(ii) (x2
+ 2x) (x2 + 2x – 11) + 24 = 0
Sol. (x2 + 2x) (x2 +
2x – 11) + 24 = 0
Substituting x2
+ 2x = m we get,
m (m – 11) + 24 = 0
m2 – 11m + 24 = 0
m2 – 8m – 3m + 24 = 0
m (m – 8) – 3 (m – 8) = 0
(m – 8) (m – 3) = 0
m – 8 = 0 or m – 3 = 0
m = 8 or m = 3
Resubstituting m = x2
+ 2x we get,
x2 + 2x =
8 or x2 + 2x = 3
x2 + 2x – 8 = 0 ....... (i)
or x2 + 2x – 3 = 0 ......(ii)
From (i), x2 +
2x – 8 = 0
x2 + 4x – 2x – 8 = 0
x(x + 4) – 2 (x + 4) = 0
(x + 4) (x – 2) = 0
x + 4 = 0 or x – 2 = 0
x = – 4 or x = 2
From (ii), x2 +
2x – 3 = 0
x2 + 3x – x – 3 = 0
x (x + 3) – 1(x + 3) = 0
(x + 3) (x – 1) = 0
x + 3 = 0 or x – 1 = 0
x = – 3 or x = 1
x = – 4 or x = 2 or x = – 3 or x = 1.
