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प्रश्न
Solve for x: (x2 - 5x)2 - 7(x2 - 5x) + 6 = 0; x ∈ R.
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उत्तर
Given equation
(x2 - 5x)2 - 7(x2 - 5x) + 6 = 0
Put x2 - 5x = y
∴ The given equation becomes
y2 - 7y + 6 = 0
⇒ y2 - 6y - y + 6 = 0
⇒ y(y - 6) -1(y - 6) = 0
⇒ y = 1, 6
But x2 - 5x = y
∴ x2 - 5x = 1
x2 - 5x - 1 = 0
Here a = 1, b = -5, c = -1
∴ x = `(-b ± sqrt(b^2 - 4ac))/(2a)`
x = `(-(-5) ± sqrt(25 + 4))/(2)`
x = `(5 ± sqrt(29))/(2)`
x2 - 5x = 6
⇒ x2 - 5x - 6 = 0
⇒ x2 - 6x + x - 6 = 0
⇒ x(x - 6) +1(x - 6) = 0
⇒ (x - 6) (x + 1) = 0
⇒ x = 6 or x = -1
Hence, the roots are -1, 6, `(5 ± sqrt(29))/(2)`.
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Complete the following activity to determine the nature of the roots of the quadratic equation x2 + 2x – 9 = 0 :
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Statement A (Assertion): If 5 + `sqrt(7)` is a root of a quadratic equation with rational co-efficients, then its other root is 5 – `sqrt(7)`.
Statement R (Reason): Surd roots of a quadratic equation with rational co-efficients occur in conjugate pairs.
