Surds Questions, Worksheets and Revision

GCSE 6 - 7GCSE 8 - 9AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022

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 number – so $\sqrt{2},\sqrt{3},\sqrt{5}$ are all surds. There are 7 key skills you need to learn when manipulating surds.

This topic will require a good understanding of:

Level 6-7 GCSE

Skill 1: Multiplying Surds

When multiplying surds you simply multiply the numbers inside the square root.

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

Example:

$\sqrt{7} \times \sqrt{2} = \sqrt{7 \times 2} = \sqrt{14}$

$2\sqrt{2} \times 3\sqrt{5} = 2\times 3 \times \sqrt{2\times5} = 6\sqrt{10}$

$\sqrt{6}^2 = \sqrt{6} \times \sqrt{6} = \sqrt{6\times6} =\sqrt{36} = 6$

Level 6-7 GCSE

Skill 2: Dividing Surds

When dividing surds you simply divide the numbers inside the square root.

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

Example:

$\dfrac{\sqrt{10}}{\sqrt{5}} = \sqrt{\dfrac{10}{5}} = \sqrt{2}$

$\dfrac{8\sqrt{12}}{2\sqrt{3}} = \dfrac{8}{2}\times\sqrt{\dfrac{12}{3}} = 4\times\sqrt{4} = 4\times 2 = 8$

Level 6-7 GCSE

Skill 3: Adding and Subtracting Surds

It is only possible to add and subtract “like” surds, this is similar to collecting like terms

$\sqrt{a} + \sqrt{a} = 2\sqrt{a}$

$5\sqrt{b} - 2\sqrt{b} = 3\sqrt{b}$

Do NOT do this:

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

Level 6-7 GCSE
Level 6-7 GCSE

Skill 4: Simplifying Surds

Surds can be simplified if the number within the surd has a square number as one of its factors.

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

We need need to think of a square number which is a factor of $28$.

$28 = \textcolor{red}{4} \times 7$

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

We know that $\sqrt{\textcolor{red}{4}} = \textcolor{red}{2}$

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

Level 6-7 GCSE

Skill 5: Double brackets and surds

We can multiply out double brackets containing surds the same way as for quadratics using FOIL, then collect like terms.

$(m+\sqrt{n}) (m+\sqrt{n})=\textcolor{red}{m^2}+\textcolor{limegreen}{m\sqrt{n}}+\textcolor{purple}{m\sqrt{n}}+\textcolor{blue}{n}\\ = \textcolor{red}{m^2}+\textcolor{maroon}{2m\sqrt{n}}+\textcolor{blue}{n}$

Example:

\begin{aligned} &(\sqrt{10} + \sqrt{3})(\sqrt{10} - \sqrt{3}) \\ &= \sqrt{10}^2 - \sqrt{3}\sqrt{10} + \sqrt{3}\sqrt{10} - \sqrt{3}^2 \\ &= 10 -\sqrt{30} + \sqrt{30} -3 \\ &= 10 - 3 \\ &= 7 \end{aligned}

Level 6-7 GCSE
Level 6-7 GCSE

Skill 6: Rationalise the denominator – Simple

Rationalising the denominator just means removing the surd from the bottom of a fraction. There are two types of question you may encounter, one harder then the other. The first type is shown below.

Example: Rationalise the denominator of the following fraction $\dfrac{\textcolor{red}{a}}{\sqrt{\textcolor{blue}{b}}}$

Simply multiply the top and bottom of the fraction by the denominator of the fraction.

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

Level 8-9 GCSE

Skill 7: Rationalise the denominator – Harder

Rationalising the denominator when there are other terms as well as the surd can be much more tricky.

Example: Rationalise the denominator of the following fraction $\dfrac{\textcolor{red}{5}}{\textcolor{blue}{3+\sqrt{5}}}$

Multiply the top and the bottom of the fraction by the denominator with the sign changed. $+$ becomes $-$ and $-$ becomes $+$.

\begin{aligned} \dfrac{\textcolor{red}{5}}{\textcolor{blue}{3+\sqrt{5}}} &= \dfrac{\textcolor{red}{5}}{\textcolor{blue}{3+\sqrt{5}}} \times \dfrac{\textcolor{limegreen}{3-\sqrt{5}}}{\textcolor{limegreen}{3-\sqrt{5}}} \\ &= \dfrac{\textcolor{red}{5}\textcolor{limegreen}{(3-\sqrt{5})}}{\textcolor{blue}{(3+\sqrt{5})}\textcolor{limegreen}{(3-\sqrt{5})}} \\ &= \dfrac{15-5\sqrt{5}}{9-3\sqrt{5} + 3\sqrt{5} -5} \\ &= \dfrac{15-5\sqrt{5}}{9 - 5} \\ &= \dfrac{15-5\sqrt{5}}{4} \end{aligned}

Level 8-9 GCSE
Level 6-7 GCSE

Example 1: Rationalising the Denominator

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

[2 marks]

This would be Type 1 so we simply need to multiply the top and bottom of the fraction by the denominator of the fraction

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

We know,

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

So,

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

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

Level 6-7 GCSE

Example 2: Rationalising Surds

Rationalise the denominator of the following fraction.

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

[4 marks]

This is a Type 2 so we need to multiply the top and the bottom of the fraction by the denominator with the sign changed. $+$ becomes $-$ and $-$ becomes $+$.

This means we must multiply by $(5-\sqrt{2})$. This is going to involve some bracket expanding. The numerator becomes

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

Now we need to multiply out the top and the bottom of the fraction, then simplify.

The Numerator:

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

The Denominator:

\begin{aligned}(5+\sqrt{2})(5-\sqrt{2})&=5^2-5\sqrt{2}+5\sqrt{2}-\sqrt{2}^2 \\ &=25-5\sqrt{2}+5\sqrt{2}- 2 \\ &= 25 -2 \\ &= 23 \end{aligned}

Therefore, we can now reform our fraction giving our final answer,

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

Level 8-9 GCSE

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

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

This can also be written as

$\dfrac{7(1+\sqrt{10})}{9}$

Level 1-3GCSEKS3

Level 6-7GCSE

Level 4-5GCSEKS3

Level 1-3GCSEKS3

Worksheet and Example Questions

(NEW) Surds - The Basics Exam Style Questions - MME

Level 6-7 GCSENewOfficial MME

(NEW) Surds - Rationalise and harder Surds Exam Style Questions - MME

Level 8-9 GCSENewOfficial MME

Level 6-7 GCSE

Level 6-7 GCSE

Level 6-7 GCSE

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