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

Mathematics
JAMB 1999

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

  • A. 3√2+4
  • B. 3√2-4
  • C. 3-2√2
  • D. 4+2√2
Correct Answer: Option A
Explanation

Start your solution by cross-multiplying,

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

[x - √2] √2  = x + √2

where √2×√2=2 

x√2 - √2 * √2 = x + √2

then collect like terms

x√2 - x = √2 + 2

and factorize accordingly to get the unknown.

x(√2 - 1) = √2 + 2

x = \(\frac{√2 + 2}{√2 - 1}\)

rationalize 

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

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

x = \(\frac{2 +  3√2 + 2}{2 - 1}\)

x =  \(\frac{3√2 + 4}{1}\)

x = 3√2 + 4


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