## Algebra: A Combined Approach (4th Edition)

$x_{1} = \sqrt{31}$ and $x_{2} = -\sqrt{31}$
Given \begin{gather*} x^2-31=0 \end{gather*} Applying difference of two squares, we have: \begin{gather*} (x+\sqrt{31})(x-\sqrt{31})=0 \end{gather*} Therefore the solutions are $x_{1} = \sqrt{31}$ and $x_{2} = -\sqrt{31}$