+26367 If x^2 + 1/x^2 = 171, then what are the possible values of x - 1/x?
\(\begin{array}{|rcll|} \hline \left( x - \dfrac{1}{x} \right)^2 &=& x^2-2x\dfrac{1}{x} + \dfrac{1}{x^2} \\ \left( x - \dfrac{1}{x} \right)^2 &=& x^2+ \dfrac{1}{x^2} -2 \quad &|\quad x^2 + \dfrac{1}{x^2} = 171 \\ \left( x - \dfrac{1}{x} \right)^2 &=&171 -2 \\ \left( x - \dfrac{1}{x} \right)^2 &=&169 \\ x - \dfrac{1}{x} &=& \pm \sqrt{169} \\ \mathbf{ x - \dfrac{1}{x} } &=& \mathbf{\pm 13} \\ \hline \end{array} \)
