Question

plese solve following euqation

Answer 1

$let \: \: \sqrt{ \frac{x}{1 - x} } = y - - - (1) \\ the \: \: equation \: \: becomes \\ y + \frac{1}{y} = \frac{13}{6} \\ 6 {y}^{2} - 13y + 6 = 0 \\ 6 {y}^{2} - 9y - 4y + 6 = 0 \\ 3y(2y - 3) - 2(2y - 3) = 0 \\ (3y - 2)(2y - 3) = 0 \\ y = \frac{2}{3} \: \: or \: \: \frac{3}{2}$
$substituting \: \: y = \frac{2}{3} \: \: in \: \: (1) \\ \frac{x}{1 - x} = \frac{4}{9} \\ 9x = 4 - 4x \\ x = \frac{4}{13}$
$substituting \: \: y = \frac{3}{2} \: \: in \: \: (1) \\ \frac{x}{1 - x} = \frac{9}{4} \\ 4x = 9 - 9x \\ x = \frac{9}{13}$
$therefore \: \: roots \: \: are \\ \frac{4}{13} \: \: and \: \: \frac{9}{13}$

Answer 2

answer is 4/13 or 9/13