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Question
Solve for x:
`(x + 1/x)^2 - (3)/(2)(x - 1/x)-4` = 0.
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Solution
Given equation
`(x + 1/x)^2 - (3)/(2)(x - 1/x)-4` = 0
Put `x - (1)/x = y, "squaring" x^2 + (1)/x^2 - 2 = y^2`
⇒ `x^2 + (1)/x^2 = y^2 + 2`
Then given rquation becomes :
`x^2 + (1)/x^2 + 2 - (3)/(2) (x - 1/x) -4` = 0
⇒ y2 + 2 + 2 - `(3)/(2)`y - 4 = 0
⇒ y2 + 4 - `(3)/(2)`y - 4 = 0
⇒ 2y2 - 3y = 0
⇒ y(2y - 3) = 0
⇒ y = 0 or 2y - 3 = 0 i.e., y = `(3)/(2)`
But `x - (1)/x = y`
Then `x - (1)/x = 0`
⇒ x2 - 1 = 0
⇒ x = ± 1
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 = ±1, 2, - `(1)/(2)`.
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