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Answer:
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Let (x2 +
x) = m
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∴ m(m – 2) = 24
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∴ m2 – 2m = 24
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∴ m2 - 2m - 24 = 0
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∴ m2 – 6m + 4m – 24 = 0
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∴ m(m – 6) + 4( m – 6 ) = 0
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∴ (m – 6) (m + 4 ) = 0
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∴ m – 6 = 0
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∴ m + 4 = 0
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∴ m = 6
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∴ m = - 4
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Re substituting m = (x2
+ x)
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∴ x2 + x = 6
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∴ x2 + x = - 4
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∴ x2 + x – 6 = 0
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∴ x 2 + x + 4 = 0
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∴ x2 + 3x – 2x – 6 = 0
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Here, a = 1 , b = 1, c = 4
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∴ x(x + 3) – 2(x + 3) = 0
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∴ b2 – 4ac = 12 – 4 (1) (4)
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∴ (x + 3 ) ( x – 2 ) = 0
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∴ b2 – 4ac = - 15
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∴ x + 3 = 0 OR x – 2 = 0
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∴ b2 – 4ac < 0
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∴ x = - 3 OR x =
2
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∴ the roots are not real but imaginary numbers
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∴ we can avoid this equation
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Solve: (x2 + x) (x2 + x – 2 ) = 24
