JAMB 1999 · UME · Q2

Find the value of $x$ if $\dfrac{\sqrt2}{x + \sqrt2} = \dfrac{1}{x - \sqrt2}$ .

Worked solution (try it first)

Cross-multiply: $\sqrt2(x - \sqrt2) = x + \sqrt2$ , so $\sqrt2x - 2 = x + \sqrt2$ .
Collect the $x$ terms: $\sqrt2x - x = 2 + \sqrt2$ , so $x(\sqrt2 - 1) = 2 + \sqrt2$ and $x = \dfrac{2 + \sqrt2}{\sqrt2 - 1}$ .
Multiply the top and bottom by $\sqrt2 + 1$ .
The bottom becomes $2 - 1 = 1$ .
Top: $(2 + \sqrt2)(\sqrt2 + 1) = 2\sqrt2 + 2 + 2 + \sqrt2$ , which is $4 + 3\sqrt2$ .
So option A.