Question

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

$\sqrt{ \frac{x}{1 - x} } + \sqrt{ \frac{1 - x}{x} } = \frac{13}{6} \\ {( \frac{ \sqrt{x} }{ \sqrt{1 - x} } + \frac{ \sqrt{1 - x} }{ \sqrt{x} }) }^{2} = {( \frac{13}{6} )}^{2} \\ \frac{ {x} }{1 - x} + \frac{1 - x}{x} + 2 = \frac{169}{36} \\ \frac{ {x}^{2} + {(1 - x)}^{2} }{x - {x}^{2} } = \frac{169 - 72}{36} \\ \frac{ {x}^{2} - 1 + 2x - {x}^{2} }{x - {x}^{2} } = \frac{97}{36} \\ \frac{ 2x - 1 }{x - {x}^{2} } = \frac{97}{36} \\ 72x - 36 = 97x - 97 {x}^{2} \\ 97 {x}^{2} - 97x + 72x - 36 = 0 \\ 97 {x}^{2} - 25x - 36 = 0$