Rationalising the Denominator

Rationalising the Denominator

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 won’t make a difference to the overall value.

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

It’s 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 – 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 \(\frac{\sqrt{2}}{\sqrt{8}\text{ }-\text{ }2} \times \frac{\sqrt{8}\text{ }+\text{ }2}{\sqrt{8}\text{ }+\text{ }2}\). Let’s 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 – 4
                                        = 4

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

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