Surds

UK Secondary (7-11)

Rationalise the denominator

Lesson

What happens when you have a fraction with surds in it? Well it's a little difficult to work with, especially if there are surds in the denominator.

$\frac{2\sqrt{3}}{5}$2√35 can be rewritten as $\frac{2}{5}\sqrt{3}$25√3, so it just becomes a simple surd with a fraction in front of it.

But we're in trouble when we have something like $\frac{8}{3\sqrt{7}}$83√7, how do we work with that??

You've probably realised by now that mathematicians are quite lazy and like to simplify things as much as they can, so they've devised a way to rewrite these complicated fractions without the surds on the bottom.

This is called **rationalising the denominator**, and it does not change the value of the fraction.

We know that when we square a surd, the answer is always going to be rational and without surds! This is the key we need to rationalise fractions such as $\frac{8}{3\sqrt{7}}$83√7.

What do you think will happen if we multiply the denominator here by $\sqrt{7}$√7? Well, the square root sign will disappear right? However, we need the fraction to still have the same value, so let's try multiplying the fraction by $\frac{\sqrt{7}}{\sqrt{7}}=1$√7√7=1, which will not change the fraction at all!

$\frac{8}{3\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{8\times\sqrt{7}}{3\times\sqrt{7}\times\sqrt{7}}$83√7×√7√7=8×√73×√7×√7

which simplifies down to

$\frac{8\sqrt{7}}{3\left(\sqrt{7}\right)^2}=\frac{8\sqrt{7}}{3\times7}$8√73(√7)2=8√73×7

This answer can again be finally simplified down to $\frac{8\sqrt{7}}{21}$8√721.

The fraction now looks completely different! But try putting it in your calculator, do you get the same value as $\frac{8}{3\sqrt{7}}$83√7?

So now we know one way of rationalising the denominator is to **multiply top and bottom by the surd in the denominator**.

The above technique is great for when there is only one term in the denominator, but it doesn't really work when we have more complicated expressions.

Let's take a look at an example: $\frac{5}{\sqrt{6}-1}$5√6−1.

If we use the squares technique and multiply the bottom by $\sqrt{6}$√6, it'll become $6-\sqrt{6}$6−√6, which has not gotten rid of the surd at all!

Let's see what happens when we multiply$\sqrt{6}-1$√6−1 by $\sqrt{6}+1$√6+1.

Does the expression $\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)$(√6−1)(√6+1) look similar to you? Of course, it's a difference of two squares!

Then the answer is a simple case of $\left(\sqrt{6}\right)^2-1^2$(√6)2−12 = $6-1=5$6−1=5.

Well what do you know, this answer doesn't have any surds!

Let's use this to transform the above fraction.

$\frac{5}{\sqrt{6}-1}\times\frac{\sqrt{6}+1}{\sqrt{6}+1}$5√6−1×√6+1√6+1 | $=$= | $\frac{5\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}$5(√6+1)(√6−1)(√6+1) |

$=$= | $\frac{5\sqrt{6}+5}{5}$5√6+55 |

We can go even further and divide top and bottom by $5$5, which gives us the final simplified answer $\frac{\sqrt{6}+1}{1}=\sqrt{6}+1$√6+11=√6+1.

Let's have a look at these worked examples.

Rationalise the denominator for the given expression: $\frac{2}{\sqrt{7}}$2√7

Rationalise the denominator for the given expression: $\frac{8\sqrt{5}}{2\sqrt{3}}$8√52√3

Simplify, expressing your answer with a rational denominator:

$\frac{9}{\sqrt{5}-7}$9√5−7

Express the following fraction in simplest surd form with a rational denominator:

$\frac{\sqrt{7}}{5\sqrt{6}+4\sqrt{2}}$√75√6+4√2