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Square numbersFactors of a number

Cube numbers Pythagoras theorem

This topic is relevant for:

Here we will learn about surds, including simplifying surds, adding and subtracting surds, multiplying surds and dividing surds and rationalising surds.

There are also surds* *worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**Surds **are numbers left as square roots that give irrational numbers.

An irrational number can’t be written as a fraction, and in decimal form is infinitely long with no recurring pattern – they would go on for ever.

Surds can be a square root, cube root, or other root and are used when detailed accuracy is required in a calculation.

For example the square root of 3 and the cube root of 2 are both surds.

For Example

\sqrt{5} \approx 2.23606 , which is an irrational number.

The square root of 5 is a surd.

When we are not able to simplify a number to remove a square root then it is a **surd **(irrational number). We use **surds** to write numbers precisely.

The decimal form of **irrational numbers** do not terminate or recur, so cannot be written exactly in decimal form. We can leave these numbers written as a **surd** which represents the value in its **exact form**.

For example,

\sqrt2 = 1.4142135623 ...If we were to use 1.41 (2dp) in our calculations they would not be accurate. We need to use it in exact form, \sqrt2, so that the full value of the number is used.

We may also come across questions that ask us to leave our answers in **exact form**.

For example,

Find the value of x. Give your answer in exact form.

We need to use Pythagoras’ theorem.

a^{2}+b^{2} =c^{2} 4^{2}+6^{2} =x^{2} 16+36 =x^{2} 52 =x^{2} x =\sqrt{52} x =2 \sqrt{13}A number that can be written as an integer (whole number) or a simple fraction is called a **rational number.** Rational numbers can be terminating decimals or recurring decimals.

E.g.

2, 100, -3, \frac{3}{4} and \frac{1}{9}.

Any number that can’t be written in this form is called an **irrational number**.

Irrational numbers in decimal form are infinite, with no recurring or repeating pattern.

E.g.

\pi is an example of an irrational number

When a root (square root, cube root or higher) gives an irrational number, it is called a **surd**. At GCSE, we are only concerned with square roots.

E.g.

\sqrt{4}=2 , which is an integer,

The square root of 4 **is not a surd.**

\sqrt{5} \approx 2.23606 , which is an infinitely long decimal with no recurring or repeating pattern, i.e. an irrational number.

The square root of 5 **is a surd.**

Get your free surds worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free surds worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREESimplify:

\sqrt{60}**Find a square number that is a factor of the number under the root.**

Square numbers are 1, 4, 9, 16, 25, …

4 is a factor of 60 (because 4 \times 15=60 ).

2**Rewrite the surd as a product of this square number and another number, then evaluate the root of the square number**.

3**Repeat if the number under the root still has square factors**.

In this case, there are no square numbers that are factors of 2 , so the surd is fully simplified.

\sqrt{60}=2\sqrt{15}**Step by step guide: Simplifying surds**

Simplify:

2\sqrt{5}+\sqrt{45}**Check whether the terms are ‘like surds’**.

The numbers under the root signs are 5 and 45 ; these are not like surds.

**If they aren’t like surds, simplify each surd as far as possible**.

2\sqrt{5} is already fully simplified; there are no square factors of 5 .

\sqrt{45} can be simplified, because 9 is a square factor of 45 .

\begin{aligned}
\sqrt{45} &=\sqrt{9 \times 5} \\\\
&=\sqrt{9} \times \sqrt{5} \\\\
&=3 \times \sqrt{5} \\\\
&=3 \sqrt{5}
\end{aligned}

**Combine the like surd terms by adding or subtracting**.

\begin{aligned} 2 \sqrt{5}+\sqrt{45} &=2 \sqrt{5}+3 \sqrt{5} \\\\ &=5 \sqrt{5} \end{aligned}

** **

**Step by step guide: Adding and subtracting surds**

** Simplify the surds if possible**.

There are no square factors of 10 , so \sqrt{10} cannot be simplified further.

9 is a square factor of 90 , so \sqrt{90} can be simplified as follows:

\begin{aligned} \sqrt{90} &=\sqrt{9 \times 10} \\\\ &=\sqrt{9} \times \sqrt{10} \\\\ &=3 \sqrt{10} \end{aligned}

There are no square factors of 2 , so \sqrt{2} cannot be simplified further.

The expression becomes:

\frac{2 \sqrt{10}-3\sqrt{10}}{\sqrt{2}}

** Use surd laws to fully simplify the numerator and denominator of the fraction**.

The numerator contains like surds, so it can be simplified further.

\frac{2\sqrt{10}-3\sqrt{10}}{\sqrt{2}}=\frac{-1\sqrt{10}}{\sqrt{2}} .

As when simplifying algebraic expressions, we would usually write -1\sqrt{10} as just -\sqrt{10}.

** Divide the numerator by the denominator**.

-\sqrt{10}\div\sqrt{2}=-\sqrt{5}

The final answer is:

-\sqrt{5}

**Step by step guide: Multiplying and dividing surds**

Expand and simplify:

\sqrt{3}(6+\sqrt{8})** Simplify the surds if possible**.

There are no square factors of 3 , so \sqrt{3} cannot be simplified further.

4 is a square factor of 8 , so \sqrt{8} can be simplified as follows:

\begin{aligned} \sqrt{8}&=\sqrt{4 \times 2}\\\\ &=\sqrt{4} \times \sqrt{2}\\\\ &=2\sqrt{2} \end{aligned}

So the expression becomes \sqrt{3}(6+2 \sqrt{2}) .

** Multiply each term inside the bracket by the term outside**.

You can do this in a table, as you would if you were expanding brackets containing algebraic terms.

\sqrt{3} \times 6=6 \sqrt{3}

\sqrt{3}\times 2\sqrt{2}=2\sqrt{6}

** Check and simplify the answer further if possible**.

The answer is:

6 \sqrt{3}+2 \sqrt{6}

This cannot be simplified further.

** Simplify any surds, if necessary**.

Root 54 will simplify:

\begin{aligned} \sqrt{54} &=\sqrt{9 \times 6} \\\\ &=\sqrt{9} \times \sqrt{6} \\\\ &=3 \sqrt{6} \end{aligned}

\frac{12}{\sqrt{54}}=\frac{12}{3 \sqrt{6}}

** Multiply both the numerator and the denominator by the surd in the denominator**.

So here we multiply the top and the bottom of the fraction by root 6.

\frac{12 \times \sqrt{6}}{3 \sqrt{6} \times \sqrt{6}}

Numerator:

12\times\sqrt{6}=12\sqrt{6}

Denominator:

3 \sqrt{6} \times \sqrt{6}=3 \times 6=18

So the full expression becomes:

\frac{12 \sqrt{6}}{18}

The denominator is now rationalised, because 18 is a rational number.

** Simplify the answer fully**.

12\div 18=\frac{12}{18}=\frac{2}{3}

So the final answer simplifies to

\frac{2 \sqrt{6}}{3}

Step by step guide: Rationalising the denominator

Rationalise the denominator:

\frac{6}{\sqrt{11}-3}** Change the sign of the expression in the denominator**.

The denominator is

(\sqrt{11}-3)

so we need

(\sqrt{11}+3)

** Multiply both the numerator and the denominator of the original fraction by this new expression**.

\frac{6 \times(\sqrt{11}+3)}{(\sqrt{11}-3)(\sqrt{11}+3)}

**Numerator**:

6 \times(\sqrt{11}+3)=6 \sqrt{11}+18

**Denominator**:

\begin{aligned} (\sqrt{11}-3)(\sqrt{11}+3) &=11+3 \sqrt{11}-3 \sqrt{11}-9 \\\\ &=11-9 \\\\ &=2 \end{aligned}

So the full expression becomes:

\frac{6 \sqrt{11}+18}{2}

The denominator is now rationalised, because 2 is a rational number.

** Simplify the answer fully**.

2 is a factor of 6 and 18, so we can divide through by 2 , leaving the answer as:

3 \sqrt{11}+9

**Step-by-step guide:** Rationalise the denominator

**Incorrectly rewriting the number under the square root sign (the radicand) as a product of any two factors**

One of these factors must be a square number in order for you to be able to simplify the surd.

**Not simplifying fully**

Always check that there are no square factors of the number under the root.

1**Remember that a root with no integer coefficient is ‘****lot’ of that surd**

As in algebra, we understand that a actually means 1a .

**Don’t mix up addition and multiplication laws**

Refer back to knowledge of algebra to help:

a+a=2a and a \times a=a^2.

**Trying to combine unlike surds**

It’s OK to leave an answer with more than one surd in it if it will not simplify further.

**Take care when expanding double brackets when using a conjugate surd expression**

Remember to multiply everything in the first bracket by everything in the second, and check that you have four terms before simplifying.

1. Simplify fully:

\sqrt{450}

3 \sqrt{50}

8 \sqrt{2}

15 \sqrt{2}

9 \sqrt{50}

In one stage:

\sqrt{450}=\sqrt{225 \times 2}=15 \sqrt{2}

In two stages:

9 \times 50=450, so rewrite \sqrt{450}=\sqrt{9 \times 50}=3 \sqrt{50}

25 \times 2=50, so rewrite 3 \sqrt{50}=3 \sqrt{25 \times 2}=3 \times 5 \times \sqrt{2}=15 \sqrt{2}

2. Simplify:

\sqrt{50}-2 \sqrt{2}

2 \sqrt{25}

3 \sqrt{2}

2 \sqrt{5}

2 \sqrt{48}

Rewrite \sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2} , then subtract like surds to get the answer.

3. Simplify fully:

\frac{\sqrt{60}}{2 \sqrt{3}}

\frac{\sqrt{5}}{2}

\frac{\sqrt{20}}{2}

2 \sqrt{20}

\sqrt{5}

Simplify \sqrt{60}=\sqrt{4 \times 15}=2 \sqrt{15}

2\sqrt{15}\div 2\sqrt{3}=\sqrt{5}

4. Expand and simplify:

4(3+\sqrt{28})

12+2\sqrt{7}

12+4\sqrt{28}

12+4\sqrt{7}

12+8\sqrt{7}

Simplify \sqrt{28}=\sqrt{4 \times 7}=2 \sqrt{7}

\begin{aligned} &4 \times 3=12 \\\\ &4 \times 2 \sqrt{7}=8 \sqrt{7} \end{aligned}

So the final answer is 12+8\sqrt{7}.

5. Expand and simplify:

(2+\sqrt{3})(2-\sqrt{3})

4-3 \sqrt{3}

1

-1

7

Multiply each term in the first bracket by each term in the second bracket to get 4+2 \sqrt{3}-2 \sqrt{3}-3. The surd terms cancel out (as with a difference of two squares), and the integer terms simplify to 1.

6. Rationalise:

\frac{4}{\sqrt{5}}

\frac{\sqrt{5}}{5}

4 \sqrt{5}

\frac{4 \sqrt{5}}{5}

Multiply the numerator and denominator by \sqrt{5}.

1. \sqrt{775}=5\sqrt{k}

where k is an integer. Find the value of k.

**(2 marks)**

Show answer

\sqrt{775}=\sqrt{25}\times\sqrt{k}

or sight of 775 \div 25

**(1)**

5\sqrt{31}

so k=31

**(1)**

2. (a) Simplify \sqrt{375}

(b) Hence, or otherwise, simplify:

\sqrt{375}+\sqrt{960}

**(4 marks)**

Show answer

(a)

\sqrt{375}=\sqrt{25}\times\sqrt{15}

**(1)**

**(1)**

(b)

\sqrt{960}=\sqrt{64} \times \sqrt{15}

**(1)**

**(1)**

3. (a) Expand and simplify

(5+\sqrt{7})(5-\sqrt{7})

(b) Hence, or otherwise, show that

\frac{36}{5+\sqrt{7}} can be written as 10-2 \sqrt{7}

**(5 marks)**

Show answer

(a)

25+5 \sqrt{7}-5 \sqrt{7}-7

Any two correct terms

**(1)**

All four correct terms

**(1)**

Simplified to 18

**(1)**

(b)

\frac{36(5-\sqrt{7})}{(5+\sqrt{7})(5-\sqrt{7})}

**(1)**

\frac{180-36 \sqrt{7}}{18} leading to 10-2 \sqrt{7} as required

**(1)**

You have now learned how to:

- Simplify surds
- Add and subtract surds
- Multiply and divide surds
- Rationalise the denominator
- Rationalise other surd expressions

- Using exact values in trigonometry
- Geometric progressions with surds

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#### GCSE Maths Papers - November 2022 Topics

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Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

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