Question

The sum of values of $x$ satisfying the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$ is

• A. $1$
• B. $\frac{5}{13}$
• C. $-\frac{5}{13}$
• D. $-1$

Answer 1

The correct option is A $1$

$\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$

For the square root's to be defined,
$\frac{x}{1-x}\ge 0$ and $\frac{1-x}{x}\ge 0$
$\Rightarrow \frac{x}{x-1}\le 0$ and $\frac{x-1}{x}\le 0$
$\Rightarrow x\in (0,1)$
Assuming
$\sqrt{\frac{x}{1-x}}=t$
So,
$t+\frac{1}{t}=\frac{13}{6}$
$\Rightarrow 6t^2-13t+6=0\Rightarrow (2t-3)(3t-2)=0\Rightarrow t=\frac{3}{2},\frac{2}{3}\Rightarrow \frac{x}{1-x}=\frac{9}{4},\frac{4}{9}\Rightarrow x=\frac{9}{13},\frac{4}{13}$
Both of them are valid solutions as $x\in (0,1)$
Hence, the sum of value of $x$ is $1$.