# Powers and Roots Worksheets, Questions and Revision

GCSE 4 - 5KS3AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022

## Powers and Roots

Powers are a shorthand way of expressing repeated multiplication. Roots are ways of reversing this. There are a total of $10$ indices rules. This page will give you the 7 easy rules to remember; there are $3$ further more complex rules which can be found in the laws of indices page.

Make sure you are happy with the following topics before continuing.

Level 4-5 GCSE KS3

## Indices Rule 1: The Multiplication Law

The multiplication law states that when you multiply similar terms, you add the powers as shown,

$a^\textcolor{red}{b} \times a^\textcolor{blue}{c} = a^{\textcolor{red}{b} + \textcolor{blue}{c}}$

This multiplication law applies to all terms with powers (positive or negative): e.g.

$x^{\textcolor{red}{-m}}\times x^\textcolor{blue}{n}=x^{({\textcolor{red}{-m})}\textcolor{blue}{+n}}=x^{\textcolor{blue}{n}\textcolor{red}{-m}}$

This works for fractional powers too. Remember when adding fractions, they must  share a common denominator.

$x^{\textcolor{red}{\frac{1}{3}}} \times$ $x^{\textcolor{blue}{\frac{1}{6}}}=$$x^{\textcolor{red}{\frac{2}{6}}+\textcolor{blue}{\frac{1}{6}}} = x^{\textcolor{black}{\frac{3}{6}}} = x^{\textcolor{black}{\frac{1}{2}}}$

Level 4-5 GCSE KS3

## Indices Rule 2: The Division Law

The division law is when you divide similar terms and in doing so, you subtract the powers:

$a^\textcolor{red}{b} \div a^\textcolor{blue}{c} = a^{\textcolor{red}{b} - \textcolor{blue}{c}}$

The division law applies to all numbers, negative numbers and fractional powers,

$x^\textcolor{red}{6}\div x^\textcolor{blue}{2}=\dfrac{x^\textcolor{red}{6}}{x^\textcolor{blue}{2}}=x^{\textcolor{red}{6} - \textcolor{blue}{2}} = x^{4}$

Level 4-5 GCSE KS3

## Indices Rule 3: Multiple Powers Law

The multiple powers law is when you raise one power to another, i.e. the power of a power. When this happens the powers are multiplied:

$\left(a^\textcolor{red}{b}\right)^\textcolor{limegreen}{c}=a^{\textcolor{red}{b}\textcolor{limegreen}{c}}$

A basic example shows how the multiple powers law works with numbers:

$\left(x^\textcolor{red}{3}\right)^\textcolor{limegreen}{2}=x^{\textcolor{red}{3}\times\textcolor{limegreen}{2}}=x^{6}$

Level 4-5 GCSE KS3
Level 4-5 GCSE KS3

## Indices Rule 4: Power $0$ Law

Anything to the power $0 = 1$

$a^\textcolor{blue}{0} = \textcolor{red}{1}$

The power $0$ law applies to everything: $100^\textcolor{blue}{0}=\textcolor{red}{1}, \quad x^\textcolor{blue}{0}=\textcolor{red}{1} \quad \pi^\textcolor{blue}{0}=\textcolor{red}{1}$

Level 4-5 GCSE KS3

## Indices Rule 5: Power $1$ Law

Anything to the power $1$ is just itself.

$\textcolor{red}{a}^\textcolor{blue}{1} = \textcolor{red}{a}$

The power $1$ law applies to everything: $\textcolor{red}{100}^\textcolor{blue}{1}=\textcolor{red}{100}, \quad \textcolor{red}{x}^\textcolor{blue}{1}=\textcolor{red}{x}, \quad \textcolor{red}{\pi}^\textcolor{blue}{1}=\textcolor{red}{\pi}$

Level 4-5 GCSE KS3

## Indices Rule 6: The $1$ Law

$1$ to the power anything $= 1$ e.g.

$\textcolor{red}{1}^\textcolor{blue}{x} =\textcolor{red}{1}$

This works for any power: $\textcolor{red}{1}^\textcolor{blue}{100} =\textcolor{red}{1}, \quad \textcolor{red}{1}^\textcolor{blue}{-5} =\textcolor{red}{1}$

Level 4-5 GCSE KS3
Level 4-5 GCSE KS3

## Indices Rule 7: The Fraction Law

The power of a fraction applies to both the top and bottom of the fraction.

$\bigg(\dfrac{\textcolor{red}{a}}{\textcolor{blue}{b}}\bigg)^\textcolor{limegreen}{c}= \dfrac{\textcolor{red}{a}^\textcolor{limegreen}{c}}{\textcolor{blue}{b}^\textcolor{limegreen}{c}}$

This also applies to mixed factions

$\bigg(2\dfrac{\textcolor{red}{3}}{\textcolor{blue}{4}}\bigg)^\textcolor{limegreen}{5} = \bigg(\dfrac{\textcolor{red}{11}}{\textcolor{blue}{4}}\bigg)^\textcolor{limegreen}{5} = \bigg(\dfrac{\textcolor{red}{11}^\textcolor{limegreen}{5}}{\textcolor{blue}{4}^\textcolor{limegreen}{5}}\bigg)$

Level 4-5 GCSE KS3

## Roots

The opposite to taking a power of some number is to take a root. Let’s consider square roots – these do the opposite of squaring. e.g.

$\textcolor{blue}{4}^\textcolor{red}{2} = \textcolor{limegreen}{16}$

$\sqrt[\textcolor{red}{2}]{\textcolor{limegreen}{16}} = \textcolor{blue}{4}$

We also have cube roots, $4$th roots, $5$th roots, etc, for when the powers are higher. e.g.

$\textcolor{blue}{2}^\textcolor{red}{3} = \textcolor{limegreen}{8}$

$\sqrt[\textcolor{red}{3}]{\textcolor{limegreen}{8}} = \textcolor{blue}{2}$

These roots use the same symbol, just with a different number in the top left to show the power, e.g. $\sqrt[3]{8}=2$

A $4$th root would be shown by $\sqrt[4]{}$, and so on.

Level 4-5 GCSE KS3
Level 4-5 GCSE KS3

## Example 1: Multiplication

Write $5p^2q^3\times3pq^4$ in its simplest form.

[2 marks]

To simplify this expression, we must recognise that it can be broken up into parts, i.e. we can write

$5p^2q^3\times3pq^4=5\times p^2\times q^3\times3\times p\times q^4$

Then, we can rearrange the terms, putting like terms together.

$5\times 3\times p^2\times p\times q^3\times q^4$

Finally using rule 1 we can multiply the following,

$5\times3=15$

$p^2\times p=p^3$

$q^3\times q^4=q^7$

This gives the final answer to be,

$15p^3q^7$

Level 4-5 GCSE KS3

## Example 2: Multiplication and Division

Work out the value of $\dfrac{3^4\times3^7}{3^8}$.

[2 marks]

First we must multiply out the top of the fraction,

$3^4\times3^7=3^{4+7}=3^{11}$

So, the calculation becomes

$\dfrac{3^{11}}{3^8}$

Next calculating the division we get,

$\dfrac{3^{11}}{3^8}=3^{11-8}=3^3$

This gives the final answer to be,

$3^3=27$.

Level 4-5 GCSE KS3

## Example Questions

we know that:

$a^a \times a^c = a^{b + c}$

so,

$a^2 \times a^3 = a^{2+3}$

$a^2 \times a^3 = a^5$

It is helpful to be able to recognise the first 15 square numbers.

In this case, we can recognise,

$12^2=144$ and $14^2=196$

Hence the calculation is simply,

$\sqrt{144}+\sqrt{196}=12+14=26$

We can rewrite the first term of the expression as,

$(3^2)^3=3^2\times3^2\times3^2$

The multiplication law tells us that,

$3^2\times3^2\times3^2=3^{2+2+2}=3^6$

This is the same result as the power-law gives,

$(3^2)^3=3^{2\times3}=3^6$

Hence, the expression now looks like,

$3^6\div3^4$

Using the division law we find,

$3^6\div3^4=3^{6-4}=3^2=9$

First considering the numerator, the laws of indices tell us,

$7^5\times7^3=7^{5+3}=7^8$

Thus the expression now is,

$\dfrac{7^8}{7^6}$

This can be simplified to,

$\dfrac{7^8}{7^6}=7^{8-6}=7^2$

Hence we are left with a simple calculation of,

$7^2=7\times7=49$

We know that,

$20^1 = 20$

and

$100^0 = 1$

So we can calculate

$20 + 1 = 21$

Level 1-3GCSE

## Worksheet and Example Questions

### (NEW) Powers and Roots Exam Style Questions - MME

Level 1-3 GCSE KS3NewOfficial MME

## Drill Questions

### Powers And Square roots - Drill Questions

Level 1-3 GCSE KS3

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