# Surds

A LevelAQAEdexcelOCR

## Surds

A surd is a number that can’t be simplified to remove a root (square root, cube root etc.), e.g. $\sqrt{2}, \sqrt{3}, \sqrt{6}$

There are 7 skills you need to learn for surds.

A Level   ## Skill 1: Adding and Subtracting Surds

You can only add and subtract “like” surds.

Example:

\begin{aligned} \sqrt{a} + \sqrt{a} &= 2 \sqrt{a} \\ 7 \sqrt{2} - 3 \sqrt{2} &= 4 \sqrt{2} \end{aligned}

Note: Don’t do this:

$\cancel{\sqrt{a} + \sqrt{b} = \sqrt{a+b}}$

A Level   ## Skill 2: Multiplying Surds

To multiply surds, you just need to multiply the numbers inside the square root.

$\sqrt{\textcolor{blue}{a}} \times \sqrt{\textcolor{red}{b}} = \sqrt{\textcolor{blue}{a}} \sqrt{\textcolor{red}{b}} = \sqrt{\textcolor{blue}{a} \textcolor{red}{b}} = \sqrt{\textcolor{blue}{a} \times \textcolor{red}{b}}$

Example:

\begin{aligned} \sqrt{2} \times \sqrt{3} &= \sqrt{2 \times 3} = \sqrt{6} \\ 2 \sqrt{3} \times 4 \sqrt{5} &= 2 \times 4 \times \sqrt{3 \times 5} = 8 \sqrt{15} \end{aligned}

A Level   ## Skill 3: Squaring Surds

Squaring surds is an extension on Skill 2. Squaring a surd gives the number inside the surd on its own.

$(\sqrt{\textcolor{red}{a}})^2 = \sqrt{\textcolor{red}{a}} \sqrt{\textcolor{red}{a}} = \textcolor{red}{a}$

Example:

\begin{aligned} (\sqrt{5})^2 &= \sqrt{5} \sqrt{5} = 5 \\ (2 \sqrt{3})^2 &= 2^2 \times (\sqrt{3})^2 \\ &= 4\times3=12 \end{aligned}

A Level   ## Skill 4: Dividing Surds

To divide surds, you just need to divide the numbers inside the square root.

$\dfrac{\sqrt{\textcolor{blue}{a}}}{\sqrt{\textcolor{red}{b}}} = \sqrt{\dfrac{\textcolor{blue}{a}}{\textcolor{red}{b}}}$

Example:

\begin{aligned} \dfrac{\sqrt{12}}{\sqrt{4}} &= \sqrt{\dfrac{12}{4}} = \sqrt{3} \\ \dfrac{6 \sqrt{10}}{2 \sqrt{5}} &= \dfrac{6}{2} \times \dfrac{\sqrt{10}}{\sqrt{5}} = 3 \times \sqrt{\dfrac{10}{5}} = 3 \sqrt{2} \end{aligned}

A Level   ## Skill 5: Simplifying Surds

You can simplify a surd if the number underneath the square root has a square number as one of its factors. There may be more than one way to simplify a surd.

Example: Write $\sqrt{20}$ in its simplest form.

Think of a square number that is a factor of $20$

$4$ is a factor of $20$:

$\textcolor{blue}{4} \times 5 = 20$

So,

\begin{aligned} \sqrt{20} &= \sqrt{\textcolor{blue}{4} \times 5} = \sqrt{\textcolor{blue}{4}} \times \sqrt{5} \\ &= \textcolor{blue}{2} \times \sqrt{5} = \textcolor{blue}{2} \sqrt{5} \end{aligned}

A Level   ## Skill 6: Surds and Double Brackets

You can multiply out double brackets containing surds using the same technique that you would for normal double brackets.

Example: Expand and simplify $(3 \sqrt{5} + 2 \sqrt{2})^2$

\begin{aligned}(3 \sqrt{5} + 2 \sqrt{2})^2&= (3 \sqrt{5} + 2 \sqrt{2})(3 \sqrt{5} + 2 \sqrt{2}) \\[1.2em]&= (3 \sqrt{5})^2 + (2 \times 3 \sqrt{5} \times 2 \sqrt{2}) + (2 \sqrt{2})^2 \\[1.2em]&= (3^2 \times \sqrt{5}^2) + (2 \times 3 \times 2 \times \sqrt{5} \times \sqrt{2}) + (2^2 \times \sqrt{2}^2) \\[1.2em]&= (9 \times 5) + (12 \times \sqrt{10}) + (4 \times 2) \\[1.2em]&= 45 + 12 \sqrt{10} + 8 \\[1.2em]&= 53 + 12 \sqrt{10}\end{aligned}

A Level   ## Skill 7: Rationalising the Denominator

To rationalise the denominator, you need to remove the surd from the bottom of the fraction. There are two different types of rationalising the denominator you will see, one easy and one hard.

Example: Rationalise the denominator of the following fraction: $\dfrac{6}{\sqrt{2}}$

Multiply the top and bottom of the fraction by the denominator.

$\dfrac{\textcolor{blue}{6}}{\textcolor{red}{\sqrt{2}}} \times \dfrac{\textcolor{red}{\sqrt{2}}}{\textcolor{red}{\sqrt{2}}} = \dfrac{6 \sqrt{2}}{2} = 3 \sqrt{2}$

Example: Rationalise the denominator of the following fraction: $\dfrac{4}{2 + \sqrt{5}}$

Multiply the top and bottom of the fraction by the denominator, with the sign changed. So, $+$ is changed to $-$ and vice versa.

\begin{aligned} \dfrac{\textcolor{blue}{4}}{\textcolor{red}{2 + \sqrt{5}}} &= \dfrac{\textcolor{blue}{4}}{\textcolor{red}{2 + \sqrt{5}}} \times \dfrac{\textcolor{limegreen}{2 - \sqrt{5}}}{\textcolor{limegreen}{2 - \sqrt{5}}} \\[1.2em] &= \dfrac{\textcolor{blue}{4} \textcolor{limegreen}{(2 - \sqrt{5})}}{\textcolor{red}{(2 + \sqrt{5})} \textcolor{limegreen}{(2 - \sqrt{5})}} \\[1.2em] &= \dfrac{8 - 4 \sqrt{5}}{4 - 2 \sqrt{5} + 2 \sqrt{5} - 5} \\[1.2em] &= \dfrac{8 - 4 \sqrt{5}}{4 - 5} \\[1.2em] &= \dfrac{8 - 4 \sqrt{5}}{-1} \\[1.2em] &= -8 + 4 \sqrt{5} \end{aligned}

A Level   ## Example Questions

$(4\sqrt{5})^2 = 4^2 \times \sqrt{5}^2 = 16 \times 5 = 80$

\begin{aligned} (3 + \sqrt{3})(2 - \sqrt{3}) &= 6 - 3 \sqrt{3} + 2 \sqrt{3} - \sqrt{3}^2 \\ &= 6 - \sqrt{3} -3 \\ &= 3 - \sqrt{3} \end{aligned}

$\dfrac{10}{\sqrt{5}} = \dfrac{10}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{10 \sqrt{5}}{5} = 2 \sqrt{5}$
\begin{aligned} \dfrac{3 + \sqrt{5}}{1 - \sqrt{5}} &= \dfrac{3 + \sqrt{5}}{1 - \sqrt{5}} \times \dfrac{1 + \sqrt{5}}{1 + \sqrt{5}} \\[1.1em] &= \dfrac{3 + 3 \sqrt{5} + \sqrt{5} + \sqrt{5}^2}{1 - \sqrt{5} + \sqrt{5} - \sqrt{5}^2} \\[1.1em] &= \dfrac{3 + 4 \sqrt{5} + 5}{1 - 5} \\[1.1em] &= \dfrac{8 + 4 \sqrt{5}}{-4} \\[1.1em] &= -2 - \sqrt{5} \end{aligned}

A Level

A Level

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