Rationalising denominators
A fraction whose denominatorThe bottom part of a fraction. For 鈪, the denominator is 8, which represents 'eighths'. is a surd can be simplified by making the denominator rational numberA number that can be written in fraction form. This includes integers, terminating decimals, repeating decimals and fractions.. This process is called rationalising the denominator.
If the denominator has just one term that is a surd, the denominator can be rationalised by multiplying the numeratorThe top part of a fraction. For 鈪 , the numerator is 5. and denominator by that surd.
Example
Rationalise the denominator of \(\frac{\sqrt{8}}{\sqrt{6}}\).
The denominator can be rationalised by multiplying the numerator and denominator by 鈭6. (This has the same effect as multiplying it by 1, as \(\frac{\sqrt{6}}{\sqrt{6}} = 1\), which does not change it apart from removing the surd from the denominator, making subsequent surd calculations easier.)
\(\frac{\sqrt{8} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \frac{\sqrt{48}}{6} = \frac{\sqrt{(16 \times 3)}}{6} = \frac{4 \sqrt{3}}{6} = \frac{2 \sqrt{3}}{3}\)
Question
Rationalise the denominator of the following:
- \(\frac{1}{\sqrt{2}}\)
- \(\frac{\sqrt{3}}{\sqrt{2}}\)
- \(\frac{5}{2 \sqrt{3}}\)
- \(\frac{\sqrt{2}}{2}\)
- \(\frac{\sqrt{6}}{2}\)
- \(\frac{5 \sqrt{3}}{6}\) Multiply the numerator and denominator by \(\sqrt{3}\). It is not wrong to multiply by \(2\sqrt{3}\) but you would need to simplify your answer at the end.
Rationalising the denominator when the denominator has a rational term and a surd
If the denominator of a fraction includes a rational number, add or subtract a surd, swap the + or 鈥 sign and multiply the numerator and denominator by this expression.
For example, if the denominator includes the bracket \((5 + 2\sqrt{3})\), then multiply the numerator and denominator by \((5 - 2\sqrt{3})\).
Example
Rationalise the denominator of \(\frac{5}{3-\sqrt{2}}\).
Rationalise the denominator by multiplying the numerator and denominator by \(3 + \sqrt{2}\).
\(\frac{5}{3-\sqrt{2}} = \frac{5(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})} = \frac{15+5\sqrt{2}}{9+3\sqrt{2}-3\sqrt{2}-2} = \frac{15+5\sqrt{2}}{7}\)
Question
Rationalise the denominator of \(\frac{11}{6-2\sqrt{5}}\) giving your answer in its simplest form.
Rationalise the denominator by multiplying the numerator and denominator by \(6 + 2\sqrt{5}\).
\(\frac{11}{6+2\sqrt{5}} = \frac{11(6+2\sqrt{5})}{(6+2\sqrt{5})(6-2\sqrt{5})} = \frac{66+22\sqrt{5}}{36+12\sqrt{5}-12\sqrt{5}-20} = \frac{66+22\sqrt{5}}{16} = \frac{33+11\sqrt{5}}{8}\)