Question

$(\sqrt{x+a} )/(\sqrt{x-a} )- (\sqrt{x-a} )/(\sqrt{x+a} )=4\sqrt{2}$
please find the value of x . step by step please help me

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\ value(s) \: of \: x \\ \\ so \: given \: equation \: is \\ \frac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} } \: - \frac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} } = 4 \sqrt{2} \\ \\ \frac{( \sqrt{x} + \sqrt{a} \: )( \sqrt{x} + \sqrt{a}) - ( \sqrt{x} - \sqrt{a} )( \sqrt{x} - \sqrt{a}) }{( \sqrt{x} - \sqrt{a} )( \sqrt{x} + \sqrt{a} )} = 4 \sqrt{2} \\ \\ \frac{x + a + 2 \sqrt{ax} - (x + a - 2 \sqrt{ax} )}{x - a} = 4 \sqrt{2} \\ \\ \frac{x + a + 2 \sqrt{ax} - x - a + 2 \sqrt{ax} }{x - a} = 4 \sqrt{2} \\ \\ \frac{2.2 \sqrt{ax} }{x - a} = 4 \sqrt{2} \\ \\ \frac{4 \sqrt{ax} }{x - a} = 4 \sqrt{2} \\ \\ \frac{ \sqrt{ax} }{x - a} = \sqrt{2}$

$squaring \: both \: sides \\ we \: get \\ \\ \frac{ax}{(x - a) {}^{2} } = 2 \\ \\ ax = 2(x - a) {}^{2} \\ \\ ax = 2(x {}^{2} + a {}^{2} - 2ax) \\ \\ 2x {}^{2} + 2a {}^{2} - 4ax - ax = 0 \\ \\ 2x {}^{2} - 4ax - ax + 2a {}^{2} = 0 \\ \\ 2x(x - 2a) - a(x - 2a) = 0 \\ \\ (2x - a)(x - 2a) = 0 \\ \\ 2x - a = 0 \: \: or \: \: x - 2a = 0 \\ \\ x = \frac{a}{2} \: \: or \: \: x = 2a$

$so \: thus \: here \\ option \: (c) \: \: is \: correct$