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Question
Solve the equation:
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`.
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Solution
Given equation
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`
Put x `-(1)/x = y, "squaring" (x - 1/x)^2 = y^2`
⇒ `x^2 + (1)/x^2 - 2 = y^2`
⇒ `x^2 + (1)/x^2 = y^2 + 2`
Now, given equation becomes
6(y2 + 2) - 25y + 12 = 0
⇒ 6y2 + 12 - 25 + 12 = 0
⇒ 6y2 - 25y + 24 = 0
⇒ 6y2 - 16y - 9y + 24 = 0
⇒ 2y(3y - 8) - 3(3y - 8) = 0
⇒ (3y - 8) (2y - 3) = 0
⇒ 3y - 8 = 0 or 2y - 3 = 0
⇒ 3y = 8 or 2y = 3
⇒ y = `(8)/(3)` or y = `(3)/(2)`
But `x - (1)/x = y`
∴ `x - (1)/x = (8)/(3)`
⇒ `(x^2 - 1)/x = (8)/(3)`
⇒ 3x2 - 3 = 8x
⇒ 3x2 - 8x - 3 = 0
⇒ 3x2 - 9x + x - 3 = 0
⇒ 3x(x - 3) + 1(x - 3) = 0
⇒ (x - 3) (3x + 1) = 0
⇒ x - 3 = 0 or 3x + 1 = 0
⇒ x = 3 or x = `(-1)/(3)`
or
`x - (1)/x = (3)/(2)`
⇒ `(x^2 - 1)/x = (3)/(2)`
⇒ 2x2 - 2 = 3x
⇒ 2x2 - 3x - 2 = 0
⇒ 2x2 - 4x + x - 2 = 0
⇒ 2x(x - 2) + 1(x - 2) = 0
⇒ (x - 2) (2x + 1) = 0
⇒ x - 2 = 0 or 2x + 1 = 0
⇒ x = 2 or x = `(-1)/(2)`
Hence, x = 3, `(-1)/(3), 2 and (-1)/(2)`.
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