Question

Please answer quick because tomorrow is my exam
$\frac{ \sqrt{ \frac{a}{b} + x } + \sqrt{ \frac{a}{b} - x} }{ \sqrt{ \frac{a}{b} + x} - \sqrt{ \frac{a}{b} - x} } = c$
then bx = ?​

Answer 1

Answer:

Use rationalization

$\frac{ \sqrt{ \frac{a}{b} + x} + \sqrt{ \frac{a}{b} - x } }{\sqrt{ \frac{a}{b} + x} - \sqrt{ \frac{a}{b} - x } } \times \frac{\sqrt{ \frac{a}{b} + x} + \sqrt{ \frac{a}{b} - x } }{\sqrt{ \frac{a}{b} + x} + \sqrt{ \frac{a}{b} - x } } \\ \\ = \frac{(\sqrt{ \frac{a}{b} + x} + \sqrt{ \frac{a}{b} - x }) {}^{2} }{(\sqrt{ \frac{a}{b} + x}) {}^{2} - (\sqrt{ \frac{a}{b} - x } ) {}^{2} } \\ \\ = \frac{ \frac{a}{b} + x + \frac{a}{b} - x + 2 \sqrt{( \frac{a}{b} + x) ( \frac{a}{b} - x)}}{ \frac{a}{b} + x - ( \frac{a}{b} - x)} \\ \\ = \frac{ \frac{2a}{b} + 2 \sqrt{ {( \frac{a}{b}) }^{2} - {x}^{2} } }{2x} \times \frac{b}{b} \\ \\ = \frac{a + \sqrt{ {a}^{2} - {b}^{2} {x}^{2} } }{bx}$