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Algebra

Question 19 of 20

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Question

Rationalise the denominator in \(\dfrac{\sqrt{3}}{\sqrt{7}-\sqrt{2}}\)

Show solution Hide solution Fully worked — 3 steps
  1. Multiply the numerator and denominator by \(\sqrt{7} + \sqrt{2}\)
    \[ \frac{\sqrt{3}}{\sqrt{7}-\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{7}-\sqrt{2}} \times \frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}} \]
  2. Expand the brackets

    Using the rule \(\sqrt{a} \times \sqrt{b} = \sqrt{(a \times b)}\):

    \[ = \frac{\sqrt{3 \times 7}+\sqrt{3 \times 2}}{7-2} \]
  3. Final answer

    Evaluating \(3 \times 7\) as \(21\), \(3 \times 2\) as \(6\), and \(7 - 2\) as \(5\):

    \[ \frac{\sqrt{3}}{\sqrt{7}-\sqrt{2}} = \frac{\sqrt{21}+\sqrt{6}}{5} \]

    That is, with the denominator rationalised, the answer is root twenty-one plus root six, all over five.