1
10 3
3 10 0
x x
x x
− −
− −
− =− − =
Let 1y x−= and 2 2y x−=The above equation becomes: 2 3 10 0y y− − =
2 5 2 10 0
( 5) 2( 5) 0
( 2)( 5) 0
y y y
y y y
y y
− + − =− + − =
+ − =2y = − 5y =
As, 1y x−=1 2x− = − 1 5x− =
1 1 1( ) ( 2)x− − −= − 1 1 1( ) (5)x− − −=1
2x = − 1
5x =
solution set = 1 1
{ , }2 5
−
Mathematics Notes for FSc Part-I Prepared by Bilal
3. 6 39 8 0x x− + =Solution:
6 39 8 0x x− + =Let 3y x= and 2 6y x=The above equation becomes: 2 9 8 0y y− + =
2 8 8 0
( 8) 1( 8) 0
( 1)( 8) 0
y y y
y y y
y y
− − + =− − − =
− − =1y = 8y =
As, 3y x=3 1x = 3 8x =
First take: 3 1x =3 1 0x − =
2( 1)( 1) 0x x x− + + =1 0x − = 2 1 0x x+ + =
1x = 1 1 4
2x
− ± −= by using quadratic formula
1 3
2x
− ± −=
1 3
2
ix
− ±=
Now take: 3 8 0x − =3 3( ) (2) 0x − =
2( 2)( 2 4) 0x x x− + + =2 0x − = 2 2 4 0x x+ + =
2x = 2 4 16
2x
− ± −=
2 12
2x
− ± −=
2 2 3
2x
− ± −=
2( 1 3)
2x
− ± −=
1 3x = − ± −
Hence, solution set = 1 3
{1,2, , 1 3}2
i− ± − ± −
Mathematics Notes for FSc Part-I Prepared by Bilal
4. 6 38 19 27 0x x− − =Solution:Let 3y x= and 2 6y x=The above equation becomes: 28 19 27 0y y− − =Use quadratic formula,Here, 8a = , 19b = − and 27c = −
2( 19) ( 19) 4(8)( 27)
2(8)y
− − ± − − −=
19 1225
16y
±=
19 35
16y
±=
19 35
16y
+= 19 35
16y
−=
54
16y = 16
16y
−=
27
8y = 1y = −
As, 3y x=3 27
8x = 3 1x = −
First take: 3 27
8x =
3 270
8x − =
3 33( ) ( ) 0
2x − =
23 3 9( )( ) 0
2 2 4
xx x− + + =
30
2x − = 2 3 9
02 4
xx + + =
3
2x =
23 3 9( ) 4(1)( )
2 2 42(1)
x− ± −
=
3 99
2 42
x− ± −
=
3 272 4
2x
− ± −=
Mathematics Notes for FSc Part-I Prepared by Bilal
3 (9)(3)2 4
2x
− ± −=
3 32 2
2
ix
− ±=
3 33
2 22
x− ± −
=
3( 1 3 )
22
ix
− ±=
3( 1 3 )
4
ix
− ±=
Now take: 1y = −3 1x = −3 1 0x + =
2( 1)( 1) 0x x x+ + + =1 0x + = 2 1 0x x+ + =
1x = − 1 1 4
2x
− ± −= by using quadratic formula
1 3
2x
− ± −=
1 3
2
ix
− ±=
Hence, solution set = 3 1 3 3( 1 3 )
{ 1, , , }2 2 4
i i− ± − ±−
5. 2 1
5 58 6x x+ =Solution:
2 1
5 58 6x x+ =
Let 1
5y x= and 2
2 5y x=The above equation becomes:
2 8 6y y+ =2 6 8 0y y− + =2 4 2 8 0
( 4) 2( 4) 0
( 2)( 4) 0
y y y
y y y
y y
− − + =− − − =
− − =
Mathematics Notes for FSc Part-I Prepared by Bilal
2 0y − = 4 0y − =2y = 4y =
As, 1
5y x=1
5 2x =1
5 4x =1
5 55( ) (2)x =1
5 55( ) (4)x =32x = 1024x =
solution set = {32,1024}
6. ( 1)( 2)( 3)( 4) 24x x x x+ + + + =Solution:
2 2
2 2
( 1)( 2)( 3)( 4) 24
( 1)( 4)( 2)( 3) 24
( 4 4)( 3 2 6) 24
( 5 4)( 5 6) 24
x x x x
x x x x
x x x x x x
x x x x
+ + + + =+ + + + =+ + + + + + =+ + + + =
Let 2 5y x x= +The above equation becomes:
2
2
( 4)( 6) 24
6 4 24 24
10 0
( 10) 0
y y
y y y
y y
y y
+ + =+ + + =+ =
+ =0y = 10 0y + =
As, 2 5y x x= +2 5 0x x+ = 2 5 10 0x x+ + =
( 5) 0x x + = 5 25 4(1)(10)
2(1)x
− ± −= Using Quadratic formula
0x = 5 0x + = 5 15
2x
− ± −=
5x = − 5 15
2
ix
− ±=
Hence, solution set = 5 15
{0, 5, }2
i− ±−
7. ( 1)( 5)( 8)( 2) 880 0x x x x− + + + − =Solution:
2 2
( 1)( 5)( 8)( 2) 880 0
( 1)( 8)( 2)( 5) 880 0
( 8 8)( 5 2 10) 880 0
x x x x
x x x x
x x x x x x
− + + + − =− + + + − =+ − − + + + − =
Mathematics Notes for FSc Part-I Prepared by Bilal
2 2( 7 8)( 7 10) 880 0x x x x+ − + + − =Let 2 7y x x= +The above equation becomes:
2
2
( 8)( 10) 880 0
10 8 80 880 0
2 960 0
y y
y y y
y y
− + − =+ − − − =+ − =
2 4 4(1)( 960)
2y
− ± − −= by using Quadratic formula
2 3844
2y
− ±=
2 62
2y
− ±=
2 62
2y
− += 2 62
2y
− −=
30y = 32y = −As, 2 7y x x= +
2 7 30x x+ = 2 7 32x x+ = −2 7 30 0x x+ − = 2 7 32 0x x+ + =
2 10 3 30 0x x x+ − − =7 49 4(1)(32)
2x
− ± −=
( 10) 3( 10) 0x x x+ − + = 7 79
2x
− ± −=
( 3)( 10) 0x x− + = 7 79
2
ix
− ±=
3 0x − = 10 0x + =3x = 10x = −
Hence, solution set = 7 79
{ 10,3, }2
i− ±−
8. ( 5)( 7)( 6)( 4) 504 0x x x x− − + + − =Solution:
2 2
2 2
( 5)( 7)( 6)( 4) 504 0
( 5)( 4)( 6)( 7) 504 0
( 4 5 20)( 7 6 42) 504 0
( 20)( 42) 504 0
x x x x
x x x x
x x x x x x
x x x x
− − + + − =− + + − − =+ − − − + − − =− − − − − =
Let 2y x x= −The above equation becomes:( 20)( 42) 504 0y y− − − =
2 42 20 840 504 0y y y− − + − =
Mathematics Notes for FSc Part-I Prepared by Bilal
2
2
62 336 0
( 62) ( 62) 4(1)(336)
2(1)
62 2500
2
y y
y
y
− + =
− − ± − −=
±=
62 50
2y
±=
62 50
2y
+= 62 50
2y
−=
56y = 6y =As, 2y x x= −
2 56x x− = 2 6x x− =2 56 0x x− − = 2 6 0x x− − =2 8 7 56 0x x x− + − = 2 3 2 6 0x x x− + − =( 8) 7( 8) 0x x x− + − = ( 3) 2( 3) 0x x x− + − =
( 7)( 8) 0x x+ − = ( 2)( 3) 0x x+ − =7 0x + = 8 0x − = 2 0x + = 3 0x − =
7x = − 8x = 2x = − 3x =Hence, solution set = { 2,3, 7,8}− −
9. ( 1)( 2)( 8)( 5) 360 0x x x x− − − + + =Solution:
2 2
2 2
( 1)( 2)( 8)( 5) 360 0
( 2 2)( 5 8 40) 360 0
( 3 2)( 3 40) 360 0
x x x x
x x x x x x
x x x x
− − − + + =− − + + − − + =− + − − + =
Let 2 3y x x= −The above equation becomes:
2
2
2
( 2)( 40) 360 0
( 40 2 80) 360 0
38 280 0
28 10 280 0
y y
y y y
y y
y y y
+ − + =− + − + =
− + =− − + =
( 28) 10( 28) 0y y y− − − =( 10)( 28) 0y y− − =
10 0y − = 28 0y − =As, 2 3y x x= −
2 3 10 0x x− − = 2 3 28 0x x− − =2 5 2 10 0x x x− + − = 2 7 4 28 0x x x− + − =( 5) 2( 5) 0x x x− + − = ( 7) 4( 7) 0x x x− + − =
( 2)( 5) 0x x+ − = ( 4)( 7) 0x x+ − =
Mathematics Notes for FSc Part-I Prepared by Bilal
2 0x + = 5 0x − = 4 0x + = 7 0x − =2x = − 5x = 4x = − 7x =
Hence, solution set = { 2,5, 4,7}− −
10. ( 1)(2 3)(2 5)( 3) 945x x x x+ + + + =Solution:( 1)(2 3)(2 5)( 3) 945x x x x+ + + + =
2 2
2 2
2 2
( 1)( 3)(2 3)(2 5) 945
( 3 3)(4 10 6 15) 945
( 4 3)(4 16 15) 945
( 4 3)(4( 4 ) 15) 945
x x x x
x x x x x x
x x x x
x x x x
+ + + + =+ + + + + + =+ + + + =+ + + + =
Let 2 4y x x= +The above equation becomes:
2
2
( 3)(4 15) 945
4 15 12 45 945 0
4 27 900 0
y y
y y y
y y
+ + =+ + + − =+ − =
227 (27) 4(4)( 900)
2(4)y
− ± − −= by using quadratic formula
27 729 14400
8y
− ± +=
27 15129
8y
− ±=
27 123
8y
− ±=
27 123
8y
− += 27 123
8y
− −=
12y = 75
4y = −
As, 2 4y x x= +
2 4 12x x+ = 2 754
4x x+ = −
2 4 12 0x x+ − = 2 754 0
4x x+ + =
2 6 2 12 0x x x+ − − =75
4 16 4(1)( )4
2x
− ± −=
( 6) 2( 6) 0x x x+ − + = 4 16 75
2x
− ± −=
Mathematics Notes for FSc Part-I Prepared by Bilal
( 2)( 6) 0x x− + = 4 59
2x
− ± −=
2 0x − = 6 0x + = 4 59
2
ix
− ±=
2x = 6x = −
Hence, solution set = 4 59
{ 6,2, }2
i− ±−
11. 2(2 7)( 9)(2 5) 91 0x x x− − + − =Solution:
2(2 7)( 9)(2 5) 91 0x x x− − + − =(2 7)( 3)( 3)(2 5) 91 0x x x x− − + + − =(2 7)( 3)( 3)(2 5) 91 0x x x x− + − + − =
2 2(2 6 7 21)(2 5 6 15) 91 0x x x x x x+ − − + − − − =2 2(2 21)(2 15) 91 0x x x x− − − − − =
Let 22y x x= −The above equation becomes:( 21)( 15) 91 0y y− − − =
2( 15 21 315) 91 0y y y− − + − =2 36 224 0y y− + =2 28 8 224 0y y y− − + =( 28) 8( 28) 0y y y− − − =
( 8)( 28) 0y y− − =8 0y − = 28 0y − =
As, 22y x x= −22 8 0x x− − = 22 28 0x x− − =
Using Quadratic formula: 22 8 7 28 0x x x− + − =( 1) 1 4(2)( 8)
2(2)x
− − ± − −= 2 ( 4) 7( 4) 0x x x− + − =
1 65
4x
±= (2 7)( 4) 0x x+ − =
2 7 0x + = 4 0x − =7
2x = − 4x =
Hence, solution set = 7 1 65
{4, , }2 4
±−
Mathematics Notes for FSc Part-I Prepared by Bilal
12. 2 2( 6 8)( 14 48) 105x x x x+ + + + =Solution:
2 2( 6 8)( 14 48) 105x x x x+ + + + =2 2( 4 2 8)( 8 6 48) 105
( 4) 2( 4)( ( 8) 6( 8)) 105
( 2)( 4)( 6)( 8) 105
x x x x x x
x x x x x x
x x x x
+ + + + + + =+ + + + + + =
+ + + + =Rearranging we have,( 2)( 8)( 4)( 6) 105x x x x+ + + + =
2 2
2 2
( 8 2 16)( 6 4 24) 105
( 10 16)( 10 24) 105
x x x x x x
x x x x
+ + + + + + =+ + + + =
Let 2 10y x x= +The above equation becomes:
2
2
2
( 16)( 24) 105
24 16 384 105 0
40 279 0
40 (40) 4(1)(279)
2(1)
40 1600 1116
2
40 484
240 22
2
y y
y y y
y y
y
y
y
y
+ + =+ + + − =+ + =
− ± −=
− ± −=
− ±=
− ±=
40 22
2y
− += 40 22
2y
− −=
9y = − 31y = −As, 2 10y x x= +
2 10 9x x+ = − 2 10 31x x+ = −2 10 9 0x x+ + = 2 10 31 0x x+ + =
2 9 9 0x x x+ + + =10 100 4(1)(31)
2x
− ± −=
( 9) 1( 9) 0x x x+ + + = 10 100 124
2x
− ± −=
( 1)( 9) 0x x+ + = 10 24
2x
− ± −=
1 0x + = 9 0x + = 10 24
2
ix
− ±=
Mathematics Notes for FSc Part-I Prepared by Bilal
1x = − 9x = − 10 2 6
2
ix
− ±=
2( 5 6 )
2
ix
− ±=
5 6x i= − ±Hence, solution set = { 1, 9 5 6 }i− − − ±
13. 2 2( 6 27)( 2 35) 385x x x x+ − − − =Solution:
2 2
2 2
( 6 27)( 2 35) 385
( 9 3 27)( 7 5 35) 385
( ( 9) 3( 9))( ( 7) 5( 7)) 385
( 3)( 9)( 5)( 7) 385
x x x x
x x x x x x
x x x x x x
x x x x
+ − − − =+ − − − + − =+ − + − + − =
− + + − =Rearranging we have,
2 2
2 2
( 3)( 5)( 9)( 7) 385
( 5 3 15)( 7 9 63) 385
( 2 15)( 2 63) 385
x x x x
x x x x x x
x x x x
− + + − =+ − − − + − =+ − + − =
Let 2 2y x x= +The above equation becomes:
2
2
2
( 15)( 63) 385
63 15 945 385
78 945 385 0
78 560 0
y y
y y y
y y
y y
− − =− − + =− + − =− + =
2( 78) ( 78) 4(1)(560)
2(1)y
− − ± − −= by using quadratic formula
78 6084 2240
2
78 3844
278 62
2
y
y
y
± −=
±=
±=
78 62
2y
+= 78 62
2y
−=
70y = 8y =As, 2 2y x x= +
2 2 70x x+ = 2 2 8x x+ =2 2 70 0x x+ − = 2 2 8 0x x+ − =
Mathematics Notes for FSc Part-I Prepared by Bilal
2 4 4(1)( 70)
2(1)x
− ± − −= 2 4 2 8 0x x x+ − − =
2 4 71
2x
− ± ×= ( 4) 2( 4) 0x x x+ − + =
2 2 71
2x
− ±= ( 2)( 4) 0x x− + =
2( 1 71)
2x
− ±= 2 0x − = 4 0x + =
1 71x = − ± 2x = 4x = −Hence, solution set = { 4,2, 1 71}− − ±
14. 2 14.2 9.2 1 0x x+ − + =Solution:
2 14.2 9.2 1 0x x+ − + =2 1
2
4.2 .2 9.2 1 0
8.2 9.2 1 0
x x
x x
− + =− + =
Let 2xy = 2 2 2 2( ) (2 ) 2x xy y= ⇒ =The above equation becomes:
2
2
8 9 1 0
8 8 1 0
8 ( 1) 1( 1) 0
(8 1)( 1) 0
y y
y y y
y y y
y y
− + =− − + =
− − − =− − =
8 1 0y − = 1 0y − =1
8y = 1y =
As, 2xy =1
28
x = 2 1x =
312 ( )
2x = 02 (2)x = 0 1a∴ =
32 (2)x −= 11( ) aa
−∴ = 0x =
3x = −Hence, solution set = { 3,0}−
15. 62 2 20 0x x− ++ − =Solution:
62 2 20 0x x− ++ − =62 2 .2 20 0x x−+ − =
Mathematics Notes for FSc Part-I Prepared by Bilal
612 .2 20 0
2x
x+ − =
Let 2xy =The above equation becomes:
164 20 0yy
+ − =
2
2
2
2
6420
64 20
20 64 0
16 4 64 0
( 16) 4( 16) 0
( 4)( 16) 0
y
y
y y
y y
y y y
y y y
y y
+ =
+ =− + =− − + =
− − − =− − =4 0y − = 16 0y − =
As, 2xy =2 4 0x − = 2 16 0x − =2 4x = 2 16x =
22 2x = 42 2x =2x = 4x = b ca a b c∴ = ⇒ = Hence, 22 2 2x x= ⇒ = , and
42 2 4x x= ⇒ =Hence, solution set = {2,4}
16. 34 3.2 128 0x x+− + =Solution:
34 3.2 128 0x x+− + =34 3.2 .2 128 0x x− + =
4 24.2 128 0x x− + =2(2 ) 24.2 128 0x x− + =
22 24.2 128 0x x− + = ( )m n mna a∴ =Let 2xy = 2 2 2 2( ) (2 ) 2x xy y= ⇒ =The above equation becomes:
2
2
24 128 0
16 8 128 0
( 16) 8( 16) 0
( 8)( 16) 0
y y
y y y
y y y
y y
− + =− − + =
− − − =− − =8 0y − = 16 0y − =8y = 16y =
As, 2xy =
Mathematics Notes for FSc Part-I Prepared by Bilal
2 8x = 2 16x =32 2x = 42 2x =
3x = 4x =Hence, solution set = {3,4}
17. 2 13 12.3 81 0x x− − + =Solution:
2 13 12.3 81 0x x− − + =2 1
2
3 .3 12.3 81 0
312.3 81 0
3
x x
xx
− − + =
− + =
Let 3xy = 2 23 xy =The above equation becomes:
2
2
2
2
12 81 03
3681
3
36 243
36 243 0
yy
y y
y y
y y
− + =
− = −
− = −− + =
2( 36) ( 36) 4(1)(243)
2(1)y
− − ± − −= by using quadratic formula
36 1296 972
2
36 324
236 18
2
y
y
y
± −=
±=
±=
36 18
2y
+= 36 18
2y
−=
27y = 9y =As, 3xy =3 27x = 3 9x =
33 3x = 23 3x =3x = 2x =
Hence, solution set = {2,3}
Mathematics Notes for FSc Part-I Prepared by Bilal
18. 21 1( ) 3( ) 4 0x x
x x+ − + − =
Solution:21 1
( ) 3( ) 4 0x xx x
+ − + − =
Let 1
y xx
= +
The above equation becomes:2
2
3 4 0
4 4 0
( 4) 1( 4) 0
( 1)( 4) 0
y y
y y y
y y y
y y
− − =− + − =
− + − =+ − =1 0y + = 4 0y − =
1y = − 4y =
As, 1
y xx
= +
11x
x+ = − 1
4xx
+ =
2 11
x
x
+ = −2 1
4x
x
+ =
2 1x x+ = − 2 1 4x x+ =2 1 0x x+ + = 2 4 1 0x x− + =
1 1 4
2x
− ± −= ( 4) 16 4
2x
− − ± −= by using quadratic formula
1 3
2x
− ± −= 4 12
2x
±=
1 3
2
ix
− ±= 4 4 3
2x
± ×=
4 2 3
2x
±=
2(2 3)
2x
±=
2 3x = ±
Hence, solution set = 1 3
{ ,2 3}2
i− ± ±
Mathematics Notes for FSc Part-I Prepared by Bilal
19. 22
1 14 0x xx x
+ − + + =
Solution:2
2
1 14 0x xx x
+ − + + =
Rearranging we have,2
2
1 14 0x x
x x+ + + − =
Add “2” to both the sides for completing the square, 2 22
1 1( ) 2x x
x x∴ + = + +
22
1 12 4 2x x
x x+ + + + − =
21 1( ) 6x x
x x+ + + =
Let 1
y xx
= +
The above equation becomes:2
2
6 0
3 2 6 0
( 3) 2( 3) 0
( 2)( 3) 0
y y
y y y
y y y
y y
+ − =+ − − =
+ − + =− + =2 0y − = 3 0y + =2y = 3y = −
As, 1
y xx
= +
12x
x+ = 1
3xx
+ = −
2 12
x
x
+ =2 1
3x
x
+ = −
2 2 1 0x x− + = 2 3 1 0x x+ + =2 4 4
2x
± −= 3 9 4
2x
− ± −=
2 0
2x
±= 3 5
2x
− ±=
1x =
Hence, solution set = 3 5
{1, }2
− ±
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20. 21 1( ) 3( ) 0x x
x x− + + =
Solution:21 1
( ) 3( ) 0x xx x
− + + =
22
1 12 3( ) 0x xx x
− + + + =
22
1 12 2 3( ) 2x x
x x− + + + + = Add “2” to both sides for completing the square
22
2
1 12 3( ) 2 2
1 1( ) 3( ) 4 0
x xx x
x xx x
+ + + + = +
+ + + − =
Let 1
y xx
= +
The above equation becomes:2
2
3 4 0
4 4 0
( 4) 1( 4) 0
( 1)( 4) 0
y y
y y y
y y y
y y
+ − =+ − − =
+ − + =− + =1 0y − = 4 0y + =1y = 4y = −
As, 1
y xx
= +
11x
x+ = 1
4xx
+ = −
2 11
x
x
+ =2 1
4x
x
+ = −
2 1x x+ = 2 1 4x x+ = −2 1 0x x− + = 2 4 1 0x x+ + =
( 1) 1 4
2x
− − ± −= (4) 16 4
2x
− ± −=
1 3
2x
± −= 4 12
2x
− ±=
1 3
2
ix
±= 4 4 3
2x
− ± ×=
4 2 3
2x
− ±=
Mathematics Notes for FSc Part-I Prepared by Bilal
2( 2 3)
2x
− ±=
2 3x = − ±
Hence, solution set = 1 3
{ , 2 3}2
i± − ±
21. 4 3 22 3 3 2 0x x x x− − − + =Solution:
4 3 22 3 3 2 0x x x x− − − + = divide both sides by 2x2
2
3 22 3 1 0x x
x x− − − + =
22
1 12( ) 3( ) 1x x
x x+ − + = add “4” both sides for completing the square
22
1 12( ) 3( ) 4 1 4x x
x x+ − + + = +
22
1 12( 2 ) 3( ) 1 4x x
x x+ + − + = + the 4 becomes “2” inside the parenthesis,
because it is multiplying by 2. 21 1
2( ) 3( ) 5x xx x
+ − + =
Let 1
y xx
= +
The above equation becomes:22 3 5 0y y− − =22 5 2 5 0
(2 5) 1(2 5) 0
( 1)(2 5) 0
y y y
y y y
y y
− + − =− + − =
+ − =1 0y + = 2 5 0y − =
1y = − 2 5y =5
2y =
As, 1
y xx
= +
11x
x+ = − 1 5
2xx
+ =
2 11
x
x
+ = −2 1 5
2
x
x
+ =
2 1 0x x+ + = 2 51
2
xx + =
Mathematics Notes for FSc Part-I Prepared by Bilal
1 1 4
2x
− ± −= 2 51
2
xx − = −
1 3
2x
− ± −=22 5
12
x x− = −
1 3
2
ix
− ±= 22 5 2x x− = −
22 5 2 0x x− + =22 4 2 0x x x− − + =
2 ( 2) 1( 2) 0x x x− − − =(2 1)( 2) 0x x− − =2 1 0x − = 2 0x − =
1
2x = 2x =
Hence, solution set = 1 1 3
{ ,2, }2 2
i− ±
22. 4 3 22 3 4 3 2 0x x x x+ − − + =Solution:
4 3 22 3 4 3 2 0x x x x+ − − + = divide both sides by 2x
22
22
22
2
3 22 3 4 0
1 12( ) 3( ) 4
1 12( 2 ) 3( ) 4 4
1 12( ) 3( ) 0
x xx x
x xx x
x xx x
x xx x
+ − − + =
+ + − =
− + + − = −
− + − =
Let 1
y xx
= −
The above equation becomes:22 3 0
(2 3) 0
y y
y y
+ =+ =
0y = 2 3 0y + =3
2y = −
As, 1
y xx
= −
10x
x− = 1 3
2xx
− = −
Mathematics Notes for FSc Part-I Prepared by Bilal
2 10
x
x
− =2 1 3
2
x
x
− = −
2 1 0x − = 2 31
2
xx − = −
2 1x = 2 31
2
xx + =
1x = ±22 3
12
x x+ =
2
2
2
2 3 2
2 3 2 0
2 4 2 0
2 ( 2) 1( 2) 0
(2 1)( 2) 0
x x
x x
x x x
x x x
x x
+ =+ − =+ − − =+ − + =
− + =2 1 0x − = 2 0x + =
1
2x = 2x = −
Hence, solution set = 1
{ 1, 2, ,}2
± −
23. 4 3 26 35 62 35 6 0x x x x− + − + =Solution:
4 3 26 35 62 35 6 0x x x x− + − + = divide both sides by 2x2
2
35 66 35 62 0x x
x x− + − + =
22
1 16( ) 35( ) 62x x
x x+ − + = − Add “12” to both sides for completing the square.
22
1 16( 2 ) 35( ) 62 12x x
x x+ + − + = − + (12 becomes 2 inside brackets, because brackets
are multiplying by 6.)21 1
6( ) 35( ) 50x xx x
+ − + = −
Let 1
y xx
= +
The above equation becomes:26 35 50 0y y− + =
2( 35) ( 35) 4(6)(50)
2(6)y
− − ± − −=
35 1225 1200
12y
± −=
Mathematics Notes for FSc Part-I Prepared by Bilal
35 25
12y
±=
35 5
12y
±=
35 5
12y
+= 35 5
12y
−=
10
3y = 5
2y =
As, 1
y xx
= +
1 10
3xx
+ = 1 5
2xx
+ =
2 1 10
3
x
x
+ =2 1 5
2
x
x
+ =
2 101
3
xx + = 2 5
12
xx + =
2 101
3
xx − = − 2 5
12
xx − = −
23 101
3
x x− = −22 5
12
x x− = −
23 10 3x x− = − 22 5 2x x− = −23 10 3 0x x− + = 22 5 2 0x x− + =
2( 10) ( 10) 4(3)(3)
2(3)x
− − ± − −=
2( 5) ( 5) 4(2)(2)
2(2)x
− − ± − −=
10 100 36
6x
± −= 5 25 16
4x
± −=
10 64
6x
±= 5 9
4x
±=
10 8
6x
±= 5 3
4x
±=
10 8
6x
+= 10 8
6x
−= 5 3
4x
+= 5 3
4x
−=
3x =1
3x = 2x =
1
2x =
Hence, solution set = 1 1
{2, ,3, }2 3
Mathematics Notes for FSc Part-I Prepared by Bilal
24. 4 22 4
6 16 10 0x x
x x− + − + =
Solution:4 2
2 4
6 16 10 0x x
x x− + − + =
4 24 2
1 16( ) 10x x
x x+ − + = −
4 24 2
1 12 6( ) 10 2x xx x
+ + − + = − +
2 2 22 2
1 1( ) 6( ) 8x x
x x+ − + = −
Let 22
1y x
x= +
The above equation becomes:2
2
6 8 0
4 2 8 0
( 4) 2( 4) 0
( 2)( 4) 0
y y
y y y
y y y
y y
− + =− − + =
− − − =− − =2 0y − = 4 0y − =2y = 4y =
As, 22
1y x
x= +
22
12x
x+ = 2
2
14x
x+ =
First take: 22
12x
x+ =
Add “2” to both sides for completing the square2
2
12 2 2xx
+ + = +
2 22
1( ) 4x
x+ =
22
12x
x+ = ±
22
12x
x+ = 2
2
12x
x+ = −
4
2
12
x
x
+ =4
2
12
x
x
+ = −
4 21 2x x+ = 4 21 2x x+ = −4 22 1 0x x− + = 4 22 1 0x x+ + =
2 2( 1) 0x − = 2 2( 1) 0x + =
Mathematics Notes for FSc Part-I Prepared by Bilal
2 1 0x − = 2 1 0x + =2 1x = 2 1x = −
1x = ± x i= ±
Now take: 22
14x
x+ =
4
2
14
x
x
+ =
4 21 4x x+ =4 24 1 0x x− + =
Let 2u x=The above equation becomes:
2 4 1 0u u− + =2( 4) ( 4) 4(1)(1)
2(1)u
− − ± − −=
4 12
2u
±=
4 4 3
2u
± ×=
4 2 3
2u
±=
2 3u = ±As, 2u x=
2 2 3x = ±
2 3x = ± ±
Hence, solution set = { 1, , 2 3}i± ± ± ±
Mathematics Notes for FSc Part-I Prepared by Bilal