01.06 Rationalising Denominators
In this lesson you will learn how to:
- Rationalise denominators of surds
Prerequisites from previous chapters:
- N/A
VIDEO LESSON
INTERACTIVE SELF-STUDY
In the previous lesson on Surds we looked at how to multiply expressions containing surds
Expand and simplify the following expressions
i. \(m\sqrt{a}\times \sqrt{a}\)
iii. \((m\sqrt{a}+n\sqrt{b})(m\sqrt{a}-n\sqrt{b})\)
Notice that these all result in rational expressions; there are no more surds.
\((m+n\sqrt{a})\) and \((m-n\sqrt{a})\) differ by the sign in between the two terms. They are called conjugate pairs
Similarly \((m\sqrt{a}+n\sqrt{b})\) and \((m\sqrt{a}-n\sqrt{b})\) are also conjugates
- We can see that when a surd involving square roots is multiplied by its conjugate, the product is a difference of two squares, which is rational.
Some fractions may contain a surd in the denominator. It is more useful to re-express the fraction so that the denominator is a rational number. This is called rationalising the denominator
In the warm up questions above, we have seen what we can multiply surds by in order to produce a rational expression
- To rationalise the denominator for a fraction in the form \(\displaystyle \frac{1}{m\sqrt{a}}\), we create an equivalent fraction by multiplying the numerator and denominator by \(\sqrt{a}\)
- To rationalise the denominator for a fraction in the form \(\displaystyle \frac{1}{m+n\sqrt{a}}\), \(\displaystyle \frac{1}{m-n\sqrt{a}}\), \(\displaystyle \frac{1}{m\sqrt{a}+n\sqrt{b}}\), or \(\displaystyle \frac{1}{m\sqrt{a}-n\sqrt{b}}\), we create an equivalent fraction by multiplying the numerator and denominator by the conjugate of the denominator
Remember to use the laws of surds \(\sqrt{ab}=\sqrt{a}\sqrt{b}\) and \(\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) to help you simplify surds
Example 1 – (GRADE D)
Express the following surds in their simplest form and rationalise their denominators
a. \(\displaystyle \frac{6}{\sqrt{2}}\)
c. \(\displaystyle \frac{1}{4\sqrt{3}}\)
d. \(\displaystyle \frac{1}{5+7\sqrt{3}}\)
e. \(\displaystyle \frac{\sqrt{6}}{9\sqrt{2}-4\sqrt{7}}\)
f. \(\displaystyle \frac{1}{(2+\sqrt{5})^2}\)
END OF LESSON!
You can now answer:
EDEXCEL Pure Year 1 Ex 1F – All Questions
OCR A Student Book 1 Ex 2B – Q5 Q6 Q7 Q10 Q15*