This guide has everything you need to know about **rationalising the denominator**. You need to be confident with simplifying surds and expanding brackets with surds.

Rationalising the denominator essentially means manipulating a fraction so there are only integers (whole numbers) in the denominator. It makes it neater and easier to work with.

**Example 1**

Rationalise the denominator of \(\frac{10}{\sqrt{45}}\).

You should always start by **simplifying** any surds. In this case, we can simplify \(\sqrt{45}\).

\(\sqrt{45}\) = \(\sqrt{9} \times \sqrt{5}\)

= \(3\sqrt{5}\)

**Exam Tip**

Writing out the square numbers can make questions on surds a lot simpler.

Now, we can put this back into the fraction:

\(\frac{10}{\sqrt{45}}\) = \(\frac{10}{3\sqrt{5}}\)

Now, we multiply by \(\frac{\sqrt{5}}{\sqrt{5}}\). In the denominator, this will **eliminate** the surd but, since \(\frac{\sqrt{5}}{\sqrt{5}}\) is equivalent to 1, it wonβt make a difference to the overall value.

\(\frac{\sqrt{10}}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}\) = \(\frac{10\sqrt{5}}{3\times 5}\)

** **= \(\frac{10\sqrt{5}}{15}\) ** **= \(\frac{2\sqrt{5}}{3}\)

Itβs worth noting that you could start by multiplying by \(\sqrt{45}\) but you would end up with much larger numbers which are harder to simplify.

**Example 2**

Show that \(\frac{\sqrt{2}}{\sqrt{8}\text{ }-\text{ }2}\) can be written in the form \(\frac{a\text{ }+\text{ }b\sqrt{2}}{2}\) where \(a\) and \(b\) are integers.

In this case, we need to multiply the numerator and denominator by (\(\sqrt{8}\) + 2). This is a similar expression to the current denominator; we have just changed the sign in the middle β this is known as the **conjugate**.

When multiplying this by (\(\sqrt{8}\) + 2), the surd expression will cancel out. This uses an idea called the **difference of two squares**.

So, we are finding \(\frac{\sqrt{2}}{\sqrt{8}\text{ }-\text{ }2} \times \frac{\sqrt{8}\text{ }+\text{ }2}{\sqrt{8}\text{ }+\text{ }2}\). Letβs look at the numerator and denominator separately.

\(\sqrt{2}\left(\sqrt{8}\text{ }+\text{ }2\right)\) = \(\sqrt{16}\text{ }+\text{ }2\sqrt{2}\)

= 4 + 2\(\sqrt{2}\)

\(\left(\sqrt{8}\text{ }-\text{ }2\right)\left(\sqrt{8}\text{ }+\text{ }2\right)\) = \(\sqrt{64}\text{ }+\text{ }2\sqrt{8}\text{ }-\text{ }2\sqrt{8}\text{ }-\text{ }4\)

= 8 β 4

= 4

\(\frac{\sqrt{2}}{\sqrt{8}\text{ }-\text{ }2} \times \frac{\sqrt{8}\text{ }+\text{ }2}{\sqrt{8}\text{ }+\text{ }2}\) = \(\frac{4\text{ }+\text{ }2\sqrt{2}}{4}\)

= \(\frac{2\text{ }+\text{ }\sqrt{2}}{2}\)** **

Finished manipulating those fractions? Then why not take advantage of our greater range of blogs **here**!? You can also **subscribe to Beyond** for access to thousands of secondary teaching resources. You can **sign up for a free account here** and take a look around **at our free KS3/GCSE Maths resources** before you subscribe too.