Rationalizing Surds - SS3 Mathematics Lesson Note

To solve surds, it is often necessary to use a method known as rationalizing the denominator. Here, the numerator and denominator of a surd by its conjugate. The conjugate of \(\sqrt{a} - \sqrt{b}\) is \(\sqrt{a} + \sqrt{b}\) so: \(\left( \sqrt{a} - \sqrt{b} \right)\left( \sqrt{a} + \sqrt{b} \right) = {(\sqrt{a})}^{2} - \left( \sqrt{b} \right)^{2} = a - b\).

Example 2 Simply \(\frac{1}{2 + \sqrt{2}}\)

Solution

\[\frac{1}{2 + \sqrt{2}} = \ \frac{1}{2 + \sqrt{2}}\ .\ \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \ \frac{2 - \sqrt{2}}{2^{2} - \left( \sqrt{2} \right)^{2}} = \ \frac{2 - \sqrt{2}}{4 - 2} = \frac{2 - \sqrt{2}}{2}\]

Example 3 Simplify \(1 - \frac{\sqrt{3}}{\sqrt{12}}\)

Solution

\[1 - \frac{\sqrt{3}}{\sqrt{12}} = \ \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12}}\]

\[\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12}}.\frac{\sqrt{12}}{\sqrt{12}} = \ \frac{\sqrt{12}(\sqrt{12} - \sqrt{3})}{\left( \sqrt{12} \right)^{2}} = \ \frac{12 - \sqrt{36}}{12} = \frac{12 - 6}{12} = \ \frac{6}{12} = \frac{1}{2}\]