Question

The roots of the equation $2x-\frac{3}{x}=1$ are
(a) $\frac{1}{2},-1$
(b) $\frac{3}{2},1$
(c) $\frac{3}{2},-1$
(d) $\frac{-1}{2},\frac{3}{2}$

Answer 1

(c) $\frac{3}{2},-1$

$$\begin{gathered} \text{Given:} \\ 2x-\frac{3}{x}=1 \\ \Rightarrow \frac{2x^2-3}{x}=1 \\ \Rightarrow 2x^2-x-3=0 \\ \Rightarrow 2x^2-3x+2x-3=0 \\ \Rightarrow x(2x-3)+1(2x-3)=0 \\ \Rightarrow (2x-3)(x+1)=0 \\ \Rightarrow x=\frac{3}{2}\text{or}x=-1 \\ \text{Thus, the roots of the equation are}\frac{3}{2}\text{and}-1. \end{gathered}$$