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

Start your solution by cross-multiplying,

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

[x - \(\sqrt{2}\)]\(\sqrt{2}\) = x + \(\sqrt{2}\)

where \(\sqrt{2}\) ×\(\sqrt{2}\) = 2

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

then collect like terms

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

and factorize accordingly to get the unknown.

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

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

rationalize

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

x = \(\frac{\sqrt{4} + \sqrt{2} + 2\sqrt{2} + 2}{\sqrt{4} + \sqrt{2} - \sqrt{2} - 1}\)

x = \(\frac{2 + 3\sqrt{2} + 2}{2 - 1}\)

x = \(\frac{3\sqrt{2} + 4}{1}\)

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

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