Surds Questions, Worksheets and Revision

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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}

Related Topics

MME

Collecting Like Terms

Level 1-3GCSEKS3
MME

The Laws of Indices

Level 6-7GCSE
MME

Powers and Roots

Level 4-5GCSEKS3
MME

Fraction Rules

Level 1-3GCSEKS3

Worksheet and Example Questions

Site Logo

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

Level 6-7 GCSENewOfficial MME
Site Logo

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

Level 8-9 GCSENewOfficial MME

Drill Questions

Site Logo

Surds 1 - Drill Questions

Level 6-7 GCSE
Site Logo

Surds 2 - Drill Questions

Level 6-7 GCSE
Site Logo

Surds Hard - Drill Questions

Level 6-7 GCSE

You May Also Like...

GCSE Maths Revision Cards

Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC.

From: £8.99
View Product

GCSE Maths Revision Guide

The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!

From: £14.99
View Product

GCSE Maths Predicted Papers 2022 (Advance Information)

GCSE Maths 2022 Predicted Papers are perfect for preparing for your 2022 Maths exams. These papers have been designed based on the new topic lists (Advance Information) released by exam boards in February 2022! They are only available on MME!

From: £5.99
View Product

Level 9 GCSE Maths Papers 2022 (Advance Information)

Level 9 GCSE Maths Papers 2022 are designed for students who want to achieve the top grades in their GCSE Maths exam. Using the information released in February 2022, the questions have been specifically tailored to include the type of level 9 questions that will appear in this year's exams.

£9.99
View Product