# Surds

https://sciencing.com/radicals-math-8565068.htmlSurds are essentially square roots of numbers that are not square. For example, 16 is a square number, if you root it you get 4. 8 is not a square number, if you root it you get $\sqrt{8}$. This is an example of a surd. Surds are sometimes referred to as ‘radicals’ (what are radicals?).

## Simplifying Surds

The trick to simplifying surds is to first consider the number within the square root. After that, see if you can identify any square factors of this number. Square factors are numbers that you can divide by (and obtain an integer result) that happen to be square numbers: 1,4,9,16,25,…

Take $\sqrt{8}$ to illustrate how to simplify a surd:

$\sqrt{8}=\sqrt{4\times 2}=\sqrt{4}\times \sqrt{2}=2\times\sqrt{2}=2\sqrt{2}$

The steps are given by the following:

- The first step is to write 8 as a product of a square number and some other number.
- Putting the square number first sometimes makes the process easier to remember.
- Subsequently, the root can be split out into two individual roots. Note that this is only true for multiplication (and division) and not addition (or subtraction).
- Next, make any simplifications.
- Finally, remove the uneccessary multiplication sign.

See Example 1 for more.

## Manipulating Surds

When manipulating or simplifying algebraic expressions involving surds it is useful to remember the following:

- $\left(\sqrt{x}\right)^2=\sqrt{x}\times \sqrt{x}=x$
- $x\times \sqrt{y}=\sqrt{y}\times x=x\sqrt{y}$
- $\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y$

The following two examples show the use of these rules in practice:

$\begin{array}{l}\left(\sqrt{3}+2\right)^2&=&\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)=\sqrt{3}\times\sqrt{3}+2\times\sqrt{3}+\sqrt{3}\times 2 +2\times 2\\&=&3+2\sqrt{3}+2\sqrt{3}+4=7+4\sqrt{3}\\(7-\sqrt{5})(4+3\sqrt{5})&=&28-4\sqrt{5}+21\sqrt{5}-5=23+17\sqrt{3}\end{array}$

See Example 2 for another demonstration of the rules in practice.

## Rationalising the Denominator

Recall that the denominator is the bottom of a fraction. It is possible that this number is a surd. Rationalising the denominator is where you remove the surd from the denominator (what does rational mean in maths?). As a result, this could mean that the surd appears on the top of the fraction, i.e. in the numerator, instead. This is OK as long as all surds are removed from the bottom. Therefore, the idea is to find an equivalent fraction where there is no surd in the denominator. Unsurprisingly, this can be achieved by multiplying or dividing the top and bottom by the same number. Inspecting the denominator will tell what number to use.

In the following simple example the denominator is root 3, this is what we should multiply the top and bottom by. It turns the denominator into a simple 3 and hence has rationalised it.

$\frac{6}{\sqrt{3}}=\frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{6\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}$

In the following more complicated example, the denominator has two terms – one rational and one irrational.

$\frac{8}{\sqrt{8}+2}=\frac{8}{2\sqrt{2}+2}=\frac{4}{\sqrt{2}+1}=\frac{4}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{4\sqrt{2}-4}{2-\sqrt{2}+\sqrt{2}-1}=\frac{4\sqrt{2}-4}{1}=4\sqrt{2}-4$

Note that we simplified the surd then the fraction in the first two steps. It is then possible to rationalise the denominator by multiplying top and bottom by the denominator with the sign changed. As you can see from this example, it causes the surd to cancel. See Example 3 for more.

## Surds Examples

Simplify the following:

- Simplify $\sqrt{27}$.
- Write $\sqrt{20}$ in the form $2\sqrt{x}$.
- Write $\sqrt{32}$ in the form $a\sqrt{2}$.
- Simplify $\sqrt{75}$.
- Simplify $\sqrt{98}$.

- $\sqrt{27}=\sqrt{9\times 3}=\sqrt{9}\times \sqrt{3}=3\times\sqrt{3}=3\sqrt{3}$
- $\sqrt{20}=\sqrt{4\times 5}=\sqrt{4}\times \sqrt{5}=2\times\sqrt{5}=2\sqrt{5}$
- $\sqrt{32}=\sqrt{16\times 2}=\sqrt{16}\times \sqrt{2}=4\times\sqrt{2}=4\sqrt{2}$
- $\sqrt{75}=\sqrt{25\times 3}=\sqrt{25}\times \sqrt{3}=5\times\sqrt{3}=5\sqrt{3}$
- $\sqrt{98}=\sqrt{49\times 2}=\sqrt{49}\times \sqrt{2}=7\times\sqrt{2}=7\sqrt{2}$

Rationalise the denominator:

$\frac{7}{\sqrt{2}-\sqrt{5}}$

$\begin{array}{l}\frac{7}{\sqrt{2}-\sqrt{5}}&=&\frac{7}{\sqrt{2}-\sqrt{5}}\times\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}\\&=&\frac{7\sqrt{2}+7\sqrt{5}}{2+\sqrt{2}\sqrt{5}-\sqrt{2}\sqrt{5}-5}\\&=&-\frac{7\sqrt{2}+7\sqrt{5}}{3}\end{array}$