Question

Solve for x:
$\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}(x\ne 0.1).$

• A. 4/13 and 10/13
• B. 4/13 and 9/13
• C. 5/13 and 10/13
• D. 6/13 and 2/13

Answer 1

The correct option is B 4/13 and 9/13

Let $\sqrt{\frac{x}{1-x}}=y$
$\Rightarrow y+\frac{1}{y}=\frac{13}{6}\Rightarrow \frac{y^2+1}{y}=\frac{13}{6}$

$\Rightarrow y^2+1=\frac{13}{6}y\Rightarrow y^2-\frac{13}{6}y+1=0$

$\Rightarrow y^2-\frac{2.13}{2.6}y+1=0$

$\Rightarrow y^2-2.y.\left(\frac{13}{12}\right)+1=0$

$\Rightarrow y^2-2.y.\left(\frac{13}{12}\right)+(\frac{13}{12}{)}^2=-1+(\frac{13}{12}{)}^2=-1+\frac{169}{144}$

$\Rightarrow (y-\frac{13}{12}{)}^2=\pm \sqrt{\frac{25}{144}}$

$\Rightarrow y-\frac{13}{12}=\pm \frac{5}{12}$

$\Rightarrow y=\frac{13}{12}\pm \frac{5}{12}$

$\Rightarrow y=\frac{13}{12}+\frac{5}{12}$ or $\Rightarrow y=\frac{13}{12}-\frac{5}{12}$

$\Rightarrow y=\frac{18}{12}ory=\frac{8}{12}$

$\text{Take y}=\frac{3}{2}$

$\Rightarrow (\sqrt{\frac{x}{1-x}}{)}^2=(\frac{2}{3}{)}^2$
$\Rightarrow \frac{x}{1-x}=\frac{4}{9}$
$\Rightarrow 9x=4-4x$
$\Rightarrow 9x+4x=4$
$\Rightarrow 13x=4$
$x=\frac{4}{13}$

$\text{Now Take y}=\frac{2}{3}$

$\Rightarrow (\sqrt{\frac{x}{1-x}}{)}^2=(\frac{3}{2}{)}^2$
$\Rightarrow \frac{x}{1-x}=\frac{9}{4}$
$\Rightarrow 4x=9-9x$
$\Rightarrow 4x+9x=9$
$\Rightarrow 13x=9$
$x=\frac{9}{13}$
$\Rightarrow x=\frac{4}{13}or\frac{9}{13}$
So, b is the correct option.