physics numericals
Equations Reducible to Quadratic Equations
Exercise 4.2
Solve the following equations:
1. x 4 − 6 x 2 + 8 = 0
Solution:
x4 − 6 x2 + 8 = 0
Let y = x 2 and y 2 = x 4
The above equation becomes: y 2 − 6 y + 8 = 0 y2 − 6 y + 8 = 0 y2 − 4 y − 2 y + 8 = 0 y ( y − 4) − 2( y − 4) = 0
( y − 2)( y − 4) = 0 y − 2 = 0 and y − 4 = 0 y=2 y=4
2
As, y = x x2 = 2 x2 = 4 x = ±2 x=± 2 solution set = { 2, − 2, 2, −2}
2. x −2 − 10 = 3 x−1
Solution:
x −2 − 10 = 3 x −1 x −2 − 3 x −1 − 10 = 0
Let y = x −1 and y 2 = x −2
The above equation becomes: y 2 − 3 y − 10 = 0 y 2 − 5 y + 2 y − 10 = 0 y ( y − 5) + 2( y − 5) = 0
( y + 2)( y − 5) = 0 y = −2 y =5
−1
As, y = x x −1 = −2 x −1 = 5
( x −1 )−1 = ( −2)−1
( x −1 ) −1 = (5)−1
1
1 x=− x=
2
5
1 1 solution set = {− , }
2 5
Mathematics Notes for FSc Part-I
Prepared by Bilal
3. x 6 − 9 x3 + 8 = 0
Solution:
x 6 − 9 x3 + 8 = 0
Let y = x 3 and y 2 = x6
The above equation becomes: y 2 − 9 y + 8 = 0 y2 − 8 y − y + 8 = 0 y ( y − 8) − 1( y − 8) = 0
( y − 1)( y − 8) = 0 y =1 y =8
3
As, y = x x3 = 1 x3 = 8
First take: x 3 = 1 x3 − 1 = 0
( x − 1)( x 2 + x + 1) = 0 x −1 = 0 x2 + x + 1 = 0
−1 ± 1 − 4 x =1 by using quadratic formula x= 2
−1 ± −3 x= 2
−1 ± 3i x= 2
3
Now take: x − 8 = 0
( x)3 − (2)3 = 0
( x − 2)( x 2 + 2 x + 4) = 0 x−2= 0 x2 + 2x + 4 = 0
−2 ± 4 − 16 x=2 x=
2
−2 ± −12 x= 2
−2 ± 2 − 3 x= 2
2(−1 ± −3) x= 2 x = −1 ± −3
Hence, solution set = {1, 2,
−1 ± 3i
, −1 ± −3}
2
Mathematics Notes for FSc Part-I
Prepared by Bilal
4. 8 x 6 − 19 x3 − 27 = 0
Solution:
Let y = x 3 and y 2 = x6
The above equation becomes: 8 y 2 − 19 y − 27 = 0
Use quadratic formula,
Here, a = 8 , b = −19 and c = −27 y= −(−19) ± (−19) 2 − 4(8)(−27)
2(8)
19 ± 1225
16
19 ± 35 y= 16
19 + 35 y= 16
54
y=
16
27 y= 8
As, y = x3
27
x3 =
8
y=
3
First take: x =
19 − 35
16
−16 y= 16 y= y = −1
x 3 = −1
27
8
27
=0
8
3
(
Exercise 4.2
Solve the following equations:
1. x 4 − 6 x 2 + 8 = 0
Solution:
x4 − 6 x2 + 8 = 0
Let y = x 2 and y 2 = x 4
The above equation becomes: y 2 − 6 y + 8 = 0 y2 − 6 y + 8 = 0 y2 − 4 y − 2 y + 8 = 0 y ( y − 4) − 2( y − 4) = 0
( y − 2)( y − 4) = 0 y − 2 = 0 and y − 4 = 0 y=2 y=4
2
As, y = x x2 = 2 x2 = 4 x = ±2 x=± 2 solution set = { 2, − 2, 2, −2}
2. x −2 − 10 = 3 x−1
Solution:
x −2 − 10 = 3 x −1 x −2 − 3 x −1 − 10 = 0
Let y = x −1 and y 2 = x −2
The above equation becomes: y 2 − 3 y − 10 = 0 y 2 − 5 y + 2 y − 10 = 0 y ( y − 5) + 2( y − 5) = 0
( y + 2)( y − 5) = 0 y = −2 y =5
−1
As, y = x x −1 = −2 x −1 = 5
( x −1 )−1 = ( −2)−1
( x −1 ) −1 = (5)−1
1
1 x=− x=
2
5
1 1 solution set = {− , }
2 5
Mathematics Notes for FSc Part-I
Prepared by Bilal
3. x 6 − 9 x3 + 8 = 0
Solution:
x 6 − 9 x3 + 8 = 0
Let y = x 3 and y 2 = x6
The above equation becomes: y 2 − 9 y + 8 = 0 y2 − 8 y − y + 8 = 0 y ( y − 8) − 1( y − 8) = 0
( y − 1)( y − 8) = 0 y =1 y =8
3
As, y = x x3 = 1 x3 = 8
First take: x 3 = 1 x3 − 1 = 0
( x − 1)( x 2 + x + 1) = 0 x −1 = 0 x2 + x + 1 = 0
−1 ± 1 − 4 x =1 by using quadratic formula x= 2
−1 ± −3 x= 2
−1 ± 3i x= 2
3
Now take: x − 8 = 0
( x)3 − (2)3 = 0
( x − 2)( x 2 + 2 x + 4) = 0 x−2= 0 x2 + 2x + 4 = 0
−2 ± 4 − 16 x=2 x=
2
−2 ± −12 x= 2
−2 ± 2 − 3 x= 2
2(−1 ± −3) x= 2 x = −1 ± −3
Hence, solution set = {1, 2,
−1 ± 3i
, −1 ± −3}
2
Mathematics Notes for FSc Part-I
Prepared by Bilal
4. 8 x 6 − 19 x3 − 27 = 0
Solution:
Let y = x 3 and y 2 = x6
The above equation becomes: 8 y 2 − 19 y − 27 = 0
Use quadratic formula,
Here, a = 8 , b = −19 and c = −27 y= −(−19) ± (−19) 2 − 4(8)(−27)
2(8)
19 ± 1225
16
19 ± 35 y= 16
19 + 35 y= 16
54
y=
16
27 y= 8
As, y = x3
27
x3 =
8
y=
3
First take: x =
19 − 35
16
−16 y= 16 y= y = −1
x 3 = −1
27
8
27
=0
8
3
(