Question

The roots of the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=2\frac{1}{6}$ are
(a) $\frac{9}{13},\frac{7}{13}$
(b) $\frac{9}{13},\frac{4}{13}$
(c) $\frac{7}{13},\frac{4}{13}$
(d) none of these

Answer 1

(b) $\frac{9}{13},\frac{4}{13}$

$$\begin{gathered} \text{Given:} \\ \sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=2\frac{1}{6} \\ \text{Let}\frac{x}{(1-x)}\text{be equal to}{y}^2. \\ \text{Then, the equation becomes:} \\ y+\frac{1}{y}=\frac{13}{6} \\ \text{⇒}\frac{y^2+1}{y}=\frac{13}{6} \\ \Rightarrow 6y^2-13y+6=0 \\ \Rightarrow 6y^2-9y-4y+6=0 \\ \Rightarrow 3y(2y-3)-2(2y-3)=0 \\ \Rightarrow (2y-3)(3y-2)=0 \\ \Rightarrow y=\frac{3}{2}\text{or}y=\frac{2}{3} \\ \therefore \frac{x}{1-x}={\left(\frac{3}{2}\right)}^2\text{or}\frac{x}{1-x}={\left(\frac{2}{3}\right)}^2 \\ \Rightarrow \frac{x}{1-x}\text{=}\frac{9}{4}\text{or}\frac{x}{1-x}=\frac{4}{9} \\ \Rightarrow 4x=9-9x\text{or 9}x=4-4x \\ \Rightarrow 13x=9\text{or 13}x=4 \\ \Rightarrow x=\frac{9}{13}\text{or}x=\frac{4}{13} \\ \mathrm{Therefore},\mathrm{the}\mathrm{re}\text{quired roots are}\frac{9}{13}\text{and}\frac{4}{13}\text{.} \end{gathered}$$