Question

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

Answer 1

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

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

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

$6=13\sqrt{x(1-x)}$

$36=169x(1-x)$

$36=169x-169x^2$

$169x^2-169x+36=0$

$169(x^2-x+\frac{36}{169})=0$

$x=\frac{1\pm \sqrt{1-\frac{144}{169}}}{2}$

$=\frac{13\pm \sqrt{169-144}}{26}$

$=\frac{13+5}{26}=\frac{18}{26},\frac{8}{26}$

$\therefore x=\frac{9}{13},\frac{4}{13}$