Question

The roots of the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$ are?

• A. $\frac{3}{13},\frac{2}{13}$
• B. $\frac{13}{6},\frac{1}{6}$
• C. $\frac{9}{13},\frac{4}{13}$
• D. None of these

Answer 1

Answer: (C)

$\frac{9}{13},\frac{4}{13}$

$Thegivenequationis\sqrt{\frac{x}{1-x}+}\sqrt{\frac{1-x}{x}}={\left(\frac{13}{6}\right)}^{2}squaringboththesidesweget\frac{x}{1-x}+\frac{1-x}{x}+2={\left(\frac{13}{6}\right)}^2\Rightarrow \frac{x}{1-x}+1+\frac{1-x}{x}+1={\left(\frac{13}{6}\right)}^2\Rightarrow \frac{1}{1-x}+\frac{1}{x}=\frac{169}{36}\Rightarrow 169x^2-169x+36=0\Rightarrow 169x^2-52x-117x+36=0\Rightarrow (13x-9)(13x-4)=0\therefore x=\frac{9}{13},\frac{4}{13}\left(Ans\right)$