Discussion
If \(x = \sqrt{3} + \sqrt{2}\), what is the value of \(\dfrac{1}{x}\) ?
Answer: C
Correct answer: \(\sqrt{3} - \sqrt{2}\)
Rationalise the denominator by multiplying above and below by the conjugate.
\[\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \sqrt{3}-\sqrt{2}\]
The denominator becomes \(a^2 - b^2\), which clears both surds at once.
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