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Algebra

Question 20 of 20

Surds

Question

Rationalise the denominator in \(\dfrac{5+\sqrt{6}}{\sqrt{3}-\sqrt{2}}\)

Show solution Hide solution Fully worked — 6 steps
  1. Multiply the numerator and denominator by \(\sqrt{3} + \sqrt{2}\)
    \[ \frac{5+\sqrt{6}}{\sqrt{3}-\sqrt{2}} = \frac{5+\sqrt{6}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \]
  2. Expand the brackets

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

    \[ = \frac{5\sqrt{3}+5\sqrt{2}+\sqrt{6 \times 3}+\sqrt{6 \times 2}}{3-2} \]
  3. Evaluate the products

    Evaluating \(6 \times 3\) as \(18\), \(6 \times 2\) as \(12\), and \(3 - 2\) as \(1\):

    \[ = \frac{5\sqrt{3}+5\sqrt{2}+\sqrt{18}+\sqrt{12}}{1} \]
  4. Split \(\sqrt{18}\) and \(\sqrt{12}\) into perfect-square factors

    Dividing by \(1\) leaves the numerator unchanged. Using the rule \(\sqrt{(a \times b)} = \sqrt{a} \times \sqrt{b}\):

    \[ = 5\sqrt{3} + 5\sqrt{2} + \left(\sqrt{9} \times \sqrt{2}\right) + \left(\sqrt{4} \times \sqrt{3}\right) \]
  5. Simplify

    Evaluating \(\sqrt{9}\) as \(3\) and \(\sqrt{4}\) as \(2\):

    \[ = 5\sqrt{3} + 5\sqrt{2} + 3\sqrt{2} + 2\sqrt{3} \]
  6. Final answer

    Collecting the like surds, using the rule \(a\sqrt{c} \pm b\sqrt{c} = (a \pm b)\sqrt{c}\):

    \[ \frac{5+\sqrt{6}}{\sqrt{3}-\sqrt{2}} = (5 + 2)\sqrt{3} + (5 + 3)\sqrt{2} \]

    That is, with the denominator rationalised, the answer is seven root three plus eight root two.