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