(x2 + x) (x2 + x – 7) + 10 = 0

(vi) (x² + x) (x² + x – 7) + 10 = 0

Sol. (x² + x) (x² + x – 7) + 10 = 0

Substituting x² + x = m we get,

m (m – 7) + 10 = 0

∴  m² – 7m + 10 = 0

∴  m² – 5m – 2m + 10 = 0

∴  m (m – 5) – 2 (m – 5) = 0

∴  (m – 5) (m – 2) = 0

∴  m – 5 = 0 or m – 2 = 0

∴  m = 5 or m = 2

Resubstituting m = x² + x we get,

x² + x = 5 ...... (i)  or x² + x = 2 ....... (ii)

From (i)

x² + x = 5

∴ x² + x – 5 = 0

Comparing with ax² + bx + c = 0 we have a = 1, b = 1, c = – 5

∴  b² – 4ac = (1)² – 4(1) (– 5)

= 1 + 20

= 21

By Formula method,

x	=	-  b ± √(b2 – 4ac)
		2a
∴ x	=	-(1)± √21
		2(1)
∴ x	=	-1 ± √21
		2
∴ x	=	-1+√21	or	-1-√21
		2		2

From (ii)

x² + x = 2

∴ x² + x – 2 = 0

∴ x² + 2x – 1x – 2 = 0

∴x (x + 2) – 1(x + 2) = 0

∴ (x + 2) (x – 1) = 0

∴x + 2 = 0 or x – 1 = 0

∴x = – 2 or x = 1

∴ x	=	-1+√21	or	-1-√21	or	-2	or	1
		2		2