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