**(ii) (x**

^{2}+ 2x) (x^{2}+ 2x – 11) + 24 = 0

**Sol.**(x

**+ 2x) (x**

^{2}**+ 2x – 11) + 24 = 0**

^{2}
Substituting x

**+ 2x = m we get,**^{2}
m (m – 11) + 24 = 0

m

**– 11m + 24 = 0**^{2}
m

**– 8m – 3m + 24 = 0**^{2}
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 = x

**+ 2x we get,**^{2}
x

**+ 2x = 8 or x**^{2}**+ 2x = 3**^{2}
x

**+ 2x – 8 = 0 ....... (i) or x**^{2}**+ 2x – 3 = 0 ......(ii)**^{2}
From (i), x

**+ 2x – 8 = 0**^{2}
x

**+ 4x – 2x – 8 = 0**^{2}
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), x

**+ 2x – 3 = 0**^{2}
x

**+ 3x – x – 3 = 0**^{2}
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.