01Grationalising

= Rationalising Denominators = toc It is easier to deal with fractions containing surds if the denominator is rational (or more commonly, an integer).

This process takes a fraction with a surd in the denominator and changes it to an equivalent fraction with a rational denominator.

Single Surds
If the fraction has a single surd as the denominator,
 * multiply the fraction by the surd from the denominator over itself and then simplify
 * this is equivalent to multiplying by 1 so the value of the fraction is not changed.

** Example 1 **
math \textbf{(a)} \quad \text{Rationalise the denominator of } \dfrac{3}{\sqrt{5}} math

math \\ \dfrac{3}{\sqrt{5}} = \dfrac{3}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} \\ \\ \\ . \quad = \dfrac{3\,\sqrt{5}}{\sqrt{25}} \qquad \qquad \textit{but } \sqrt{25}=5 \textit{ so :}\\ \\ \\ . \quad = \dfrac{3\,\sqrt{5}}{5} math

math \textbf{(b)} \quad \text{Write } \dfrac{7\,\sqrt{2}}{4\,\sqrt{7}} \; \text{ as a fraction with a rational denominator} math

math \\ \dfrac{7\,\sqrt{2}}{4\,\sqrt{7}} = \dfrac{7\,\sqrt{2}}{4\,\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} \\ \\ \\ . \qquad = \dfrac{7\,\sqrt{14}}{4\,\sqrt{49}} \qquad \qquad \textit{but }\sqrt{49} = 7 \textit{ so :}\\ \\ \\ . \qquad = \dfrac{7 \, \sqrt{14}}{28} \qquad \qquad \textit{cancel the 7 with the 28}\\ \\ \\ . \qquad = \dfrac{\sqrt{14}}{4} math

Binomial Denominator
If the denominator is a binomial (has 2 terms), then you need to create the ** conjugate surd ** of the denominator.

The conjugate surd of a binomial is the same expression but swap the sign between the two terms.

math \\ \textbf{Eg : } 3 + \sqrt{2} \; \Longrightarrow \; 3 - \sqrt{2} \\ \\ \textbf{Eg : } \sqrt{5} - 3\,\sqrt{7} \; \Longrightarrow \; \sqrt{5} + 3\,\sqrt{7} math

To rationalise the denominator of a fraction with a binomial denominator, multiply by the conjugate surd over itself and simplify

** Example 2 **
math \text{Write } \dfrac{5}{2 + \sqrt{3}} \; \text{ as a fraction with a rational denominator} math

math \\ \dfrac{5}{2 + \sqrt{3}} = \dfrac{5}{2 + \sqrt{3}} \times \dfrac{2- \sqrt{3}}{2 - \sqrt{3}} \\ \\ \\ . \qquad \quad = \dfrac{5 \big(2- \sqrt{3} \big)}{ \big(2+\sqrt{3} \big)\big( 2 - \sqrt{3} \big)} math

math \\ . \qquad \quad = \dfrac{5 \big(2 - \sqrt{3} \big)}{ 4 - 2\,\sqrt{3} +2\,\sqrt{3} -\sqrt{9}} \\ \\ \\ . \qquad \quad = \dfrac{5 \big( 2 - \sqrt{3} \big)}{4 - 3} \\ \\ \\ . \qquad \quad = \dfrac{5 \big( 2 - \sqrt{3} \big)}{1} \\ \\ \\ . \qquad \quad = 5 \big( 2 - \sqrt{3} \big) math

For another site that explains this process, go here: MathIsFun

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