 Fsc Part 1 Mathematics (Complete Solution)

# Q4 Solve the Following equations

8x6 - 19x3 - 27 = 0
Solution:
Let x3 = t and Put in the given equation
8t2 - 19t - 27 = 0
Comparing the given equation with at2 + bt + c = 0, we get
a = 8,
b = -19,
c = -27

By Quardratic formula, we have
t =
-b ±
 √ b2 - 4ac
2a
Putting the values of a, b, and c in the formula

t =
-(-19) ±
 √ (-19)2 - 4(8)(-27)
2(8)
t =
19 ±
 √ 361 + 864
16
t =
19 ±
 √ 1225
16
t =
19 ±
 √ (35)2
16
t =
 19 ± 35
16

Either
t =
 19 - 35
16
OR
t =
 19 + 35
16
t =
 -16 16
t =
 54 16
 t = -1
t =
 27 8

By letting x3 = t, Putting the value of t, we get
 x3 = -1 ------------------ (i)
x3 =
 27 8
------------------ (ii)

By equation (i)      x3 = -1
x3 + 1 = 0
x3 + 13 = 0
(x + 1)(x2 - x + 1) = 0
Either         x + 1 = 0         OR         x2 - x + 1 = 0
x = -1
x =
-(-1) ±
 √ (-1)2 - 4(1)(1)
2(1)
x =
1 ±
 √ -3
2

By equation (ii)
x3 =
 27 8

x3 -
 27 8
= 0
x3 - (
 3 2
)
3
= 0
(x -
 3 2
) (x2 +
 3 2
x +
 9 4
)    = 0
Either
x -
 3 2
= 0
OR
x2 +
 3 2
x +
 9 4
= 0
x =
 3 2
Multiply by 4 on both sides
4x2 + (4)
 3 2
x + (4)
 9 4
= 0
 4x2 + (2)(3)x + 9 = 0
4x2 + 6x + 9 = 0

x =
-6 ±
 √ (6)2 - 4(4)(9)
2(4)
x =
-6 ±
 √ 36 - 144
8
x =
-6 ±
 √ -108
8
x =
-6 ±
 √ -3(36)
8
x =
-6 ± 6
 √ -3
8
x =
6 ( -1 ±
 √ -3
)
8
x =
3 ( -1 ±
 √ -3
)
4

Hence the Solution Set of given equation is
{ -1 ,
 3 2
,
1 ±
 √ -3
2
,
3 ( -1 ±
 √ -3
)
4
}