Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 635: 20

{$\dfrac{-1}{2},\dfrac{1}{3}$}

Since, $u^2-u-6=0$ Factorize the expression as follows: $(u-3)(u+2)=0$ or, $u=${$-2,3$} Replace $u$ with $x^{-1}$, we have $x^{-1}=-2 \implies \dfrac{1}{x}=-2$ $x=\dfrac{-1}{2}$ and $x^{-1}=3 \implies \dfrac{1}{x}=3$ $x=\dfrac{1}{3}$ Hence, our solution set is $x=${$\dfrac{-1}{2},\dfrac{1}{3}$}