Simplifying Surds (OCR GCSE Maths)

Revision Note

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Maths

A surd is the square root of a non-square integer
Using surds lets you leave answers in exact form
- e.g. $5 \sqrt{2}$ rather than $7.071067812$

Surd and not surd, A Level & AS Level Pure Maths Revision Notes

Multiplying surds
- You can multiply numbers under square roots together
- eg. $\sqrt{3} \times \sqrt{5} = \sqrt{3 \times 5} = \sqrt{15}$
Dividing surds
- You can divide numbers under square roots
- eg. $\sqrt{21} \div \sqrt{7} = \sqrt{21 \div 7} = \sqrt{3}$
Factorising surds
- You can factorise numbers under square roots
- eg. $\sqrt{35} = \sqrt{5 \times 7} = \sqrt{5} \times \sqrt{7}$
Adding or subtracting surds is very like adding or subtracting letters in algebra – you can only add or subtract multiples of “like” surds
- eg. $3 \sqrt{5} + 8 \sqrt{5} = 11 \sqrt{5} o r 7 \sqrt{3} - 4 \sqrt{3} = 3 \sqrt{3}$
- Be very careful here, you can not add or subtract numbers under square roots
- Think about $\sqrt{9} + \sqrt{4} = 3 + 2 = 5$
  - It is not equal to $\sqrt{9 + 4} = \sqrt{13} = 3.60555 \dots$

If you are working on an exam question and your calculator gives you an answer as a surd, leave the value as a surd throughout the rest of your calculations to make sure you do not lose accuracy throughout your questions

To simplify a surd, separate out a square factor and square root it
- Look for the greatest square number that is a factor of the number you are simplifying
- eg. $\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4 \times \sqrt{3} = 4 \sqrt{3}$

2-1-2-surds-simplify

You can collect like terms with surds like you do with letters in algebra
Understanding how to simplify surds can help reduce expressions and collect like terms
- e.g. simplify $\sqrt{32} + \sqrt{8}$ by simplifying each part separately
  $\begin{array}{rcl} \sqrt{32} + \sqrt{8} & = & \sqrt{16} \times \sqrt{2} + \sqrt{4} \times \sqrt{2} \\ = & 4 \sqrt{2} + 2 \sqrt{2} \\ = & 6 \sqrt{2} \end{array}$
An important skill is multiplying double brackets containing surds
- This can be done in the same way as multiplying out double brackets algebraically and simplifying
- The property ${(\sqrt{a})}^{2} = a$ can be used to simplify the expression, once expanded

When simplifying surds, use the fact that the one, non square factor will be the same in each part to help you find the correct, highest square factor

Write $\sqrt{54} - \sqrt{24}$ in the form $p \sqrt{q}$ where $p$ and $q$ are integers and $q$ has no square factors.

Simplify both surds separately by finding the highest square number that is a factor of each of them

9 is a factor of 54, so $\sqrt{54} = \sqrt{9 \times 6} = 3 \sqrt{6}$

4 is a factor of 24, so $\sqrt{24} = \sqrt{4 \times 6} = 2 \sqrt{6}$

Simplify the whole expression by collecting the like terms

$\sqrt{54} - \sqrt{24} = 3 \sqrt{6} - 2 \sqrt{6} = \sqrt{6}$

$\sqrt{54} - \sqrt{24} = \sqrt{6}$

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