Find the value of x if \(\frac{\sqrt{2}}{x + \sqrt{2}}\) = \(\frac{1}{x - \sqrt{2}}\)
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Correct Answer: Option A
Explanation:
\(\frac{\sqrt{2}}{x + 2}\) = 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
\(\frac{\sqrt{2}}{x + 2}\) = 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