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}

Other Topics

;