Powers and Roots Worksheets | Questions and Revision | MME

# Powers and Roots Worksheets, Questions and Revision

Level 4 Level 5

## What you need to know

### Powers and Roots

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

$6^\textcolor{red}{3}=6\times 6\times 6,$

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

$a^5=a\times a\times a\times a\times a.$

This topic will cover the basics of powers and roots, for more complex question types visit the laws of indices page.

### Take Note:

Memorising the square numbers up to 15 is useful for the exam. The first 15 squared numbers are:

$\textcolor{red}{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225}$.

When we raise a number to the power of 3 we say it is being cubed. 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 five:

$\textcolor{red}{1, 8, 27, 64, 125}$

Common mistakes include forgetting the following rules

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

for any value of $a$.

### 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. Since we know that $4^2$ is 16, we also know that the square root of 16 is 4.

We also have cube roots, 4th roots, 5th roots, etc, for when the powers are higher. They do the same thing, for example $2^3$ is 8, so we 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 show the power $\sqrt[3]{8}=2$

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

### Negative Powers

We can have negative powers. 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 $b$

See if you can spot a pattern in the powers below, this will hopefully help with the understanding of negative powers

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

And so on. For more on negative indices revise the rules of indices topic.

### Fractional Powers

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. 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 is the definition of a square root, so the only possible number that $7^{\frac{1}{2}}$ could be is $\sqrt{7}$.

### Example 1: Powers

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

Firstly, let’s work out $3^4$. We know that

$3^4=3\times 3\times 3\times 3 =81$.

Then,

$7^2=7\times 7=49$

Then, adding the results together, we get

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

### Example 2: Negative Powers

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

$4^{-3}=\frac{1}{64}$.

### Example 3: Fractional Powers and Roots

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

So, we know that $9^{\frac{3}{2}}$ is equal to $\sqrt[2]{9}^3$

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

$3^3=3\times 3\times 3 = 9$

### Example Questions

Considering the first term,

$8^2=8\times 8=64$

Next, we have

$5^3=5\times 5\times 5=25\times5=125$

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

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,

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

Firstly, as $3^2=9$, the inverse operation gives, $\sqrt{9}=3$

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

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

Multiplying our two answers together, we get

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

This expression can be rewritten as,

$\sqrt4\times\sqrt4^3$

Given we know that $\sqrt4=2$ , this becomes,

$2\times2^3$

Hence,

$2\times2^3=2\times8=16$

As it is a negative power we can rewrite this as,

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

Now, we can work out the denominator, which we will write as,

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

We know that $\sqrt[3]{8}=2$. So this simplifies to,

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

Counting up in powers of 2: 4, 8, 16, 32 – we see that 32 is the 5th power of 2, so

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

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

Level 1-3

Level 1-3

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