Numbers: Numbers and number relationships

# Unit 5: Simplify surds

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

• Rationalise fractions with surd denominators.
• Add, subtract, multiply and divide simple surds.

## What you should know

Before you start this unit, make sure you can:

• Identify rational and irrational numbers. Work through Unit 1: Identify and work with rational and irrational numbers if you need to revise rational and irrational numbers.
• Apply all the exponent laws and write powers using rational exponents. If you need to revise exponent laws and rational exponents, go over Units 2, 3 and 4 within this topic.
• Simplify these expressions:
1. $\scriptsize {{2}^{2x}}{{.3}^{x}}{{.2}^{x-1}}$
2. $\scriptsize \sqrt[3]{{{x}^{2}}.y}\times {{x}^{\displaystyle \frac{1}{3}}}.{{y}^{\displaystyle \frac{2}{3}}}$
3. $\scriptsize \sqrt[8]{\sqrt[4]{{{a}^{32}}{{b}^{64}}}}$

Solutions

1. .
\scriptsize \begin{align} & {{2}^{2x}}{{.3}^{x}}{{.2}^{x-1}}={{2}^{2x+x-1}}{{.3}^{x}} \\ & ={{2}^{3x-1}}{{.3}^{x}} \end{align}
2. .
\scriptsize \begin{align} & \sqrt[3]{{{x}^{2}}.y}\times {{x}^{\displaystyle \frac{1}{3}}}.{{y}^{\displaystyle \frac{2}{3}}}={{x}^{\displaystyle \frac{2}{3}}}\cdot {{y}^{\displaystyle \frac{1}{3}}}\times {{x}^{\displaystyle \frac{1}{3}}}.{{y}^{\displaystyle \frac{2}{3}}} \\ &={{x}^{\displaystyle \frac{1}{3}+\displaystyle \frac{2}{3}}}.{{y}^{\displaystyle \frac{1}{3}+\displaystyle \frac{2}{3}}} \\ &=x.y \end{align}
3. .
\scriptsize \begin{align} & \sqrt[8]{\sqrt[4]{{{a}^{32}}{{b}^{64}}}}=\sqrt[8]{{{({{a}^{32}})}^{\displaystyle \frac{1}{4}}}{{({{b}^{64}})}^{\displaystyle \frac{1}{4}}}} \\ &=\sqrt[8]{{{(a)}^{\displaystyle \frac{32}{4}}}{{(b)}^{\displaystyle \frac{64}{4}}}} \\ &=\sqrt[8]{{{a}^{8}}{{b}^{16}}} \\ &={{a}^{\displaystyle \frac{8}{8}}}{{b}^{\displaystyle \frac{16}{8}}} \\ & =a{{b}^{2}} \end{align}

## Introduction

We saw in Unit 4 of this topic that $\scriptsize {{a}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{a}$. We also learnt that $\scriptsize \sqrt[n]{a}$ is called the nth root of $\scriptsize a$ and is known as a radical or root. Let us recap what the parts that make up a radical are called.

Remember we write fractional exponents as roots. The number in the numerator of the fraction becomes the power of the base inside the root, and the number in the denominator of the fraction becomes the index.

Some radicals can be written as rational numbers (positive or negative whole numbers or fractions), for example $\scriptsize \sqrt{64}=8$ or $\scriptsize \sqrt[3]{\displaystyle \frac{8}{27}}=\displaystyle \frac{2}{3}$.

But, some radicals cannot be written as rational numbers and we can only work out a rough approximation of their value. So it is best to leave them in radical form, for example $\scriptsize \sqrt{55}$ or $\scriptsize \sqrt[3]{\displaystyle \frac{1}{3}}$. We call these irrational roots, surds. In other words, surds are roots that cannot be reduced to a whole number or fraction.

Leaving a root in surd form is easier and more accurate than writing and rounding off the decimal value. However, there are methods we can use to simplify surds.

You will learn various ways to simplify surds in this unit.

## Simplest surd form

The most basic way to simplify surds is to rewrite the radicand as a product of factors that can be further simplified. So, we need to find factors that are perfect nth roots in order for them to give ‘nice’ answers when we take them out of the radical sign. For example, $\scriptsize \sqrt{80}$ can be written as $\scriptsize \sqrt{8\times 10}$ or $\scriptsize \sqrt{16\times 5}$. Can you see that $\scriptsize \sqrt{16\times 5}$ is the better option because we can take the $\scriptsize \sqrt{16}=4$ so $\scriptsize \sqrt{16\times 5}=4\sqrt{5}$.

The product rule for simplifying roots is $\scriptsize \sqrt[n]{ab}=\sqrt[n]{a}\times \sqrt[n]{b}$.

Here is an example for you to work through.

### Example 5.1

Simplify: $\scriptsize \sqrt[3]{54}$

Solution

First, write the radicand as a product of prime factors.

$\scriptsize \sqrt[3]{{54}}=\sqrt[3]{{27\times 2}}=\sqrt[3]{{3\times 3\times 3\times 2}}=\sqrt[3]{{{{3}^{3}}\cdot 2}}$

Next, use $\scriptsize \sqrt[n]{ab}=\sqrt[n]{a}\times \sqrt[n]{b}$ to simplify further.

$\scriptsize \sqrt[3]{{54}}=\sqrt[3]{{{{3}^{3}}}}\times \sqrt[3]{2}$

Lastly, find the nth roots where possible.

$\scriptsize \sqrt[3]{{{3}^{3}}}\times \sqrt[3]{2}=3\times \sqrt[3]{2}=3\sqrt[3]{2}$

### Exercise 5.1

Simplify these radical expressions:

1. $\scriptsize \sqrt{300}$
2. $\scriptsize \sqrt{50{{a}^{6}}{{b}^{5}}}$
3. $\scriptsize \sqrt[3]{3}\times \sqrt[3]{9}$
4. $\scriptsize \sqrt{50a}\times \sqrt{2a}$
5. $\scriptsize \sqrt[3]{2}\times \sqrt{3}$

The full solutions are at the end of the unit.

The quotient rule for simplifying roots is $\scriptsize \sqrt[n]{\displaystyle \frac{a}{b}}=\displaystyle \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

Just as we can rewrite the nth root of a product as a product of nth roots, so too can we rewrite the nth root of a quotient as a quotient of nth roots. We can separate the numerator and denominator of a fraction under a radical so that we can take their roots separately and vice versa.

The rule above is shown in Example 5.2 below.

### Example 5.2

Simplify:

1. $\scriptsize \sqrt{\displaystyle \frac{5}{49}}$
2. $\scriptsize \displaystyle \frac{\sqrt{234{{a}^{11}}b}}{\sqrt{26{{x}^{7}}y}}$

Solutions

1. Rewrite this as a quotient of two radical expressions.
$\scriptsize \sqrt{\displaystyle \frac{5}{49}}=\displaystyle \frac{\sqrt{5}}{\sqrt{49}}\text{ }$
Simplify the denominator. The numerator is already in its simplest or surd form.
$\scriptsize =\displaystyle \frac{\sqrt{5}}{7}\text{ }$
2. Combine the numerator and denominator into one radical expression.
$\scriptsize \displaystyle \frac{\sqrt{234{{x}^{11}}y}}{\sqrt{26{{x}^{7}}y}}=\sqrt{\displaystyle \frac{234{{x}^{11}}y}{26{{x}^{7}}y}}\text{ }$
Simplify the fraction within the radicand and then simplify the root.
\scriptsize \begin{align} & \sqrt{9{{x}^{11-7}}{{y}^{0}}}=\sqrt{9{{x}^{4}}} \\ & =3{{x}^{2}} \end{align}

### Adding and subtracting surds

We can add or subtract radical expressions only when they have the same radicand and the same index, for example $\scriptsize \sqrt{3}+2\sqrt{3}$ is $\scriptsize 3\sqrt{3}$. Adding and subtracting radicals is similar to collecting like terms. Just as we cannot simplify $\scriptsize x+y$ so too we cannot simplify $\scriptsize \sqrt{3}+\sqrt[3]{3}$. Even though they have the same radicand the indexes are different so we cannot add them.

### Example 5.3

Simplify:

1. $\scriptsize 5\sqrt{3}+2\sqrt{3}$
2. $\scriptsize 6\sqrt{20}-\sqrt{5}$
3. $\scriptsize \sqrt[3]{x}+2\sqrt[3]{x}$

Solutions

1. $\scriptsize 5\sqrt{3}+2\sqrt{3}=7\sqrt{3}$
2. .
\scriptsize \begin{align} & 6\sqrt{20}-\sqrt{5}=6\sqrt{4\times 5}-\sqrt{5} \\ & =6\sqrt{4}\times \sqrt{5}-\sqrt{5} \\ & =6(2)\sqrt{5}-\sqrt{5} \\ & =12\sqrt{5}-\sqrt{5} \\ & =11\sqrt{5} \end{align}
3. $\scriptsize \sqrt[3]{x}+2\sqrt[3]{x}=3\sqrt[3]{x}$

Work through this exercise to assess your understanding of surds.

### Exercise 5.2

Simplify:

1. $\scriptsize \sqrt{12}+\sqrt{3}$
2. $\scriptsize \sqrt{32}-\sqrt{98}$
3. $\scriptsize {{\left( \sqrt{20}-\sqrt{5} \right)}^{2}}$

The full solutions are at the end of the unit.

### Note

If you would like to see a summary of simplifying surds you can watch the following videos online if you have an internet connection.

Adding and subtracting surds (Duration: 01.57)

## Rationalising denominators

When an expression with radicals is written in simplest form, it will not contain any root in the denominator. We can remove radicals from the denominators of fractions by rationalising the denominator.

What happens to the value of $\scriptsize \sqrt{2}$ if we multiply it by $\scriptsize \displaystyle \frac{\sqrt{2}}{\sqrt{2}}$? Can you see that $\scriptsize \displaystyle \frac{{\sqrt{2}}}{{\sqrt{2}}}$ is the same as $\scriptsize 1$ ?

We know that multiplying by one does not change the value of an expression. So if we multiply $\scriptsize \sqrt{2}$ by $\scriptsize \displaystyle \frac{\sqrt{2}}{\sqrt{2}}$ its value will not change.

Let’s see how this works:

\scriptsize \begin{align} & \displaystyle \frac{1}{\sqrt{2}}\times \displaystyle \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{\sqrt{2}}{{{(\sqrt{2})}^{2}}} \\ & =\displaystyle \frac{\sqrt{2}}{2} \end{align}

Now there is a rational number in the denominator. This fraction is in simplest form as the denominator does not contain a radical. This is the technique we use to rationalise denominators.

### Take note!

To remove radicals from the denominators of fractions, multiply by the form of $\scriptsize 1$ that will eliminate the radical.

For a denominator containing a single term, multiply by the radical in the denominator over itself.

For example, in $\scriptsize \displaystyle \frac{2}{a\sqrt{b}}$ the denominator is $\scriptsize a\sqrt{b}$, so we must multiply by $\scriptsize \displaystyle \frac{\sqrt{b}}{\sqrt{b}}$ to write the surd in its simplest form.

To rationalise a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator.

Remember that the conjugate of an expression of the form $\scriptsize a+b\sqrt{c}$ is $\scriptsize a-b\sqrt{c}$. We find the conjugate by changing the sign of the radical portion of the expression.

For example, in $\scriptsize \displaystyle \frac{2}{a+b\sqrt{c}}$ the denominator is $\scriptsize a+b\sqrt{c}$, so we must multiply by $\scriptsize \displaystyle \frac{a-b\sqrt{c}}{a-b\sqrt{c}}$ to rationalise the denominator.

### Example 5.4

Rewrite the following in simplest surd form:

1. $\scriptsize \displaystyle \frac{\sqrt{2}}{2\sqrt{3}}$
2. $\scriptsize \displaystyle \frac{3}{1-2\sqrt{5}}$
3. $\scriptsize \displaystyle \frac{4}{\sqrt{6}-2}$

Solutions

1. The radical in the denominator is $\scriptsize \sqrt{3}$. To simplify, multiply by $\scriptsize \displaystyle \frac{\sqrt{3}}{\sqrt{3}}$.
\scriptsize \begin{align} & \displaystyle \frac{\sqrt{2}}{2\sqrt{3}}\times \displaystyle \frac{\sqrt{3}}{\sqrt{3}}=\displaystyle \frac{\sqrt{2}\times \sqrt{3}}{2{{\left( \sqrt{3} \right)}^{2}}} \\ & =\displaystyle \frac{\sqrt{2}\cdot \sqrt{3}}{2\cdot 3} \\ & =\displaystyle \frac{\sqrt{6}}{6} \\ \end{align}
2. The denominator is $\scriptsize 1-2\sqrt{5}$. Begin by finding the conjugate of the denominator by writing the denominator and changing the sign between the terms. The conjugate of $\scriptsize 1-2\sqrt{5}$ is $\scriptsize 1+2\sqrt{5}$. Therefore, we multiply the fraction by $\scriptsize \displaystyle \frac{1+2\sqrt{5}}{1+2\sqrt{5}}$. By multiplying the denominator by its conjugate, you will find that that it will always lead to the result of expanding a difference of two squares.
\scriptsize \begin{align} & \displaystyle \frac{3}{1-2\sqrt{5}}\times \displaystyle \frac{1+2\sqrt{5}}{1+2\sqrt{5}}=\displaystyle \frac{3(1+2\sqrt{5})}{(1+2\sqrt{5})(1-2\sqrt{5})} \\ & =\displaystyle \frac{3+6\sqrt{5}}{1-4{{\left( \sqrt{5} \right)}^{2}}} \\ & =\displaystyle \frac{3+6\sqrt{5}}{1-4\left( 5 \right)} \\ & =\displaystyle \frac{3+6\sqrt{5}}{-19} \end{align}
3. .
\scriptsize \begin{align} & \displaystyle \frac{4}{\sqrt{6}-2}\times \displaystyle \frac{4}{\sqrt{6}+2}=\displaystyle \frac{16}{(\sqrt{6}-2)(\sqrt{6}+2)} \\ & =\displaystyle \frac{16}{{{(\sqrt{6})}^{2}}-{{(2)}^{2}}} \\ & =\displaystyle \frac{16}{6-2} \\ & =4 \end{align}

Now, do this exercise to check your understanding of rationalising denominators.

### Exercise 5.3

Simplify each expression:

1. $\scriptsize \displaystyle \frac{\sqrt{8}}{1-\sqrt{2}}$
2. $\scriptsize \displaystyle \frac{\sqrt{12}}{1-\sqrt{3x}}$
3. $\scriptsize \displaystyle \frac{\sqrt{27}}{2+2\sqrt{3}}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to add and subtract surds.
• How to multiply and divide surds.
• How to rationalise denominators to simplify surds.

# Unit 5: Assessment

#### Suggested time to complete: 20 minutes

1. Simplify:
1. $\scriptsize \displaystyle \frac{\sqrt{20}+\sqrt{5}+\sqrt{45}}{\sqrt{80}}$
2. $\scriptsize \displaystyle \frac{\sqrt{20}+\sqrt{80}+\sqrt{125}-\sqrt{5}}{2\sqrt{5}+3\sqrt{5}}$
3. $\scriptsize \displaystyle \frac{3\sqrt{12}+4\sqrt{75}+-\sqrt{27}}{\sqrt{75}}$
2. Simplify by rationalising the denominator:
1. $\scriptsize \displaystyle \frac{3}{3-\sqrt{2}}$
2. $\scriptsize \displaystyle \frac{3}{2\sqrt{7}-7}$
3. $\scriptsize \displaystyle \frac{4}{2\sqrt{3}-3\sqrt{2}}$
3. Prove that: $\scriptsize \displaystyle \frac{\sqrt{125}+\sqrt{80}-\sqrt{20}}{\sqrt{180}+\sqrt{5}}=1$

The full solutions are at the end of the unit.

# Unit 5: Solutions

### Exercise 5.1

1. .
\scriptsize \begin{align} & \sqrt{300}=\sqrt{3\times {{10}^{2}}} \\ & =\sqrt{3}\times \sqrt{{{10}^{2}}} && \text{ Rewrite as product of radical expressions} \\ & =10\sqrt{3} \end{align}
2. .
\scriptsize \begin{align} & \sqrt{50{{a}^{6}}{{b}^{5}}}=\sqrt{25\times 2{{a}^{6}}{{b}^{5}}} \\ & =\sqrt{25{{a}^{6}}}\times \sqrt{2{{b}^{5}}}\text{ } \\ & \text{=5}{{\text{a}}^{3}}\sqrt{2{{b}^{5}}} \end{align}
3. .
\scriptsize \begin{align} & \sqrt[3]{3}\times \sqrt[3]{9}=\sqrt[3]{3\times 9} && \text{ Express the product as a single radical expression}\text{.} \\ & =\sqrt[3]{27} \\ & =3 \end{align}
4. .
\scriptsize \begin{align} & \sqrt{50a}\times \sqrt{2a}=\sqrt{50a\times 2a} \\ & =\sqrt{100{{a}^{2}}} \\ & =10a \end{align}
5. .
\scriptsize \begin{align} & \sqrt[3]{2}\times \sqrt{3} && \text{ Since the index of the roots are different, you cannot simplify any further}\text{.} \end{align}

Back to Exercise 5.1

### Exercise 5.2

1. .
\scriptsize \begin{align} & \sqrt{12}+\sqrt{3}=\sqrt{4\cdot 3}+\sqrt{3} \\ & =2\sqrt{3}+\sqrt{3} \\ & =3\sqrt{3} \end{align}
2. .
\scriptsize \begin{align} & \sqrt{32}-\sqrt{98}=\sqrt{16\cdot 2}-\sqrt{49\cdot 2} \\ & =4\sqrt{2}-7\sqrt{2} \\ & =-3\sqrt{2} \end{align}
3. .
\scriptsize \begin{align} & {{\left( \sqrt{20}-\sqrt{5} \right)}^{2}}={{\left( \sqrt{4\cdot 5}-\sqrt{5} \right)}^{2}} \\ & ={{\left( 2\sqrt{5}-\sqrt{5} \right)}^{2}} \\ & ={{\left( \sqrt{5} \right)}^{2}} \\ & =5 \end{align}

Back to Exercise 5.2

### Exercise 5.3

1. .
\scriptsize \begin{align} & \displaystyle \frac{\sqrt{8}}{1-\sqrt{2}}\times \displaystyle \frac{1+\sqrt{2}}{1+\sqrt{2}}=\displaystyle \frac{\sqrt{4\cdot 2}\left( 1+\sqrt{2} \right)}{\left( 1+\sqrt{2} \right)\left( 1-\sqrt{2} \right)} \\ & =\displaystyle \frac{2\sqrt{2}\left( 1+\sqrt{2} \right)}{1-{{\left( \sqrt{2} \right)}^{2}}} \\ & =\displaystyle \frac{2\sqrt{2}+2{{\left( \sqrt{2} \right)}^{2}}}{1-2} \\ & =\displaystyle \frac{2\sqrt{2}+2(2)}{-1} \\ & =-2\sqrt{2}-4 \end{align}
2. .
\scriptsize \begin{align} & \displaystyle \frac{\sqrt{12}}{1-\sqrt{3x}}\times \displaystyle \frac{1+\sqrt{3x}}{1+\sqrt{3x}}=\displaystyle \frac{\sqrt{4\cdot 3}(1+\sqrt{3x})}{(1-\sqrt{3x})(1+\sqrt{3x})} && \text{ Multiply by the conjugate} \\ & =\displaystyle \frac{2\sqrt{3}(1-\sqrt{3x})}{1-{{(\sqrt{3x})}^{2}}} \\ & =\displaystyle \frac{2\sqrt{3}-2\sqrt{3}(\sqrt{3x})}{1-3x} \\ & =\displaystyle \frac{2\sqrt{3}-2{{(\sqrt{3})}^{2}}\sqrt{x}}{1-3x} \\ & =\displaystyle \frac{2\sqrt{3}-6\sqrt{x}}{1-3x} \end{align}
3. .
\scriptsize \begin{align} & \displaystyle \frac{\sqrt{27}}{2+2\sqrt{3}}\times \displaystyle \frac{2-2\sqrt{3}}{2-2\sqrt{3}}=\displaystyle \frac{\sqrt{9\cdot 3}\left( 2-2\sqrt{3} \right)}{\left( 2+2\sqrt{3} \right)\left( 2-2\sqrt{3} \right)} \\ & =\displaystyle \frac{3\sqrt{3}\left( 2-2\sqrt{3} \right)}{4-4(3)} \\ & =\displaystyle \frac{6\sqrt{3}-6(3)}{-8} && \text{ Divide by common factor of }2 \\ & =\displaystyle \frac{3\sqrt{3}-9}{-4} \end{align}

Back to Exercise 5.3

### Unit 5: Assessment

1. .
1. .
\scriptsize \begin{align} & \displaystyle \frac{\sqrt{20}+\sqrt{5}+\sqrt{45}}{\sqrt{80}}=\displaystyle \frac{\sqrt{4\cdot 5}+\sqrt{5}+\sqrt{9\cdot 5}}{\sqrt{16\cdot 5}} \\ & =\displaystyle \frac{2\sqrt{5}+\sqrt{5}+3\sqrt{5}}{4\sqrt{5}} \\ & =\displaystyle \frac{6\sqrt{5}}{4\sqrt{5}} \\ & =\displaystyle \frac{6}{4} \end{align}
2. .
\scriptsize \begin{align} & \displaystyle \frac{\sqrt{20}+\sqrt{80}+\sqrt{125}-\sqrt{5}}{2\sqrt{5}+3\sqrt{5}}=\displaystyle \frac{\sqrt{4\cdot 5}+\sqrt{16\cdot 5}+\sqrt{25\cdot 5}-\sqrt{5}}{5\sqrt{5}} \\ & =\displaystyle \frac{2\sqrt{5}+4\sqrt{5}+5\sqrt{5}-\sqrt{5}}{5\sqrt{5}} \\ & =\displaystyle \frac{10\sqrt{5}}{5\sqrt{5}} \\ & =2 \end{align}
3. .
\scriptsize \begin{align} & \displaystyle \frac{3\sqrt{12}+4\sqrt{75}-2\sqrt{27}}{\sqrt{75}}=\displaystyle \frac{3\sqrt{4\cdot 3}-4\sqrt{25\cdot 3}-2\sqrt{9\cdot 3}}{\sqrt{25\cdot 3}} \\ & =\displaystyle \frac{6\sqrt{3}+20\sqrt{3}-6\sqrt{3}}{5\sqrt{3}} \\ & =\displaystyle \frac{20\sqrt{3}}{5\sqrt{3}} \\ & =4 \\ \end{align}
2. .
1. .
\scriptsize \begin{align} & \displaystyle \frac{3}{3-\sqrt{2}}\times \displaystyle \frac{3+\sqrt{2}}{3+\sqrt{2}}=\displaystyle \frac{3\left( 3+\sqrt{2} \right)}{3-{{\left( \sqrt{2} \right)}^{2}}} \\ & =\displaystyle \frac{9+3\sqrt{2}}{3-2} \\ & =9+3\sqrt{2} \end{align}
2. .
\scriptsize \begin{align} & \displaystyle \frac{3}{2\sqrt{7}-7}\times \displaystyle \frac{2\sqrt{7}+7}{2\sqrt{7}+7}=\displaystyle \frac{3\left( 2\sqrt{7}+7 \right)}{{{\left( 2\sqrt{7} \right)}^{2}}-{{\left( 7 \right)}^{2}}} \\ & =\displaystyle \frac{3\left( 2\sqrt{7}+7 \right)}{28-49} \\ & =\displaystyle \frac{3\left( 2\sqrt{7}+7 \right)}{-21} && \text{ Divide by common factor of }3 \\ & \text{=}\displaystyle \frac{2\sqrt{7}+7}{-7} \end{align}
3. .
\scriptsize \begin{align} & \displaystyle \frac{4}{2\sqrt{3}-3\sqrt{2}}\times \displaystyle \frac{2\sqrt{3}+3\sqrt{2}}{2\sqrt{3}+3\sqrt{2}}=\displaystyle \frac{4\left( 2\sqrt{3}+3\sqrt{2} \right)}{{{\left( 2\sqrt{3} \right)}^{2}}-{{\left( 3\sqrt{2} \right)}^{2}}} \\ & =\displaystyle \frac{4\left( 2\sqrt{3}+3\sqrt{2} \right)}{(4\cdot 3)-(9\cdot 2)} \\ & =\displaystyle \frac{4\left( 2\sqrt{3}+3\sqrt{2} \right)}{-6} & \text{ Divide by common factor of }2 \\ & =\displaystyle \frac{2\left( 2\sqrt{3}+3\sqrt{2} \right)}{-3} \\ & =\displaystyle \frac{4\sqrt{3}+6\sqrt{2}}{-3} \end{align}
3. .
To prove that $\scriptsize \displaystyle \frac{\sqrt{125}+\sqrt{80}-\sqrt{20}}{\sqrt{180}+\sqrt{5}}=1$:\scriptsize \begin{align} & \displaystyle \frac{\sqrt{125}+\sqrt{80}-\sqrt{20}}{\sqrt{180}+\sqrt{5}}=\displaystyle \frac{\sqrt{25\cdot 5}+\sqrt{16\cdot 5}-\sqrt{4\cdot 5}}{\sqrt{36\cdot 5}+\sqrt{5}} \\ & =\displaystyle \frac{5\sqrt{5}+4\sqrt{5}-2\sqrt{5}}{6\sqrt{5}+\sqrt{5}} \\ & =\displaystyle \frac{7\sqrt{5}}{7\sqrt{5}} \\ & =1 \\ & \therefore \displaystyle \frac{\sqrt{125}+\sqrt{80}-\sqrt{20}}{\sqrt{180}+\sqrt{5}}=1 \end{align}

Back to Unit 5: Assessment

## License

National Curriculum (Vocational) Mathematics Level 2 by Natashia Bearam-Edmunds is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.