Learning Goal
Syllabus Requirement
Students need to learn
the use and manipulation of surds, with the below point being highlighted in particular:

Be able to rationalise denominators.
Study Notes
A surd is an unresolved ??n??^{th} root. So, for example, ??\sqrt{5}?? is a surd and ??\sqrt{4}?? is a surd  although the latter can be written so that it is not a surd because ??\sqrt{4} = 2?? and ??2?? is not a surd.
A simple surd is one with just a single term, such as ??\sqrt{3}?? or ??\sqrt[3]{8}??.
A compound surd is the sum of two or
more simple surds or a simple surd and a rational number, such as ??\sqrt{3} + 2?? or ??\sqrt{3} + \sqrt{2}??.
The conjugate surd to a compound surd, with two terms, ??\sqrt{a} + \sqrt{b}?? or ??c + \sqrt{d}?? is given
by ??\sqrt{a}  \sqrt{b}?? and ??c  \sqrt{d}??, respectively.
There are 4 basic laws that govern surd operations (namely addition, subtraction, multiplication and division) and they are listed in the information box below:
The 4 basic laws of surds.
Law 1: Addition of surds
??a \sqrt{b} + c \sqrt{b} = (a + c) \sqrt{b}??
Law 2: Subtraction of surds
??a \sqrt{b}  c \sqrt{b} = (a  c) \sqrt{b}??
Law 3: Multiplication of surds
??\sqrt{a} \times \sqrt{b} = \sqrt{ab}??
Law 4: Division of surds
?? \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}??
You should also be aware of one common mistake, that students often make:
Common mistake
Mistake 1: Addition of surds
??\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}??
Make sure that you understand how this statement is different to Law 1.
One method that is often examined and is specifically mentioned on the syllabus is rationalising the denominator.
Remember, you should always rationalise the denominator if you want to give the answer in its simplest form.
Rationalising the denominator
Method 1: Simple surd in denominator
$$\frac{x}{\sqrt{y}}$$
If there is just a simple surd in the denominator, such as ??\sqrt{y}??, as above, then simply
multiply both the numerator and denominator by that surd to remove it. This rationalises the denominator
as shown below:
$$\frac{x}{\sqrt{y}}$$
$$= \frac{x \times \sqrt{y}}{\sqrt{y} \times \sqrt{y}}$$
$$= \frac{x \sqrt{y}}{y}$$
Method 2: Compound surd in denominator
$$\frac{x}{\sqrt{y} + \sqrt{z}}$$
If there is a compound surd with two terms in the denominator, such as ??\sqrt{y} + \sqrt{z}??, then simply
multiply both the numerator and denominator by the conjugate surd to remove it. This rationalises the
denominator as shown below:
$$\frac{x}{\sqrt{y} + \sqrt{z}}$$
$$=\frac{x \times (\sqrt{y}  \sqrt{z})}{(\sqrt{y} + \sqrt{z})(\sqrt{y}  \sqrt{z})}$$
$$=\frac{x \times (\sqrt{y}  \sqrt{z})}{\sqrt{y}\sqrt{y} + \sqrt{y}\sqrt{z}  \sqrt{y}\sqrt{z}  \sqrt{z}\sqrt{z}}$$
$$=\frac{x \times (\sqrt{y}  \sqrt{z})}{y  z}$$
Worked Examples
We'll now run through some examples of each of these rules in turn, with some more advanced examples thereafter.
Law 1
Example 1
Question
Simplify the following expression ?? 3 \sqrt{11} + 2 \sqrt{11} ??.
Show Solution
Worked Solution
??3 \sqrt{11} + 2 \sqrt{11} = (3 + 2)\sqrt{11} = 5\sqrt{11}??
Law 2
Example 2
Question
Simplify the following expression ?? 9\sqrt{13}  3 \sqrt{13} ??.
Show Solution
Worked Solution
??9\sqrt{13}  3 \sqrt{13} = (93)\sqrt{13} = 6\sqrt{13}??
Law 3
Example 3
Question
Simplify the following expression ?? 12\sqrt{108}??.
Show Solution
Worked Solution
??12\sqrt{108} = 12\sqrt{4 \times 27} = 12 \sqrt{4}\sqrt{27} = 12 \times 2 \sqrt{27} = 24\sqrt{27} = 24 \sqrt{9 \times 3} = 24 \sqrt{9} \sqrt{3} = 24 \times 3 \sqrt{3} = 72\sqrt{3}??
Law 4
Example 4
Question
Simplify the following expression ?? \sqrt{4} \sqrt{\frac{9}{4}}??.
Show Solution
Worked Solution
??\sqrt{4} \sqrt{\frac{9}{4}} = \sqrt{4} \frac{\sqrt{9}}{\sqrt{4}} = \sqrt{9}??
Multiple Laws
Example 5
Question
Simplify the following expression ?? 12\sqrt{24}  6 \sqrt{48} ??.
Show Solution
Worked Solution
Apply Law 3 first, before applying Law 2:
??12\sqrt{24}  6 \sqrt{54} = 12\sqrt{4 \times 6}  6\sqrt{9 \times 6} = 12 \sqrt{4} \sqrt{6}  6 \sqrt{9} \sqrt{6} = 12 \times 2 \sqrt{6}  6 \times 3 \sqrt{6} = 24 \sqrt{6}  18 \sqrt{6} = 6\sqrt{6}??
Rationalising the Denominator
Example 6
Question
Rationalise the following expression ?? \frac{3}{2 + \sqrt{5}} ??.
Show Solution
Worked Solution
?? \frac{3}{2 + \sqrt{5}} = \frac{3}{2 + \sqrt{5}} \times \frac{(2  \sqrt{5})}{(2  \sqrt{5})} = \frac{3 \times 2  3 \times \sqrt{5}}{2 \times 2  2 \sqrt{5} + 2 \sqrt{5}  \sqrt{5}\sqrt{5}} = \frac{6  3\sqrt{5}}{4  5} = \frac{6  3\sqrt{5}}{1} = 3\sqrt{5}  6??
Exam Questions
The table below contains every exam question that has been asked on this topic, this includes normal papers from both January
and June sittings, International papers and Specimen papers.
To see an exam question and solution simply click on the load question icon (), the question will appear below the table and
the solution can be shown by clicking the "Show Solution" button that also appears.
The full exam paper and mark schemes are also available for download by clicking on the download icons in each
row of the table ().
Load Question 
Exam Board 
Subject 
Paper 
Year 
Month 
Module 
Question No. 
Parts 
Total Marks 
Exam Paper 
Mark Scheme 

Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
5 
(a), (b) 
6 



Edexcel 
Maths 
Standard 
2006 
June 
Core 1 
6 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2007 
January 
Core 1 
2 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2007 
June 
Core 1 
1 
n/a 
2 



Edexcel 
Maths 
Standard 
2008 
January 
Core 1 
3 
n/a 
4 



Edexcel 
Maths 
Standard 
2009 
January 
Core 1 
3 
n/a 
2 



Edexcel 
Maths 
Standard 
2009 
June 
Core 1 
1 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2010 
January 
Core 1 
2 
(a), (b) 
6 



Edexcel 
Maths 
Standard 
2010 
June 
Core 1 
1 
n/a 
2 



Edexcel 
Maths 
Standard 
2011 
January 
Core 1 
3 
n/a 
4 



Edexcel 
Maths 
Standard 
2012 
January 
Core 1 
2 
(a), (b) 
6 



Edexcel 
Maths 
Standard 
2012 
June 
Core 1 
3 
n/a 
5 



Edexcel 
Maths 
Standard 
2013 
January 
Core 1 
3 
(a), (b) 
6 



Edexcel 
Maths 
Standard 
2013 
June 
Core 1 
1 
n/a 
4 



Edexcel 
Maths 
International 
2013 
June 
Core 1 
2 
n/a 
4 


Edexcel  Maths  Standard  2006  January  Core 1  Question 5
Question
5. (a)
Write ??\sqrt{45}?? in the form ??a\sqrt{5}??, where ??a?? is an integer. [1]
5. (b)
Express ??\frac{2(3+\sqrt{5})}{(3\sqrt{5})}?? in the form ??b + c\sqrt{5}??, where ??b?? and ??c?? are integers. [5]
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Show Solution
Worked Solution
5. (a)
Start by noting that the expression you are trying to achieve ??a\sqrt{5}?? contains a ??\sqrt{5}??.
Therefore, we will use Law 3, which states:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
To rewrite ??\sqrt{45}?? as a product of two surds, one being ??\sqrt{5}??:
$$\sqrt{45}$$
$$=\sqrt{5 \times 9}$$
$$=\sqrt{5} \times \sqrt{9}$$
$$=\sqrt{5} \times 3$$
$$=3\sqrt{5}$$
For completeness, we should make it clear that ??a = 3??.
5. (b)
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??3  \sqrt{5}?? so the conjugate surd is ??3 + \sqrt{5}??. We can now multiply the numerator
and denominator by the conjugate surd:
$$\frac{2(3+\sqrt{5})}{(3\sqrt{5})}$$
$$=\frac{2(3+\sqrt{5}) \times (3 + \sqrt{5})}{(3\sqrt{5}) \times (3 + \sqrt{5})}$$
$$=\frac{2(3 \times 3 + 3 \times \sqrt{5} + 3 \times \sqrt{5} + \sqrt{5} \times \sqrt{5})}{3 \times 3 + 3 \times \sqrt{5} + 3 \times (\sqrt{5}) + \sqrt{5} \times (\sqrt{5})}$$
$$=\frac{2(9 + 6\sqrt{5} + 5)}{9  5}$$
$$=\frac{18 + 12\sqrt{5} + 10}{4}$$
$$=\frac{28 + 12\sqrt{5}}{4}$$
$$=7 + 3\sqrt{5}$$
For completeness, we should make it clear that ??b=7?? and ??c=3??.
Edexcel  Maths  Standard  2006  June  Core 1  Question 6
Question
6. (a)
Expand and simplify ??(4 + \sqrt{3})(4  \sqrt{3})??. [2]
6. (b)
Express ??\frac{26}{4 + \sqrt{3}}?? in the form ??a + b\sqrt{3}??, where ??a?? and ??b?? are integers. [2]
Hide Question
Show Solution
Worked Solution
6. (a)
You may notice that the terms in each bracket together form conjugate surds, this means that there won't
be any surds in the answer  which is a useful check (if you get an answer with surds left in, it means you've made
a mistake or not simplified the answer enough).
Multiplying out the brackets gives:
$$(4 + \sqrt{3})(4  \sqrt{3})$$
$$= 4 \times 4 + 4 \times \sqrt{3} + 4 \times (\sqrt{3}) + \sqrt{3} \times (\sqrt{3})$$
$$= 16 + 4 \sqrt{3}  4 \sqrt{3}  3$$
$$= 13$$
6. (b)
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??4 + \sqrt{3}?? so the conjugate surd is ??4  \sqrt{3}??. We can now multiply the numerator
and denominator by the conjugate surd:
$$\frac{26}{4 + \sqrt{3}}$$
$$=\frac{26 \times (4  \sqrt{3})}{(4 + \sqrt{3}) \times (4\sqrt{3})}$$
$$=\frac{26 \times 4  26\sqrt{3}}{4 \times 4 + 4 \times \sqrt{3} + 4 \times (\sqrt{3}) + \sqrt{3} \times (\sqrt{3})}$$
$$=\frac{104  26\sqrt{3}}{16  3}$$
$$=\frac{104  26\sqrt{3}}{13}$$
$$=8  2\sqrt{3}$$
For completeness, we should make it clear that ??a=8?? and ??b=2??.
Note that you could (and should!) have used the answer from part (a) to avoid having to expand the same expression
twice.
Edexcel  Maths  Standard  2007  January  Core 1  Question 2
Question
2. (a)
Express ??\sqrt{108}?? in the form ??a\sqrt{3}??, where ??a?? is an integer. [1]
2. (b)
Express ??(2  \sqrt{3})^{2}?? in the form ??b + c\sqrt{3}??, where ??b?? and ??c?? are integers to be found. [3]
Hide Question
Show Solution
Worked Solution
2. (a)
Start by noting that the expression you are trying to achieve ??a\sqrt{3}?? contains a ??\sqrt{3}??.
Therefore, we will use Law 3, which states:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
To rewrite ??\sqrt{108}?? as a product of two surds, one being ??\sqrt{3}??:
$$\sqrt{108}$$
$$=\sqrt{3 \times 36}$$
$$=\sqrt{3} \times \sqrt{36}$$
$$=\sqrt{3} \times 6$$
$$=6\sqrt{3}$$
For completeness, we should make it clear that ??a = 6??.
6. (b)
Expanding the brackets leads directly to the answer:
$$(2  \sqrt{3})^{2}$$
$$=(2\sqrt{3})(2\sqrt{3})$$
$$=2 \times 2 + 2 \times (\sqrt{3}) + 2 \times (\sqrt{3}) + (\sqrt{3}) \times (\sqrt{3})$$
$$=4  4\sqrt{3} + 3$$
$$=7  4\sqrt{3}$$
For completeness, we should make it clear that ??b=7?? and ??c=4??.
Edexcel  Maths  Standard  2007  June  Core 1  Question 1
Question
1.
Simplify ??(3 + \sqrt{5})(3  \sqrt{5})??. [2]
Hide Question
Show Solution
Worked Solution
1.
You may notice that the terms in each bracket together form conjugate surds, this means that there won't
be any surds in the answer  which is a useful check (if you get an answer with surds left in, it means you've made
a mistake or not simplified the answer enough).
Multiplying out the brackets gives:
$$(3 + \sqrt{5})(3  \sqrt{5})$$
$$= 3 \times 3 + 3 \times \sqrt{5} + 3 \times (\sqrt{5}) + \sqrt{5} \times (\sqrt{5})$$
$$= 9  5$$
$$= 4$$
Edexcel  Maths  Standard  2008  January  Core 1  Question 3
Question
3.
Simplify ??\frac{5  \sqrt{3}}{2 + \sqrt{3}}??, giving your answer in the form ??a+b\sqrt{3}??, where ??a??
and ??b?? are integers. [4]
Hide Question
Show Solution
Worked Solution
3.
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??2 + \sqrt{3}?? so the conjugate surd is ??2  \sqrt{3}??. We can now multiply the numerator
and denominator by the conjugate surd:
$$\frac{5  \sqrt{3}}{2 + \sqrt{3}}$$
$$=\frac{(5\sqrt{3})(2\sqrt{3})}{(2+\sqrt{3})(2\sqrt{3})}$$
$$=\frac{5 \times 2 + 2 \times (\sqrt{3}) + 5 \times (\sqrt{3})}{2 \times 2 + 2 \times \sqrt{3} + 2 \times (\sqrt{3}) + \sqrt{3} \times (\sqrt{3})}$$
$$=\frac{10  7\sqrt{3} + 3}{4  3}$$
$$=\frac{13  7\sqrt{3}}{1}$$
$$=13  7\sqrt{3}$$
For completeness, we should make it clear that ??a=13?? and ??b=7??.
Edexcel  Maths  Standard  2009  January  Core 1  Question 3
Question
3.
Expand and simplify ??(\sqrt{7} + 2)(\sqrt{7}  2)??. [2]
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Show Solution
Worked Solution
3.
You may notice that the terms in each bracket together form conjugate surds, this means that there won't
be any surds in the answer  which is a useful check (if you get an answer with surds left in, it means you've made
a mistake or not simplified the answer enough).
Multiplying out the brackets gives:
$$(\sqrt{7} + 2)(\sqrt{7}  2)$$
$$= \sqrt{7} \times \sqrt{7} + \sqrt{7} \times 2 + \sqrt{7} \times (2) + 2 \times (2)$$
$$= 7  4$$
$$= 3$$
Edexcel  Maths  Standard  2009  June  Core 1  Question 1
Question
1.
Simplify:
1. (a)
??(3\sqrt{7})^{2}??. [1]
1. (b)
??(8 + \sqrt{5})(2  \sqrt{5})??. [3]
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Show Solution
Worked Solution
1. (a)
Simply expand the brackets:
$$(3\sqrt{7})^{2}$$
$$=3^{2} \times \sqrt{7}^{2}$$
$$= 9 \times 7$$
$$= 63$$
1. (b)
Again, like the first part of the question, we simply need to expand the brackets:
$$(8 + \sqrt{5})(2  \sqrt{5})$$
$$= 8 \times 2 + 2 \times \sqrt{5} + 8 \times (\sqrt{5}) + \sqrt{5} \times (\sqrt{5})$$
$$= 16  6\sqrt{5}  5$$
$$= 11  6\sqrt{5}$$
Edexcel  Maths  Standard  2010  January  Core 1  Question 2
Question
2. (a)
Expand and simplify ??(7 + \sqrt{5})(3  \sqrt{5})??. [3]
2. (b)
Express ??\frac{7 + \sqrt{5}}{3 + \sqrt{5}}?? in the form ??a + b\sqrt{5}??, where ??a?? and ??b?? are integers. [3]
Hide Question
Show Solution
Worked Solution
2. (a)
Multiplying out the brackets gives:
$$(7 + \sqrt{5})(3  \sqrt{5})$$
$$= 7 \times 3 + 3 \times \sqrt{5} + 7 \times (\sqrt{5}) + \sqrt{5} \times (\sqrt{5})$$
$$= 21  4 \sqrt{5}  5$$
$$= 16  4 \sqrt{5}$$
2. (b)
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??3 + \sqrt{5}?? so the conjugate surd is ??3  \sqrt{5}??. We can now multiply the numerator
and denominator by the conjugate surd:
$$\frac{7+\sqrt{5}}{3 + \sqrt{5}}$$
$$=\frac{(7+\sqrt{5})(3  \sqrt{5})}{(3 + \sqrt{5})(3  \sqrt{5})}$$
$$=\frac{7 \times 3 + 3 \times \sqrt{5} + 7 \times (\sqrt{5}) + \sqrt{5} \times (\sqrt{5})}{3 \times 3 + 3 \times \sqrt{5} + 3 \times (\sqrt{5}) + \sqrt{5} \times (\sqrt{5})}$$
$$=\frac{21  4\sqrt{5}  5}{9  5}$$
$$=\frac{16  4\sqrt{5}}{4}$$
$$=4  \sqrt{5}$$
For completeness, we should make it clear that ??a=4?? and ??b=1??.
Note that you could (and should!) have used the answer from part (a) to avoid having to expand the same expression
twice.
Edexcel  Maths  Standard  2010  June  Core 1  Question 1
Question
1.
Write ??\sqrt{75}  \sqrt{27}?? in the form ??k\sqrt{x}??, where ??k?? and ??x?? are integers. [2]
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Worked Solution
1.
Let's start by simplifying each of the surds in the question as much as possible, using Law 3, which
states that:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
Starting with ??\sqrt{75}?? the aim is to try and find a square number, or square numbers? that can be
brought outside of the square root sign:
$$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}$$
We can do the same for ??\sqrt{27}??:
$$\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}$$
Now we can simplify the original expression:
$$\sqrt{75}  \sqrt{27}$$
$$=5\sqrt{3}  3\sqrt{3}$$
$$=2\sqrt{3}$$
For completeness, we should make it clear that ??k=2?? and ??x=3??.
Edexcel  Maths  Standard  2011  January  Core 1  Question 3
Question
3.
Simplify ??\frac{52\sqrt{3}}{\sqrt{3}1}??, giving your answer in the form ??p + q\sqrt{3}??,
where ??p?? and ??q?? are rational numbers. [4]
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Worked Solution
3.
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??\sqrt{3}1?? so the conjugate surd is ??\sqrt{3}+1??. We can now multiply the numerator
and denominator by the conjugate surd:
$$\frac{52\sqrt{3}}{\sqrt{3}1}$$
$$=\frac{(52\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}1)(\sqrt{3}+1)}$$
$$=\frac{5 \times \sqrt{3} + 5 \times 1 + (2\sqrt{3})\times\sqrt{3} + (2\sqrt{3}) \times 1}{\sqrt{3} \times \sqrt{3} + 1 \times \sqrt{3} + (1)\times \sqrt{3} + 1 \times (1)}$$
$$=\frac{5\sqrt{3} + 5 6  2\sqrt{3}}{31}$$
$$=\frac{3\sqrt{3}1}{2}$$
$$=\frac{1}{2} + \frac{3}{2}\sqrt{3}$$
For completeness, we should make it clear that ??p=\frac{1}{2}?? and ??q=\frac{3}{2}??.
Edexcel  Maths  Standard  2012  January  Core 1  Question 2
Question
2. (a)
Simplify ??\sqrt{32} + \sqrt{18}?? giving your answer in the form ??a\sqrt{2}??, where
??a?? is an integer. [2]
2. (b)
Simplify ??\frac{\sqrt{32} + \sqrt{18}}{3+\sqrt{2}}?? giving your answer in the form ??b\sqrt{2} + c??, where ??b?? and ??c?? are integers. [4]
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Worked Solution
2. (a)
We know from the question that we need to convert both surds into some coefficient multiplied
by ??\sqrt{2}??. Law 3 states that:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
Using Law 3, we can rewrite the expression as:
$$\sqrt{32} + \sqrt{18}$$
$$=\sqrt{2 \times 16} + \sqrt{2 \times 9}$$
$$=\sqrt{2} \times \sqrt{16} + \sqrt{2} \times \sqrt{9}$$
$$=4\sqrt{2} + 3\sqrt{2}$$
$$=7\sqrt{2}$$
For completeness, we should make it clear that ??a=7??.
2. (b)
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??3 + \sqrt{2}?? so the conjugate surd is ??3  \sqrt{2}??. We can now multiply the numerator
and denominator by the conjugate surd, after using the answer from part (a) to simplify the numerator:
$$\frac{\sqrt{32} + \sqrt{18}}{3 + \sqrt{2}}$$
$$=\frac{7\sqrt{2}}{3 + \sqrt{2}}$$
$$=\frac{(7\sqrt{2})(3\sqrt{2})}{(3+\sqrt{2})(3\sqrt{2})}$$
$$=\frac{7\sqrt{2} \times 3 + 7\sqrt{2} \times (\sqrt{2})}{3 \times 3 + 3 \times \sqrt{2} + 3 \times (\sqrt{2}) + \sqrt{2} \times (\sqrt{2})}$$
$$=\frac{21\sqrt{2}  14}{92}$$
$$=\frac{21\sqrt{2}  14}{7}$$
$$3\sqrt{2}  2$$
For completeness, we should make it clear that ??b=3?? and ??c=2??.
Edexcel  Maths  Standard  2012  June  Core 1  Question 3
Question
3.
Show that ??\frac{2}{\sqrt{12}\sqrt{8}}?? can be written in the form ??\sqrt{a} + \sqrt{b}??, where ??a??
and ??b?? are integers. [5]
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Worked Solution
3.
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??\sqrt{12}  \sqrt{8}?? so the conjugate surd is ??\sqrt{12} + \sqrt{8}??. We can now multiply the numerator
and denominator by the conjugate surd:
$$\frac{2}{\sqrt{12}\sqrt{8}}$$
$$=\frac{(2)(\sqrt{12}+\sqrt{8})}{(\sqrt{12}\sqrt{8})(\sqrt{12}+\sqrt{8})}$$
$$=\frac{2(\sqrt{12}+\sqrt{8}) }{\sqrt{12} \times \sqrt{12} + \sqrt{12} \times \sqrt{8} + \sqrt{12} \times (\sqrt{8}) + \sqrt{8} \times (\sqrt{8}) } $$
$$=\frac{2(\sqrt{12}+\sqrt{8}) }{12  8}$$
$$=\frac{2(\sqrt{12}+\sqrt{8}) }{4}$$
Now, we can use Law 3, which states that:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
And this allows us to break ??\sqrt{12}?? and ??\sqrt{8}?? up:
$$\frac{2(\sqrt{12}+\sqrt{8}) }{4}$$
$$=\frac{2(\sqrt{3 \times 4}+\sqrt{2 \times 4}) }{4}$$
$$=\frac{2(\sqrt{3} \times \sqrt{4} + \sqrt{2} \times \sqrt{4})}{4}$$
$$=\frac{2(2\sqrt{3} + 2\sqrt{2})}{4}$$
$$=\frac{4\sqrt{3} + 4\sqrt{2}}{4} $$
$$=\sqrt{3} + \sqrt{2}$$
For completeness, we should make it clear that ??a=3?? and ??b=2?? (or vice versa, ??a=2?? and ??b=3??).
Edexcel  Maths  Standard  2013  January  Core 1  Question 3
Question
3. (a)
Express ??(5\sqrt{8})(1+\sqrt{2})?? in the form ??a+b\sqrt{2}??, where ??a?? and ??b?? are integers. [3]
3. (b)
Express ??\sqrt{80} + \frac{30}{\sqrt{5}}?? in the form ??c\sqrt{5}??, where ??c?? is an integer. [3]
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Worked Solution
3. (a)
Multiplying out the brackets gives:
$$(5\sqrt{8})(1+\sqrt{2})$$
$$= 5 \times 1 + 5 \times \sqrt{2} + 1 \times (\sqrt{8}) + \sqrt{2} \times (\sqrt{8})$$
$$= 5 + 5\sqrt{2}  \sqrt{8}  \sqrt{2}\sqrt{8}$$
Now, using Law 3 which states that:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
We can break down ??\sqrt{8}??:
$$5 + 5\sqrt{2}  \sqrt{8}  \sqrt{2}\sqrt{8}$$
$$= 5 + 5\sqrt{2}  \sqrt{4 \times 2}  \sqrt{2}\sqrt{4 \times 2}$$
$$= 5 + 5\sqrt{2}  \sqrt{4}\sqrt{2}  \sqrt{2}\sqrt{4}\sqrt{2}$$
$$= 5 + 5\sqrt{2}  2\sqrt{2}  2\sqrt{2}\sqrt{2}$$
$$= 5 + 3\sqrt{2}  2 \times 2$$
$$= 5 + 3\sqrt{2}  4$$
$$= 1 + 3\sqrt{2}$$
For completeness, we should make it clear that ??a=1?? and ??b=3??.
3. (b)
We know that the expression we are targetting contains only terms in ??\sqrt{5}??, so note that Law 3 states:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
Which allows us to break down ??\sqrt{80}??:
$$\sqrt{80} + \frac{30}{\sqrt{5}}$$
$$=\sqrt{16 \times 5} + \frac{30}{\sqrt{5}}$$
$$=\sqrt{16}\sqrt{5} + \frac{30}{\sqrt{5}}$$
$$= 4\sqrt{5} + \frac{30}{\sqrt{5}}$$
Next, rationalise the denominator of the fraction by multiplying both numerator and denominator by ??\sqrt{5}??:
$$ 4\sqrt{5} + \frac{30}{\sqrt{5}}$$
$$= 4\sqrt{5} + \frac{30 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}$$
$$= 4\sqrt{5} + \frac{30\sqrt{5}}{5}$$
$$= 4\sqrt{5} + 6\sqrt{5}$$
$$= 10\sqrt{5}$$
For completeness, we should make it clear that ??c=10??.
Edexcel  Maths  Standard  2013  June  Core 1  Question 1
Question
1.
Simplify ??\frac{7 + \sqrt{5}}{\sqrt{5}1}??, giving your answer in the form ??a+b\sqrt{5}??, where ??a??
and ??b?? are integers. [4]
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Worked Solution
1.
Start by noting that we have an irrational denominator and the answer we are trying to reach does not. We
are therefore going to have to rationalise the denominator. To do this, we will multiply the numerator and
denominator by the conjugate surd of the denominator.
The denominator is ??\sqrt{5}  1?? so the conjugate surd is ??\sqrt{5} + 1??. We can now multiply the numerator
and denominator by the conjugate surd:
$$\frac{7 + \sqrt{5}}{\sqrt{5}1}$$
$$=\frac{ (7 + \sqrt{5}) (\sqrt{5} + 1) }{ (\sqrt{5} 1) (\sqrt{5} + 1) }$$
$$=\frac{7 \times \sqrt{5} + 1 \times 7 + \sqrt{5} \times \sqrt{5} + 1 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5} + 1 \times \sqrt{5} + \sqrt{5} \times (1) + 1 \times (1)}$$
$$=\frac{7\sqrt{5} + 7 + 5 + \sqrt{5}}{51}$$
$$=\frac{12 + 8\sqrt{5}}{4}$$
$$3 + 2\sqrt{5}$$
For completeness, we should make it clear that ??a=3?? and ??b=2??.
Edexcel  Maths  International  2013  June  Core 1  Question 2
Question
2.
Express ??\frac{15}{\sqrt{3}}  \sqrt{27}?? in the form ??k\sqrt{3}??, where ??k?? is an integer. [4]
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Worked Solution
2.
We know that the expression we are targetting contains only terms in ??\sqrt{3}??, so note that Law 3 states:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
Which allows us to break down ??\sqrt{27}??:
$$\frac{15}{\sqrt{3}}  \sqrt{27}$$
$$=\frac{15}{\sqrt{3}}  \sqrt{9 \times 3}$$
$$=\frac{15}{\sqrt{3}}  \sqrt{9}\sqrt{3}$$
$$=\frac{15}{\sqrt{3}}  3\sqrt{3}$$
Next, rationalise the denominator of the fraction by multiplying both numerator and denominator by ??\sqrt{3}??:
$$\frac{15}{\sqrt{3}}  3\sqrt{3}$$
$$=\frac{15 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}  3\sqrt{3}$$
$$=\frac{15\sqrt{3}}{3}  3\sqrt{3}$$
$$=5\sqrt{3}  3 \sqrt{3}$$
$$2\sqrt{3}$$
For completeness, we should make it clear that ??k=2??.
Exam Tips

If you're stuck on a question involving surds, there are two things that you can try:
Rationalise any denominators.
Remove any square factors from inside surds (e.g. ??\sqrt{12}?? can be written as ??2\sqrt{3}?? using Law 3).

Exam questions are often simple applications of the above laws, for example, most often taking the form of rewriting a given expression in another form, through rationalising the denominator, such as:
Rewrite ??\frac{4}{1+\sqrt{3}}?? in the form ??a + b\sqrt{3}??, where ??a?? and ??b?? are integers.
Rewrite ??\frac{8}{\sqrt{7}+3}?? in the form ??a + b\sqrt{7}??, where ??a?? and ??b?? are integers.

Sometimes exam questions involve rewriting a given expression in another form, through the use of Law 3, such as:
Rewrite ??\sqrt{20} + \sqrt{5}?? in the form ??a\sqrt{b}??, where ??a?? and ??b?? are integers.
Rewrite ??\sqrt{32}  \sqrt{8}?? in the form ??a\sqrt{b}??, where ??a?? and ??b?? are integers.

Remember, there is no calculator allowed, so the examiners won’t ask you to complete complicated arithmetic. When looking to break apart a surd using Law 3, start by checking whether each square number (e.g. ??4??, ??9??, ??16??, etc.) is a factor, if it is remove it:
??\sqrt{200}?? becomes ??\sqrt{100 \times 2} = \sqrt{100}\sqrt{2} = 10\sqrt{2}??.
??\sqrt{48}?? becomes ??\sqrt{16 \times 3} = \sqrt{16}\sqrt{3} = 4\sqrt{3}??.

You should learn and become familiar with the laws above. None of the information from this page is in the materials given to you in the exam ('Mathematical Formulae'). Although
surds are sometimes used in other questions, very rarely will a question involve the surd laws on this page and another part of the material.