# Powers and Roots Worksheets, Questions and Revision

Level 4 Level 5
Navigate topic

## What you need to know

Powers (or indices) are a shorthand way of expressing repeated multiplication. For example

$6^3=6\times 6\times 6,$

where the small number 3 positioned slightly above and to the right of the 6 is the power/index, and it tells us how many sixes are in the multiplication. This idea of indices works the same with algebra:

$a^4=a\times a\times a\times a\times a.$

When we raise a number to the power of 2, we say it is being squared. For example, $4^2$ is “4 squared”. Furthermore, a square number is the result of squaring a whole number – in this case, $4^2=4\times 4=16$, so 16 is a square number. You are expected to know the first 15 square numbers by heart. For reference, they are

$1, 4, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225$.

Similarly, when we raise a number to the power of 3 we say it is being cubed, and the result of cubing a number is cube number. For example, $2^3=2\times 2\times 2=8$, so 8 is a cube number. You don’t have to memorise the cube numbers, but it helps to be familiar with the first few: $1, 8, 27, 64, 125$. There are no fancy terms for higher powers, we just say “raising to the 4th power/power of 4” or “raising to the 5th power/power of 5” and so on.

Example: Work out $3^4+7^2$.

Firstly, let’s work out $3^4$. We know that $3^4=3\times 3\times 3\times 3$, so we can work this out by splitting it up, multiplying the first two 3s together and the second two 3s together, and then finally multiplying the results together. So, we get

$3\times 3\times 3\times 3=(3\times 3)\times(3\times3)=(9)\times(9),$

and $9\times 9=81$, so this bit is done. Then, $7^2=7\times 7$ which we know to be the seventh square number: 49. Then, adding the results together, we get

$3^4+7^2=81+49=130$.

The opposite to taking a power of some number (in the same way that subtraction is the opposite of addition) is to take a root. Let’s consider square roots – these do the opposite of squaring. Since we know that $4^2$ is 16, we also know that the square root of 16 is 4 –  it takes us from the result back to starting number. Square roots are denoted with a $\sqrt{}$ symbol as such:

$\sqrt{16}=4$,

with the number being rooted on the inside. Furthermore, we also have cube roots, 4th roots, 5th roots, etc, for when the powers are higher. They do the same thing, for example we’ve seen that $2^3$ is 8, so we now also know that the cube root of 8 is 2. These roots use the same symbol, just with an extra number in the top left to denote the associated power, such as

$\sqrt[3]{8}=2$

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

Powers aren’t just limited to the positive whole numbers. For example, we have that

$a^1=1\text{ and }a^0=1$,

for any value of $a$. Furthermore, we can have negative indices. For example

$5^{-1}=\dfrac{1}{5},\text{ and }5^{-2}=\dfrac{1}{5^2}=\dfrac{1}{25}$

In general, the result of a negative power is “1 over that number to the positive power”, i.e.

$a^{-b}=\dfrac{1}{a^b}$,

for any value of $a$ or $b$. If you aren’t sure why this all makes sense, think of this: if you’re taking powers of 2, like 4, 8, 16, 32, etc, then we know that each time the power increases by 1, the result is multiplied by another 2. So, if we go in the other direction, we can say that each time the power is decreased by 1, the result is divided by 2. So, $2^3=8$ and $2^2=4$, so then to get $2^1$ we must have to divide the result by 2 again, so $2^1=2$. Keep going with this and you get

$2^1=2,\text{ }2^0=1,\text{ }2^{-1}=\dfrac{1}{2},\text{ }2^{-2}=\dfrac{1}{4}$

And so on.

Example: Work out $4^{-3}$.

We now know that $4^{-3}$ is equal to $\frac{1}{4^3}$. We also know that

$4^3=4\times 4\times 4=16\times 4=64$.

So, we get that $4^{-3}=\frac{1}{64}$.

For the higher students only, there are also fractional indices. The rule for these is as follows:

$x^{\frac{a}{b}}=\sqrt[b]{x^a}=\sqrt[b]{x}^a$.

Notice how the power of $a$ can be inside or outside the bracket. These are both correct and will both give the same result, but sometimes one will be easier than other. To have an idea of why this is what a fractional power means, consider the power of a half. The laws of indices tell us

$7^{\frac{1}{2}}\times 7^{\frac{1}{2}}=7^1=7$

So, $7^{\frac{1}{2}}$ is some number that, when multiplied by itself, gives 7 as the answer. This the definition of a square root, so the only possible number that $7^{\frac{1}{2}}$ could be is $\sqrt{7}$.

Example: Work out $9^{\frac{3}{2}}$.

So, we know that $9^{\frac{3}{2}}$ is equal to $\sqrt[2]{9^3}$ and $\sqrt[2]{9}^3$. We’re going to opt for the second one, since the first one will require us to work out $9^3$ and square root the answer, which sounds like no fun at all.

So, to work out $\sqrt[2]{9}^3$, we first have to square root 9, which is easy enough – the square root of 9 is 3. So, $\sqrt[2]{9}^3$ becomes $3^3$, which is

$3^3=3\times 3\times 3$.

Fortunately, multiplying by 3 is not so bad: $3\times 3=9$ and $9\times 3=27$, so the answer to $3^3$ (and therefore the whole question) is 27.

## GCSE Maths Revision Cards

• All major GCSE maths topics covered
• Higher and foundation
• All exam boards - AQA, OCR, Edexcel, WJEC.

### Example Questions

Firstly, $8^2=8\times 8=64$. Next up, we have

$5^3=5\times 5\times 5$

Well, $5\times 5=25$, and $25\times 5=125$. Now, we must add the result of these powers together, and get

$8^2+5^3=64+125=189$.

#### Is this a topic you struggle with? Get help now.

Firstly, given that $3^2=9$, we must have that $\sqrt{9}=3$. So, that leaves $6^{-2}$. This becomes the following fraction

$6^{-2}=\dfrac{1}{6^2}$

We know that $6^2=6\times 6=36$, so

$6^{-2}=\dfrac{1}{36}$.

Then, multiplying our two answers together, we get

$\sqrt{9}\times 6^{-2}=3\times\dfrac{1}{36}=\dfrac{3}{36}=\dfrac{1}{12}$

$\dfrac{1}{12}$ is the simplest form of this fraction so it is the final answer.

#### Is this a topic you struggle with? Get help now.

So, as the hint suggests, we will deal with the minus sign first. This is to say that, we know that

$8^{-\dfrac{5}{3}}=\dfrac{1}{8^{\dfrac{5}{3}}}$

Now, we can work out the denominator. So, we know that

$8^{\dfrac{5}{3}}=\sqrt[3]{8^5}=\sqrt[3]{8}^5$

We’re going to do the latter (to avoid having to do $8^5$). Firstly, observing that $2^3=8$, we must have that $\sqrt[3]{8}=2$. So,

$\sqrt[3]{8}^5=2^5$

Fortunately, powers of 2 are not too troubling. Counting up powers of 2: 4, 8, 16, 32 – we see that 32 is the 5th power of 2, so

$\sqrt[3]{8}^5=32$.

Therefore, the final answer to the question is

$8^{-\dfrac{5}{3}}=\dfrac{1}{32}$.

Level 1-3

## Powers and Roots Worksheets and Revision

Powers And Square roots (1)
Level 1-3
Powers And Square roots (2)
Level 1-3
Powers And Square roots (3)
Level 1-3
Square and cube of numbers (1)
Level 1-3
Square and cube of numbers (2)
Level 1-3
Square and cube of numbers (3)
Level 1-3

## Powers and Roots Teaching Resources

### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.