Paper 1 · surds / rationalisation

Express $\dfrac{6}{3 + \sqrt{5} + \sqrt{14}}$ in the form $a + b\sqrt{c} + d\sqrt{e}$ .

1 + \tfrac{3}{5}\sqrt{5} - \tfrac{1}{5}\sqrt{70}

1 - \tfrac{3}{5}\sqrt{5} + \tfrac{1}{5}\sqrt{70}

3 + \sqrt{5} - \sqrt{14}

1 + \tfrac{3}{5}\sqrt{5} + \tfrac{1}{5}\sqrt{70}

Explanation: Multiplying by

\tfrac{3+\sqrt5-\sqrt{14}}{3+\sqrt5-\sqrt{14}}

makes the denominator

(3+\sqrt5)^2-14=6\sqrt5

, leaving

\tfrac{3+\sqrt5-\sqrt{14}}{\sqrt5}=\tfrac{3\sqrt5+5-\sqrt{70}}{5}=1+\tfrac35\sqrt5-\tfrac15\sqrt{70}

Derived from ZIMSEC Mathematics Paper 1, November 2014, Q1

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