Surds Questions | Worksheets and Revision | MME

Surds Questions, Worksheets and Revision

Level 6-7
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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:

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

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

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

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

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

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 deliminator 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 6-7

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 deliminator 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}

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

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
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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}

Maths Exam Worksheets

Worksheets and Exam Questions

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

Level 6-7 New Official MME
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(NEW) Surds - Rationalise and harder Surds Exam Style Questions - MME

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Drill Questions

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Surds 1 - Drill Questions

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Surds 2 - Drill Questions

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Surds Hard - Drill Questions

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