Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 635: 21

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

Since, $2u^2-7u+3=0$ Factorize the expression as follows: $(2u-1)(u-3)=0$ or, $u=${$\dfrac{1}{2},3$} Replace $u$ with $x^{-1}$, we have $x^{-1}=\dfrac{1}{2} \implies \dfrac{1}{x}=\dfrac{1}{2}$ $x=2$ and $x^{-1}=3 \implies \dfrac{1}{x}=3$ $x=\dfrac{1}{3}$ Hence, our solution set is $x=${$\dfrac{1}{3},2$}