Find the value of x if \(\frac{\sqrt{2}}{x + \sqrt{2}}\) = \(\frac{1}{x - \sqrt{2}}\) 

A.

3\(\sqrt{2}\) + 4

B.

3\(\sqrt{2}\) - 4

C.

3 - 2\(\sqrt{2}\)

D.

4 + 2\(\sqrt{2}\)

Correct answer is A

\(\frac{\sqrt{2}}{x + 2}\) = x - \(\frac{1}{\sqrt{2}}\)

x\(\sqrt{2}\) (x - \(\sqrt{2}\)) = x + \(\sqrt{2}\) (cross multiply)

x\(\sqrt{2}\) - 2 = x + \(\sqrt{2}\) 

= x\(\sqrt{2}\) - x 

= 2 + \(\sqrt{2}\)

x (\(\sqrt{2}\) - 1) = 2 + \(\sqrt{2}\)

= \(\frac{2 + \sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}\)

x = \(\frac{2 \sqrt{2} + 2 + 2 + \sqrt{2}}{2 - 1}\) 

= 3\(\sqrt{2}\) + 4