Fsc Part 1 Mathematics (Complete Solution)

# Q6 Solve the Following equations (x + 1)(x + 2)(x + 3)(x + 4) = 24

(x + 1)(x + 2)(x + 3)(x + 4) = 24

Solutions

As
 1 + 2 + 3 + 4 2
=
 10 2
= 5, So we can write

(x + 1)(x + 4)(x + 2)(x + 3) = 24
[(x + 1)(x + 4)(x + 2)(x + 3)] - 24 = 0
[(x2 + 4x + x + 4)(x2 + 3x + 2x + 6)] - 24 = 0
[(x2 + 5x + 4)(x2 + 5x + 6)] - 24 = 0

Let x2 + 5x = t and Put in the equation

[(t + 4)(t + 6)] - 24 = 0
t2 + 6t + 4t + 24 - 24 = 0
t2 + 10t = 0
t(t + 10) = 0

 Either t = 0 OR t + 10 = 0 By letting x2 + 5x = t, Putting the value of t, we get x2 + 5x = 0 x2 + 5x + 10 = 0 x(x + 5) = 0 --------(i) x2 + 5x + 10 = 0 ----------- (ii)

By equation (i)      x(x + 5) = 0

x(x + 5) = 0
 Either x = 0 OR x + 5 = 0 x = -5

By equation (ii)      x2 + 5x + 10 = 0

x2 + 5x + 10 = 0
By Quadratic Formula
t =
-5 ±
 √ 52 - 4(1)(10)
2(1)
t =
-5 ±
 √ 25 - 40
2
t =
-5 ±
 √ -15
2

Hence the Solution Set of given equation is
{ 0, -5 ,
-5 -
 √ -15
2
,
-5 +
 √ -15
2
}

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