National 5 Maths
Surds

We need to factorise 12, making sure that one of the factors is a square number.

So 6\(\times\)2 doesn't help but 4\(\times\)3 does.

$$ \begin{eqnarray} \sqrt{12} &=& \sqrt{4\times3}\\[6pt] &=& \sqrt4\ \times\sqrt3\\[6pt] &=& 2\sqrt3 \end{eqnarray} $$

Note that the 3 is not under the root sign, so it stays out.

When factorising 72, your first thought might be 9\(\times\)8 because that's in your times tables and 9 is a square number. However, 8 still has a square factor, so 72 = 9\(\times\)8 = 9\(\times\)4\(\times\)2 = 36\(\times\)2 and now we can see that 36 is actually the biggest square factor of 72. Always look for the largest square factor.

$$ \begin{eqnarray} 3\sqrt{72}\ &=&\ 3\sqrt{36\times2}\\[6pt] &=&\ 3\times\sqrt{36}\times\sqrt2\\[6pt] &=&\ 3\times6\times\sqrt2\\[6pt] &=&\ 18\sqrt2 \end{eqnarray} $$

Here, the strategy is to simplify each term and then collect like terms.

$$ \begin{eqnarray} 2\sqrt{20} + 6\sqrt{45}\ &=&\ 2\sqrt{4\times5}\ \ +\ \ 6\sqrt{9\times5}\\[6pt] &=&\ 2\sqrt4\sqrt5\ \ +\ \ 6\sqrt9\sqrt5\\[6pt] &=&\ 2\times2\sqrt5\ \ +\ \ 6 \times3\sqrt5\\[6pt] &=&\ 4\sqrt5\ \ +\ \ 18\sqrt5\\[6pt] &=&\ 22\sqrt5 \end{eqnarray} $$

We could do \( \sqrt{8} \times \sqrt{12} = \sqrt{96} \) and simplify from there, but it's usually easier to simplify before multiplying. This keeps the numbers small.

$$ \begin{eqnarray} \sqrt{8} \times \sqrt{12} &=&\ \sqrt{4\times 2} \times \sqrt{4\times 3} \\[6pt] &=&\ 2\sqrt 2 \times 2\sqrt 3 \\[6pt] &=&\ 4\sqrt6 \end{eqnarray} $$

One way of tackling this is to simplify each surd first:

$$ \begin{eqnarray} \small\frac{4\sqrt{20}}{\sqrt8}\normalsize\ \ &=&\ \small\frac{4\sqrt{4\times5}}{\sqrt{4\times2}}\normalsize\\[6pt] &=&\ \small\frac{4\times2\sqrt5}{2\sqrt2}\normalsize\\[6pt] &=&\ \small\frac{4\sqrt5}{\sqrt2}\normalsize\\[6pt] &=&\ \small\frac{4\sqrt5}{\sqrt2}\normalsize\times\small\frac{\sqrt2}{\sqrt2}\normalsize\ \ \ \ \ \small\textsf{(because } \small\frac{\sqrt2}{\sqrt2}=1)\normalsize\\[6pt] &=&\ \small\frac{4\sqrt{10}}{2}\normalsize\ \ \ \ \ \small\textsf{(because } \sqrt2\sqrt2=2)\normalsize\\[6pt] &=&\ 2\sqrt{10} \end{eqnarray} $$

Alternatively, we might choose not to simplify first:

$$ \begin{eqnarray} \small\frac{4\sqrt{20}}{\sqrt8}\normalsize\ \ &=&\ \small\frac{4\sqrt{20}}{\sqrt8}\normalsize\ \times\ \small\frac{\sqrt8}{\sqrt8}\normalsize\ \ \ \ \small\textsf{(because }\ \frac{\sqrt8}{\sqrt8}\ =1)\normalsize\\[6pt] &=&\ \small\frac{4\sqrt{160}}{8}\normalsize\ \ \ \ \small\textsf{(because }\ \sqrt8\sqrt8\ =\ 8)\normalsize\ \ \\[6pt] &=&\ \small\frac{\sqrt{160}}{2}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{16\times10}}{2}\normalsize\\[6pt] &=&\ \small\frac{4\sqrt{10}}{2}\normalsize\\[6pt] &=&\ 2\sqrt{10}\ \end{eqnarray} $$

$$ \begin{eqnarray} &\ & \sqrt{40}+4\sqrt{10}+\sqrt{90} \\[6pt] &=&\ \sqrt{4\times 10}+4\sqrt{10}+\sqrt{9\times 10}\\[6pt] &=&\ 2\sqrt{10}+4\sqrt{10}+3\sqrt{10}\\[6pt] &=&\ 9\sqrt{10} \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{9}{\sqrt{6}}\normalsize\ \ &=&\ \small\frac{9}{\sqrt{6}}\normalsize\ \times\ \small\frac{\sqrt{6}}{\sqrt{6}}\normalsize\\[6pt] &=&\ \small\frac{9\sqrt{6}}{6}\normalsize\\[6pt] &=&\ \small\frac{3\sqrt{6}}{2}\normalsize \end{eqnarray} $$

This 3-mark question was more awkward than usual for an SQA exam paper, because the \(40\) inside the square root sign in the denominator has a square factor. You therefore have a choice of method: to rationalise and then simplify, or to simplify first and then rationalise.

Method 1: Rationalise first.

$$ \begin{eqnarray} \small\frac{\sqrt{2}}{\sqrt{40}}\normalsize\ \ &=&\ \small\frac{\sqrt{2}}{\sqrt{40}}\normalsize\ \times\ \small\frac{\sqrt{40}}{\sqrt{40}}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{2}\sqrt{40}}{40}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{80}}{40}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{16\times 5}}{40}\normalsize\\[6pt] &=&\ \small\frac{4\sqrt{5}}{40}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{5}}{10}\normalsize \end{eqnarray} $$

Method 2: Simplify first.

$$ \begin{eqnarray} \small\frac{\sqrt{2}}{\sqrt{40}}\normalsize\ \ &=&\ \small\frac{\sqrt{2}}{\sqrt{4\times 10}}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{2}}{2\sqrt{10}}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{2}}{2\sqrt{10}}\normalsize \times\ \small\frac{\sqrt{10}}{\sqrt{10}}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{20}}{20}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{4\times 5}}{20}\normalsize\\[6pt] &=&\ \small\frac{2\sqrt{5}}{20}\normalsize\\[6pt] &=&\ \small\frac{\sqrt{5}}{10}\normalsize \end{eqnarray} $$

We usually prefer the second method, but in this case it didn't make the working shorter, because some extra simplifying was also needed at the end.

This question combined expanding brackets and simplifying surds.

$$ \begin{eqnarray} &\ & \sqrt{10}\left(\sqrt{10}-\sqrt{2}\right)+8\sqrt{5} \\[6pt] &=&\ 10-\sqrt{20}+8\sqrt{5}\\[6pt] &=&\ 10-\sqrt{4\times 5}+8\sqrt{5}\\[6pt] &=&\ 10-2\sqrt{5}+8\sqrt{5}\\[6pt] &=&\ 10+6\sqrt{5} \end{eqnarray} $$

This was quite a simple 2-mark question.

$$ \begin{eqnarray} \small\frac{12}{\sqrt{15}}\normalsize\ \ &=&\ \small\frac{12}{\sqrt{15}}\normalsize\ \times\ \small\frac{\sqrt{15}}{\sqrt{15}}\normalsize\\[6pt] &=&\ \small\frac{12\sqrt{15}}{15}\normalsize\\[6pt] &=&\ \small\frac{4\sqrt{15}}{5}\normalsize \end{eqnarray} $$

This was a particularly simple 2-mark surd question.

$$ \begin{eqnarray} \sqrt{75} - \sqrt{3}\ &=&\ \sqrt{25\times3}\ \ -\ \ \sqrt{3}\\[6pt] &=&\ 5\sqrt3\ \ -\ \ \sqrt3\\[6pt] &=&\ 4\sqrt3\\[6pt] \end{eqnarray} $$