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Algebra

Question 16 of 20

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Question

Rationalise the denominator in \(\dfrac{4\sqrt{2}}{\sqrt{5}}\left(\sqrt{2} + \sqrt{8}\right)\)

Show solution Hide solution Fully worked — 6 steps
  1. Expand the bracket

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

    \[ \frac{4\sqrt{2}}{\sqrt{5}}\left(\sqrt{2} + \sqrt{8}\right) = \frac{4(\sqrt{2})^{2} + 4\sqrt{2 \times 8}}{\sqrt{5}} \]
  2. Simplify the numerator

    \((\sqrt{2})^{2} = 2\), so \(4(\sqrt{2})^{2} = 8\), and \(2 \times 8 = 16\):

    \[ = \frac{8 + 4\sqrt{16}}{\sqrt{5}} \]
  3. Evaluate \(4\sqrt{16}\)

    \(\sqrt{16} = 4\), so \(4\sqrt{16} = 16\):

    \[ = \frac{8 + 16}{\sqrt{5}} \]
  4. Add \(8\) and \(16\)
    \[ = \frac{24}{\sqrt{5}} \]
  5. Multiply the numerator and denominator by \(\sqrt{5}\)

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

    \[ = \frac{24}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \]
  6. Final answer
    \[ \frac{4\sqrt{2}}{\sqrt{5}}\left(\sqrt{2} + \sqrt{8}\right) = \frac{24\sqrt{5}}{5} \]

    That is, with the denominator rationalised, the answer is twenty-four times the square root of five, over five.