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Algebra

Question 18 of 20

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Question

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

Show solution Hide solution Fully worked — 4 steps
  1. Multiply the numerator and denominator by \(\sqrt{3} - 2\)
    \[ \frac{7}{\sqrt{3}+2} = \frac{7}{\sqrt{3}+2} \times \frac{\sqrt{3}-2}{\sqrt{3}-2} \]
  2. Expand the brackets

    In the numerator \(7(\sqrt{3} - 2) = 7\sqrt{3} - 14\), and in the denominator \((\sqrt{3} + 2)(\sqrt{3} - 2) = 3 - 4\):

    \[ = \frac{7\sqrt{3}-14}{3-4} \]
  3. Evaluate the denominator

    Evaluating \(3 - 4\) as \(-1\):

    \[ = \frac{7\sqrt{3}-14}{-1} \]
  4. Final answer

    Dividing the numerator by \(-1\):

    \[ \frac{7}{\sqrt{3}+2} = -7\sqrt{3} + 14 \]

    That is, with the denominator rationalised, the answer is minus seven root three, plus fourteen.