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
By Quadratic formula
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

By Quadratic formula
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
}

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