Fsc Part 1 Mathematics (Complete Solution)

Q10
Solve the Following equations (x + 1)(2x + 3)(2x + 5)(x + 3) = 945

(x + 1)(2x + 3)(2x + 5)(x + 3) = 945

Solutions


The given equation can be written as
(x + 1)2(x +
3
2
)2(x +
5
2
)(x + 3) - 945 = 0
As
1 +
3
2
+
5
2
+ 3
2
=
16
2
2
= 4, So we can write as

[(x + 1)(x + 3)] [2(x +
3
2
)2(x +
5
2
)] - 945 = 0
(x2 + 3x + x + 3) 4(x2 +
5
2
x +
3
2
x +
15
4
) - 945 = 0
(x2 + 4x + 3) 4(x2 + 4x +
15
4
) - 945 = 0

Let x2 + 4x = t       and Put in the equation

(t + 3) 4(t +
15
4
) - 945 = 0
(t + 3) (4t + 15) - 945 = 0
4t2 + 15t + 12t + 45 - 945 = 0
4t2 + 27t - 900 = 0

By Quadratic Formula, we have

t =
-27 ±
(27)2 - 4(4)(-900)
2(4)
t =
-27 ±
729 + 14400
8
t =
-27 ±
15129
8
t =
-27 ±
(123)2
8
t =
-27 ± 123
8

Either      
t =
-27 - 123
8
     OR      
t =
-27 + 123
8
t =
-150
8
t =
96
8
t =
-75
4
t = 12

By Letting t = x2 + 4x,       Putting the value of t, we get
x2 + 4x =
-75
4
x2 + 4x = 12
4x2 + 16x = -75 x2 + 4x = 12
4x2 + 16x + 75 = 0 --------------------- (i) x2 + 4x - 12 = 0 --------------------- (ii)

By Equation (i)       4x2 + 16x + 75 = 0

Using the Quadratic formula, we have

x =
-16 ±
(16)2 - 4(4)(75)
2(4)
x =
-16 ±
256 - 1200
8
x =
-16 ±
-944
8
x =
-16 ±
(16)(-59)
8
x =
-16 ± 4
-59
8
x =
4(-4 ±
-59
)
8
x =
-4 ±
-59
2

By Equation (ii)       x2 + 4x - 12 = 0

Using the Quadratic formula, we have

x =
-4 ±
(4)2 - 4(1)(-12)
2(1)
x =
-4 ±
16 + 48
2
x =
-4 ±
64
2
x =
-4 ± 8
2

Either      
x =
-4 - 8
2
     OR      
x =
-4 + 8
2
x =
-12
2
x =
4
2
x = -6
x = 2


Hence the Solution Set of given equation is
{ -6, 2 ,   
-4 -
-59
2
,   
-4 +
-59
2
}

Other Topics

;