(x2 + 2x) (x2 + 2x – 11) + 24 = 0

(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.