 Fsc Part 1 Mathematics (Complete Solution)

# Q8 Solve the Following equations (x - 5)(x - 7)(x + 6)(x + 4) - 504 = 0

(x - 5)(x - 7)(x + 6)(x + 4) - 504 = 0

Solutions

As
 -5 - 7 + 6 + 4 2
=
 -2 2
= -1, So we can write as

[(x - 5)(x + 4)] [(x - 7)(x + 6)] - 504 = 0
(x2 + 4x - 5x - 20) (x2 + 6x - 7x - 42) - 504 = 0
(x2 - x - 20) (x2 - x - 42) - 504 = 0

Let x2 - x = t       and Put in the equation

(t - 20)(t - 42) - 504 = 0
t2 - 42t - 20t + 840 - 504 = 0
t2 - 62t + 336 = 0

Using the quadratic formula, we have

t =
-(-62) ±
 √ (-62)2 - 4(1)(336)
2(1)
t =
62 ±
 √ 3844 - 1344
2
t =
62 ±
 √ 2500
2
t =
62 ±
 √ (50)2
2
t =
 62 ± 50
2

Either
t =
 62 - 50
2
OR
t =
 62 + 50
2
t =
 12 2
t =
 112 2
 t = 6
 t = 56

By Letting t = x2 - x,       Putting the value of t, we get
x2 - x = 6 x2 - x = 56
x2 - x - 6 = 0 --------------(i) x2 - x - 56 = 0 --------------(ii)

By Equation (i)       x2 - x - 6 = 0

Using the Quadratic formula, we have

x =
-(-1) ±
 √ (-1)2 - 4(1)(-6)
2(1)
x =
1 ±
 √ 1 + 24
2
x =
1 ±
 √ 5
2
x =
 1 ± 5
2
Either
x =
 1 - 5
2
OR
x =
 1 + 5
2
x =
 -4 2
x =
 6 2
 x = -2
 x = 3

By Equation (ii)       x2 - x - 56 = 0

Using the Quadratic formula, we have

x =
-(-1) ±
 √ (-1)2 - 4(1)(-56)
2(1)
x =
1 ±
 √ 1 + 224
2
x =
1 ±
 √ 225
2
x =
1 ±
 √ (15)2
2
x =
 1 ± 15
2
Either
x =
 1 - 15
2
OR
x =
 1 + 15
2
x =
 -14 2
x =
 16 2
 x = -7
 x = 8

Hence the Solution Set of given equation is {-7, -2, 3, 8}