Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 635: 15

{$0,\pm \sqrt 3$}

Since, $u^2-u=2$ or, $u^2-u+2=0$ Factorize the expression as follows: $(u-2)(u+1)=0$ or, $u=${$-1,2$} Replace $u$ with $x^2-1$, we have $x^2-1=-1 \implies x=0$ and $x^2-1=2 \implies x^2=3$ or, $x=\pm \sqrt 3$ Hence, our solution set is $x=${$0,\pm \sqrt 3$}