Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 635: 2

{$\pm 2, \pm 3$}

Since, $u^2-13u+36=0$ Factorize the expression as follows: $(u-4)(u-9)=0$ or, $u=${$4,9$} Replace $u$ with $x^2$, we have $x^2=4$ and $x^2=9$ This implies that $x=\pm 2, \pm 3$ Hence, our solution set is $x=${$\pm 2, \pm 3$}