# Simplifying Surds – A Level Maths Revision

As you make the vast step up to advanced study in maths, you will need a firm grasp on the fundamentals of simplifying surds.

Before you dive right into the nitty-gritty of worked examples, Beyond: Advanced can offer you the following resources (great for teachers and students, alike).

### Simplifying Surds – Background

The name “surd” comes from the Latin surdus, which translates as “mute” or “unspoken”. This used to apply to all irrational numbers but, now, it refers to a specific type of irrational number: those that can be written as the square root of an integer.

You will have come across Surds at GCSE; at A level it is important that you understand and can manipulate surds.

To simplify surds, there are two rules that you need to be able to use: $\sqrt{ab}=\sqrt{a} \times \sqrt{b}$ $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

## Simplifying Surds – Example Content

Example Question 1:

Simplify the surd $\sqrt{75}$.

When simplifying surds, we need to find a factor that is a square number. In this case, we can see that 25 is a factor of 75, so we can write: $\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25}\sqrt{3} = 5\sqrt{3}$

Example Question 2:

Simplify the surd $\sqrt{\frac {7}{81}}$.

The square root of a fraction can be calculated by finding the square root of both the denominator and numerator: $\sqrt{\frac{7}{81}} = \frac{\sqrt{7}}{\sqrt{81}} = \frac{\sqrt{7}}{9}$

Example Question 3:

Find $(3\sqrt{7} - 4\sqrt{6})^2$. Give your answer as a surd in its simplest form.

It is important to remember that $(a + b)^2 = (a + b)(a + b)$. We can expand brackets that include surds in exactly the same way as we would expand algebraic brackets: $(3\sqrt{7} - 4\sqrt{6})^2 = (3\sqrt{7} - 4\sqrt{6})(3\sqrt{7} - 4\sqrt{6})$ $= (3\sqrt{7})^2 - 2 \times (4\sqrt{6}) \times (3\sqrt{7}) + (4\sqrt{6})^2$ $= 63 - 24\sqrt{42} + 96$ $= 159 - 24\sqrt{42}$

Example Question 4:

Express $\sqrt{294} - \sqrt{150}$ in the form $k\sqrt{x}$ where $k$ and $x$ are integers.

It is hard to see how to simplify $\sqrt{294}$, so let’s start by looking at $\sqrt{150}$. You may be able to spot that 150 is a multiple of 25, so we can write: $\sqrt{150} = \sqrt{6 \times 25} = 5\sqrt{6}$

This then helps us simplify $\sqrt{294}$ as we can assume that 6 is a factor: $\sqrt{294} = \sqrt{6 \times 49} = 7\sqrt{6}$

Now, we can simplify $\sqrt{294} - \sqrt{150}$: $\sqrt{294} - \sqrt{150} = 7\sqrt{6} - 5\sqrt{6} = 2\sqrt{6}$

When a surd is written on the denominator of a fraction, we rationalise the denominator to make it simpler. We do this by multiplying the numerator and denominator by an expression that will simplify the surd on the denominator.

Example Question 5:

Simplify $\frac{5}{\sqrt{32}}$.

In this case, we start by simplifying $\sqrt{32}$: $\sqrt{32}=\sqrt{16 \times 2} = 4\sqrt{2}$

We now have: $\frac{5}{\sqrt{32}} = \frac{5}{4\sqrt{2}}$

We can rationalise the denominator by multiplying the numerator and denominator by $\sqrt{2}$.

We can do this because $\sqrt{2} \times \sqrt{2} = 2$, which removes the root, and $\frac{\sqrt{2}}{\sqrt{2}}=1$, so our multiplication does not affect the value of the fraction. $\frac{5}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{4\sqrt{2} \times \sqrt{2}}=\frac{5\sqrt{2}}{8}$

We could have rationalised this surd by multiplying the numerator and denominator by $\sqrt{32}$. However, by simplifying the surd first, we make sure our fraction is simplified and make the multiplication easier.

Example Question 6:

Simplify $\frac{4}{5 \: - \: 2\sqrt{3}}$.

To do this, we consider $5 + 2\sqrt{3}$. This is called the conjugate of $5 - 2\sqrt{3}$; it can be found by simply changing the sign of the surd. To help understand how this works, think back to the difference of two squares: $(a+b)(a-b)=a^2 - b^2$

When we expand $(5 + 2\sqrt{3})(5 - 2\sqrt{3})$, we get: $(5 + 2\sqrt{3})(5 - 2\sqrt{3}) = 5^2 - 5 \times 2\sqrt{3} + 5 \times 2\sqrt{3} - (2\sqrt{3})^2$ $= 25 - 4 \times 3$ $= 13$

We can now use this to remove the surd on the denominator: $\frac{4}{5 \: - \: 2\sqrt{3}} \times \frac{5 \: + \: 2\sqrt{3}}{5 \: + \: 2\sqrt{3}} = \frac{4(5 \: + \: 2\sqrt{3})}{(5 \: - \: 2\sqrt{3})(5 \: + \: 2\sqrt{3})}$ $= \frac{20 \: + \: 8\sqrt{3}}{13}$

So, to rationalise the denominator of a surd, we multiply the numerator and denominator by the conjugate of the denominator.

Example Question 7:

Simplify $\frac{4 \: + \: \sqrt{7}}{9 \: + \: 2\sqrt{7}}$.

This time, our conjugate is $9 - 2\sqrt{7}$; we multiply our numerator and denominator by this. Since the numerator is also a surd, you may find it helpful to multiply them separately: $(4+\sqrt{7})(9-2\sqrt{7}) = 36 + 9\sqrt{7} - 8\sqrt{7} - 14$ $= 22 + \sqrt{7}$ $(9 + 2\sqrt{7})(9 - 2\sqrt{7}) = 81 + 18\sqrt{7} - 18\sqrt{7} - 28$ $= 53$

Bringing these together gives: $\frac{4 \: + \: \sqrt{7}}{9 \: + \: 2\sqrt{7}} = \frac{4 \: + \: \sqrt{7}}{9 \: + \: 2\sqrt{7}} \times \frac{9 \: - \: 2\sqrt{7}}{9 \: - \: 2\sqrt{7}}$ $= \frac{22 \: + \: \sqrt{7}}{53}$

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