Rationalising Denominators (Edexcel IGCSE Maths)

Spec point — 1.4B

Rationalising denominators

What does rationalising the denominator mean?

• If a fraction has a denominator containing a surd then it has an irrational denominator

  • E.g. $\frac{4}{\sqrt{5}}$ or $\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}}$

• The fraction can be rewritten as an equivalent fraction, but with a rational denominator

  • E.g. $\frac{4\sqrt{5}}{5}$ or $\frac{\sqrt{6}}{3}$

• The numerator may contain a surd, but the denominator is rationalised

How do I rationalise simple denominators?

• If the denominator is a surd:

  • Multiply the top and bottom of the fraction by the surd on the denominator

    • $\frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times\frac{\sqrt{b}}{\sqrt{b}}$
    • This is equivalent to multiplying by 1, so does not change the value of the fraction
    • $\sqrt{b}\times\sqrt{b}=b$ so the denominator is no longer a surd

  • Multiply the fractions as you would usually, and simplify if needed

    • $\frac{a\sqrt{b}}{b}$

How do I rationalise harder denominators?

• If the denominator is an expression containing a surd:

  • For example $\frac{2}{3+\sqrt{5}}$

  • Multiply the top and bottom of the fraction by the expression on the denominator, but with the sign changed

    • $\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}}\times\frac{3-\sqrt{5}}{3-\sqrt{5}}$
    • This is equivalent to multiplying by 1, so does not change the value of the fraction

  • Multiply the fractions as you would usually (use brackets to help)

    • $\frac{2}{3 + \sqrt{5} } = \frac{2 ( 3 - \sqrt{5} )}{( 3 + \sqrt{5} ) ( 3 - \sqrt{5} )}$

  • Expand any brackets, and simplify

    • $\frac{2}{3 + \sqrt{5} } = \frac{6 - 2 \sqrt{5} }{( 3 \times 3 ) + 3 \sqrt{5} - 3 \sqrt{5} - ( \sqrt{5} \sqrt{5} )} = \frac{6 - 2 \sqrt{5} }{9 - ( 5 )}$

  • You can use the difference of two squares to expand the denominator quickly

    • $( a + \sqrt{b} ) ( a - \sqrt{b} ) = a^{2} - ( \sqrt{b} )^{2} = a^{2} - b$
    • This is what makes the denominator rational

  • Simplify

    • $\frac{2}{3+\sqrt{5}}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$

Exam Hint If your answer still has a surd on the bottom, go back and check your working!

Worked Example Write $\frac{4}{\sqrt{6}-2}$ in the form $p+q\sqrt{r}$ where $p,q$and $r$ are integers and $r$ has no square factors.

Answer:

Multiply the top and bottom of the fraction by the expression on the denominator, but with the sign changed

$\frac{4}{\sqrt{6}-2}=\frac{4}{\sqrt{6}-2}\times\frac{\sqrt{6}+2}{\sqrt{6}+2}$

Multiply the fractions as you would usually

$\frac{4}{ \sqrt{6} - 2} = \frac{4 ( \sqrt{6} + 2 )}{( \sqrt{6} - 2 ) ( \sqrt{6} + 2 )}$

Expand the brackets The denominator can be expanded using the difference of two squares

$\frac{4}{ \sqrt{6} - 2} = \frac{4 \sqrt{6} + 8}{( \sqrt{6} )^{2} - ( 2 )^{2}} = \frac{4 \sqrt{6} + 8}{6 - 4}$

Simplify

$\frac{4}{\sqrt{6}-2}=\frac{4\sqrt{6}+8}{2}=2\sqrt{6}+4$

Write in the form given in the question

$4+2\sqrt{6}\,p=4\,q=2\,r=6$