Fsc Part 1 Mathematics (Complete Solution)

Q9
Solve the Following equations (x - 1)(x - 2)(x - 8)(x + 5) + 360 = 0

(x - 1)(x - 2)(x - 8)(x + 5) + 360 = 0

Solutions

As
-1 - 2 - 8 + 5
2
=
-6
2
= -3, So we can write as

[(x - 1)(x - 2)] [(x - 8)(x + 5)] + 360 = 0
(x2 - 2x - x + 2)(x2 + 5x - 8x - 40) + 360 = 0
(x2 - 3x + 2)(x2 - 3x - 40) + 360 = 0
(x2 - 3x + 2)(x2 - 3x + 2 - 42) + 360 = 0

Let x2 - 3x + 2 = t       and Put in the equation

(t)(t - 42) + 360 = 0
t2 - 42t + 360 = 0

Using the quadratic formula, we have

t =
-(-42) ±
(-42)2 - 4(1)(360)
2(1)
t =
42 ±
1764 - 1440
2
t =
42 ±
324
2
t =
42 ± 18
2

Either      
t =
42 - 18
2
     OR      
t =
42 + 18
2
t =
24
2
t =
60
2
t = 12 t = 30

By Letting t = x2 - 3x + 2,       Putting the value of t, we get
x2 - 3x + 2 = 12 x2 - 3x + 2 = 30
x2 - 3x + 2 - 12 = 0 x2 - 3x + 2 - 30 = 0
x2 - 3x - 10 = 0 -------------------- (i) x2 - 3x - 28 = 0 -------------------- (ii)

By Equation (i)       x2 - 3x - 10 = 0

Using the Quadratic formula, we have

x =
-(-3) ±
(-3)2 - 4(1)(-10)
2(1)
x =
3 ±
9 + 40
2
x =
3 ±
49
2
x =
3 ± 7
2
Either      
x =
3 - 7
2
     OR      
x =
3 + 7
2
x =
-4
2
x =
10
2
x = -2
x = 5

By Equation (ii)       x2 - 3x - 28 = 0

Using the Quadratic formula, we have

x =
-(-3) ±
(-3)2 - 4(1)(-28)
2(1)
x =
3 ±
9 + 112
2
x =
3 ±
121
2
x =
3 ± 11
2
Either      
x =
3 - 11
2
     OR      
x =
3 + 11
2
x =
-8
2
x =
14
2
x = -4
x = 7



Hence the Solution Set of given equation is {-4, -2, 5, 7}

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