Surds Questions | Worksheets and Revision | MME

# Surds Questions, Worksheets and Revision

Level 6 Level 7

## What you need to know

### Surds

A surd is a square root number that doesn’t give a whole number answer e.g. $\sqrt{3}$

More generally, we get a surd when we take the square root of a number that isn’t a square – so $\sqrt{2},\sqrt{3},\sqrt{5}$ are all surds. Like many things in maths, surds can be simplified. The process of simplifying surds is based on this multiplication rule:

$\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$

This topic will require a good understanding of:

### Take Note:

With surds, like with algebra, we often miss out the multiplication symbol and write $2\sqrt{7}$ instead of $2\times\sqrt{7}$.

If a question asks you to leave your answer in surd form, you should always simplify it. If it’s a question in a calculator paper, your calculator will handle surds for you.

Surds are irrational numbers, and we don’t like to have irrational numbers on the bottom of a fraction. So, by cleverly applying that multiplication rule mentioned above, we can make the denominator rational. You do this when questions ask you to rationalise the denominator.

### Example 1: Simplifying Surds

Write $\sqrt{28}$ in simplified surd form.

To do this, we need to think of a square number (bigger than 1) that goes into 28 in order to split up the surd. 4 goes into 28 seven times, so we get

$\sqrt{28}=\sqrt{4\times7}=\sqrt{4}\times\sqrt{7}$

This can then simplify because we purposefully chose a number that could be square rooted. This gives us

$\sqrt{28}=2\times\sqrt{7}=2\sqrt{7}$

This is now in what we call surd form, which means it can’t be simplified any further. To check if it can be simplified further, see if there are any square numbers that go into 7 (since 7 is now the number inside the square root). Clearly, there aren’t, so we’re done.

### Example 2: Rationalising the Denominator

Rationalise the denominator of $\dfrac{3}{\sqrt{5}}$.

As we know, you can multiply the top and bottom of a fraction by the same number without changing the fraction. In this case, we must multiply top and bottom by whatever surd is on the bottom of the fraction, here that’s $\sqrt{5}$. So, the numerator becomes

$3\times\sqrt{5}=3\sqrt{5}$

The denominator becomes

$\sqrt{5}\times\sqrt{5}=\sqrt{5\times5}=\sqrt{25}$

But we know the square root of 25 – it’s 5. So, the fraction becomes

$\dfrac{3}{\sqrt{5}}=\dfrac{3\sqrt{5}}{\sqrt{25}}=\dfrac{3\sqrt{5}}{5}$

The denominator no longer involves a surd, only a 5 – which is a rational number – and so we have successfully rationalised the denominator

### Example 3: Rationalising Surds

Rationalise the denominator of the following fraction.

$\dfrac{8}{5+\sqrt{2}}$

To rationalise the denominator of this fraction, we are going to multiply top and bottom by $(5-\sqrt{2})$. This is going to involve some bracket expanding. The numerator becomes

$8\times(5-\sqrt{2})=8\times5+8\times(-\sqrt{2})=40-8\sqrt{2}$

Now perform double bracket expansion. Remember you can always use the FOIL method. The denominator becomes

$(5+\sqrt{2})(5-\sqrt{2})=5\times5+5\times(-\sqrt{2})+5\times\sqrt{2}-\sqrt{2}\times\sqrt{2}$

So, we have the denominator

$25-5\sqrt{2}+5\sqrt{2}-\sqrt{2\times2}$

Remember: the aim is to end up with no surds on the bottom. The first term is 25, so that’s not a problem. Now, the middle two terms are $-\sqrt{2}$ and $5\sqrt{2}$. Since the first one is just the negative of the second one, these two terms cancel each other out. Then, finally, the last term is $-\sqrt{2\times2}=-sqrt{4}=-2$, which is another rational number. The denominator becomes

$25+0-2=23$

Therefore, the fraction and final answer is

$\dfrac{40-8\sqrt{2}}{23}$

### Example Questions

We are looking for a square number that goes into 75. There is one: 25. Specifically, $72=25\times3$

Using the multiplication rule, we can write,

$\sqrt{75}=\sqrt{3\times25}=\sqrt{3}\times\sqrt{25}$

The square root of 25 is 5, so this becomes,

$\sqrt{3}\times\sqrt{25}=\sqrt{3}\times5=5\sqrt{3}$

Thus, the answer in simplified surd form is $5\sqrt{3}$

$\sqrt{63}=\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}=3\sqrt7$

Using, $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$, the expresion can be simplified to,

$\sqrt{\dfrac{3}{16}}=\dfrac{\sqrt3}{\sqrt16}=\dfrac{\sqrt3}{4}$

We will multiply the top and bottom of this fraction by the surd on the bottom: $\sqrt{3}$

Doing so, we get,

$\dfrac{12}{\sqrt{3}}=\dfrac{12\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$

The numerator is just $12\sqrt{3}$. Using the multiplication rule, the denominator is

$\sqrt{3}\times\sqrt{3}=\sqrt{3\times3}=\sqrt{9}=3$

Therefore, the fraction is,

$\dfrac{12\sqrt{3}}{3}$

However, this is not in its simplest form. We can cancel a factor of 3 from the top and bottom and get,

$\dfrac{12\sqrt{3}}{3}=\dfrac{4\sqrt{3}}{1}=4\sqrt{3}$

We will multiply top and bottom of this fraction by $(\sqrt{10}+1)$. So, the numerator becomes

$7\times(\sqrt{10}+1)=7\sqrt{10}+7$

Then, using FOIL, the denominator becomes

$(\sqrt{10}-1)(\sqrt{10}+1)=\sqrt{10}\times\sqrt{10}+1\times\sqrt{10}-1\times\sqrt{10}-1\times1$

Completing each multiplication, including applying the multiplication law to the first term, we get

$\sqrt{10\times10}+\sqrt{10}-\sqrt{10}-1$

The first term is $\sqrt{10\times10}=\sqrt{100}=10$. So, denominator finally becomes

$10-1=9$

Thus, the fraction is

$\dfrac{7\sqrt{10}+7}{9}$

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