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If \(x^2 + \frac{1}{x^2} = 171\), then find all possible values of \(x - \frac{1}{x}\)

 Jul 13, 2020
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Let \(t = x - \dfrac1x\).

 

\(t^2 + 2 = x^2 + \dfrac1{x^2} = 171\)

 

\(t^2 = 169\)

 

\(t = \pm 13\)

 

\(x - \dfrac1x = \pm13\)

 

Therefore there are two possible values.

 Jul 13, 2020

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