N6 - Powers and roots | AQA GCSE Mathematics

Powers compress repeated multiplication, while roots undo it. You will work with squares, cubes and higher powers, recognise useful number facts, and distinguish a root from the power that produced it. The final part estimates unfamiliar powers and roots for Higher tier.

Positive integer powers and squares

Can $64$ be arranged both as a square of counters and as a cube of small blocks? The two arrangements need different side lengths, but they contain the same number. Powers and roots will tell us both lengths.

Start smaller: four rows of four counters contain $4 \times 4 = 16$ counters. We write this as $4^2 = 16$ , read “four squared”. The raised $2$ counts the factors of $4$ ; it does not mean multiply $4$ by $2$ .

You already know

Multiplication combines equal groups. Brackets tell you which number an operation acts on: $(-3)^2$ squares the whole negative number, while $-3^2$ means the negative of $3^2$ .

With three equal factors, $4^3 = 4 \times 4 \times 4 = 64$ , read “four cubed”. A fourth power has four factors, and so on. The base can be a decimal, a fraction, a negative number or zero.

Base and index

In $a^n$ , $a$ is the base and $n$ is the index or exponent. For a positive integer $n$ , it means a product of $n$ factors, each equal to $a$ . For example, $3^4 = 3 \times 3 \times 3 \times 3$ , not $3 \times 4$ .

A power with index $1$ contains one factor, so $a^1 = a$ . You will meet zero and negative indices in the next lesson; they need a different interpretation.

Positive integer powers

a^n=\underbrace{a\times a\times\cdots\times a}_{n\text{ factors}}

Here $n$ is a positive integer. In particular, $a^2$ is a square and $a^3$ is a cube.

Not on the formulae sheet: learn it.

Square facts make roots and later calculations much quicker. Learn the squares through $15^2$ , rather than counting from scratch each time.

Number	Square	Number	Square	Number	Square
$1$	$1$	$6$	$36$	$11$	$121$
$2$	$4$	$7$	$49$	$12$	$144$
$3$	$9$	$8$	$64$	$13$	$169$
$4$	$16$	$9$	$81$	$14$	$196$
$5$	$25$	$10$	$100$	$15$	$225$

These square facts are not on the formulae sheet: learn them. The next examples keep the base fixed and change the number of factors.

Worked example 1: squaring a number

Question. Work out $6^2$ . Non-calculator.

Plan. Write two factors of $6$ , then multiply.

\begin{aligned} 6^2&=6\times6 && \text{two factors} \\ &=36 && \text{multiplication fact} \end{aligned}

Why two factors? The index is $2$ ; the base $6$ tells us what each factor is.

Answer. $36$ .

Check. A square of six rows of six contains $36$ counters.

Change only the index from $2$ to $3$ . This adds one factor, rather than adding $1$ to the answer.

Worked example 2: cubing the same number

Question. Work out $6^3$ . Non-calculator.

Plan. Use the square already known, then multiply by one more $6$ .

\begin{aligned} 6^3&=6\times6\times6 && \text{three factors} \\ &=36\times6 && \text{first two factors} \\ &=216 && \text{multiply} \end{aligned}

Why multiply by $6$ ? Increasing the index by $1$ adds a factor of the base, not a fixed amount.

Answer. $216$ .

Check. $216 \div 6 = 36$ , the square from example 1.

The size of the base matters. For a number between $0$ and $1$ , repeated multiplication makes the result smaller: $0.5^2 = 0.25$ and $0.5^3 = 0.125$ .

Watch out

Wrong: $5^3 = 15$ . That is $5 \times 3$ . Right: $5^3 = 5 \times 5 \times 5 = 125$ . Write the factors if the notation feels uncertain.

Keep the same distinction in a slightly larger cube.

Square roots, cube roots and higher roots

A root reverses a power: if $4^2 = 16$ , the non-negative square root of $16$ is $4$ .

[DIAGRAM: asset_name: Squaring and taking a square root; asset_slug: aqa8300_n03_square_root; caption: A square with side 4 has area 16, so the square root of 16 is 4; description: A four-by-four array of unit squares. Arrows connect the side length 4 to the area 16 by squaring, and connect the area back to the side length by taking its square root.]

The little index on a root sign tells you which power to undo. $\sqrt[3]{27} = 3$ , because $3^3 = 27$ . The square-root sign is the exception: its little $2$ is normally left out.

Real roots

A root of order $n$ is a real number whose power with index $n$ gives the number inside the root. For an even root, the root symbol means the non-negative root: $\sqrt{25} = 5$ , even though both $5$ and $-5$ square to $25$ . An odd root can be negative: $\sqrt[3]{-8} = -2$ .

Every real square is non-negative. So $\sqrt{-9}$ has no real value. By contrast, $\sqrt{0} = 0$ . A root is not division by the index: $\sqrt[3]{27}$ is $3$ , not $9$ .

Undoing a positive power

For a positive integer $n$ and $a \geqslant 0$ ,

\sqrt[n]{a^n}=a

For odd $n$ , this also holds for negative $a$ . For even $n$ , a negative input returns its non-negative magnitude, for example $\sqrt{(-4)^2}=4$ .

Not on the formulae sheet: learn it.

Read the entire root before looking for a number fact. The root bar covers everything whose root is being taken.

Worked example 3: using a square fact backwards

Question. Work out $\sqrt{144}$ . Non-calculator.

Plan. Find the non-negative number whose square is $144$ .

\begin{aligned} 12\times12&=144 && \text{known square} \\ \sqrt{144}&=12 && \text{undo the square} \end{aligned}

Why not $-12$ ? The square-root symbol asks for the non-negative value.

Answer. $12$ .

Check. Squaring the answer gives $12^2 = 144$ .

Change only the root index: a cube root needs a three-factor fact instead of a square fact.

Worked example 4: using a cube fact backwards

Question. Work out $\sqrt[3]{125}$ . Non-calculator.

Plan. Find the number whose cube is $125$ .

\begin{aligned} 5\times5\times5&=125 && \text{known cube} \\ \sqrt[3]{125}&=5 && \text{undo the cube} \end{aligned}

Why three factors? The small $3$ belongs to the root sign and tells us which power to undo.

Answer. $5$ .

Check. $5^3 = 125$ .

Higher roots follow the same meaning: $\sqrt[4]{81} = 3$ , because $3^4 = 81$ , and $\sqrt[5]{32} = 2$ , because $2^5 = 32$ . A higher root is a different operation from dividing or taking repeated square roots indiscriminately.

Watch out

Wrong: $\sqrt{49} = \pm7$ . Right: $\sqrt{49} = 7$ . If a question instead asks for all numbers that square to $49$ , both $7$ and $-7$ are answers. Read what is being asked.

Use a power fact to justify a root, even if you spot its answer immediately.

Recognising powers of 2, 3, 4, 5 and 10

Recognising a power lets you read a large number as a short multiplication story.

The table begins each sequence. Move right by multiplying by the base again, rather than adding it. These facts are not on the formulae sheet: learn the familiar ones and build the others.

Base	First power	Second	Third	Fourth	Fifth
$2$	$2$	$4$	$8$	$16$	$32$
$3$	$3$	$9$	$27$	$81$	$243$
$4$	$4$	$16$	$64$	$256$	$1024$
$5$	$5$	$25$	$125$	$625$	$3125$

One number can belong to more than one row. $64 = 8^2 = 4^3 = 2^6$ : it really can be both a square and a cube. For the opening arrangements, the square has side $8$ , and the cube has side $4$ .

Worked example 5: recognising a power

Question. Write $256$ as a power of $4$ . Non-calculator.

Plan. Multiply by $4$ repeatedly, keeping track of the number of factors.

\begin{aligned} 4^2&=16 && \text{two factors} \\ 4^3&=64 && \text{multiply by }4 \\ 4^4&=256 && \text{multiply by }4\text{ again} \end{aligned}

Why count the factors? $256$ is the value; the requested index records how many fours made it.

Answer. $256 = 4^4$ .

Check. $256 \div 4 \div 4 \div 4 \div 4 = 1$ .

Change only the base being searched for: use $5$ instead of $4$ .

Worked example 6: building the next familiar power

Question. Write $625$ as a power of $5$ . Non-calculator.

Plan. Start with the known cube of $5$ , then extend it.

\begin{aligned} 5^3&=125 && \text{known cube} \\ 125\times5&=625 && \text{one more factor} \\ 625&=5^4 && \text{four factors altogether} \end{aligned}

Why start at $125$ ? Recognising a nearby power avoids rebuilding the whole sequence from $5$ .

Answer. $625 = 5^4$ .

Check. $625 \div 5 = 125 = 5^3$ .

Powers of $10$ have an especially visible pattern. $10^3 = 1000$ , and $10^6 = 1\,000\,000$ : a million is a thousand thousands. For a positive integer index, the index counts the zeros after the $1$ .

Watch out

Wrong: $10^6 = 60$ . Right: $10^6$ contains six factors of $10$ , giving $1\,000\,000$ . The same index means six factors for any base, but it counts written zeros only for base $10$ .

You can now recognise the number and name the base it comes from.

Estimating powers and roots (Higher)

When a number is unfamiliar, put it between two familiar numbers and use their powers or roots to trap the answer.

For example, $4^2 = 16$ and $5^2 = 25$ . Since $20$ lies between $16$ and $25$ , its square root lies between $4$ and $5$ . It is not exactly halfway: equal steps in a number do not give equal steps in its square.

Comparing positive powers and roots

For positive numbers $a < b$ and a positive integer $n$ ,

a^n<b^n\qquad\text{and}\qquad\sqrt[n]{a}<\sqrt[n]{b}

Nearby convenient inputs give estimates; testing closer inputs refines them.

Not on the formulae sheet: learn it.

The positivity condition matters. Squaring does not keep every negative-number comparison in the same order: $-5 < -4$ , but $25 > 16$ .

Worked example 7: estimating an unfamiliar power

Question. Without a calculator, estimate $4.98^3$ , and decide whether your estimate is above or below the exact value.

Plan. Replace $4.98$ by the nearby whole number $5$ , then compare the positive bases.

\begin{aligned} 4.98&\approx5 && \text{nearby convenient base} \\ 4.98^3&\approx5^3=125 && \text{cube the estimate} \\ 4.98^3&<125 && 4.98<5 \end{aligned}

Why cube the rounded base? Repeated multiplication uses the base three times, so the approximation must replace all three factors.

Answer. About $125$ ; this is an overestimate.

Check. $4^3 < 4.98^3 < 5^3$ , so the value is between $64$ and $125$ and close to the upper end.

Change only the direction of the operation: estimate a root by testing powers of possible answers.

Worked example 8: refining an unfamiliar root

Question. Estimate $\sqrt{20}$ to one decimal place. Non-calculator.

Plan. Locate the root between whole numbers, then compare it with the halfway point between candidate tenths.

\begin{aligned} 4^2&=16<20<25=5^2 && \text{root between }4\text{ and }5 \\ 4.4^2&=19.36<20<20.25=4.5^2 && \text{root between }4.4\text{ and }4.5 \\ 4.45^2&=19.8025<20 && \text{root above halfway} \end{aligned}

Why test $4.45$ ? Values at or above this midpoint round to $4.5$ , whereas values below it round to $4.4$ .

Answer. $\sqrt{20} \approx 4.5$ to one decimal place.

Check. $4.5^2 = 20.25$ is close to $20$ . It is an estimate, so use $\approx$ , not $=$ .

The same comparison works at any positive scale: $0.2^3 = 0.008$ and $0.3^3 = 0.027$ , so $0.2 < \sqrt[3]{0.015} < 0.3$ . For a fourth root, compare fourth powers rather than squares.

Watch out

Wrong: $\sqrt{20} = 4.5$ because $20$ is halfway between $16$ and $25$ . It is not halfway, and the square-root scale is not linear. Right: test nearby powers and mark an estimate with $\approx$ .

Refine only as far as the required accuracy needs. A whole-number bracket is enough when the question asks for consecutive integers.

Quick Recap

A positive integer index counts equal factors: $a^3 = a \times a \times a$ .
A root undoes a power. The square-root symbol returns the non-negative root; odd roots can be negative.
Know squares through $15^2$ , recognise familiar powers of $2$ , $3$ , $4$ , $5$ , and remember $1000 = 10^3$ and $1\,000\,000 = 10^6$ .

The opening $64$ counters make an $8$ -by- $8$ square, or a cube with side $4$ , because $64 = 8^2 = 4^3$ .

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N6 - Powers and roots

Positive integer powers and squares

Square roots, cube roots and higher roots

Recognising powers of 2, 3, 4, 5 and 10

Estimating powers and roots (Higher)

Quick Recap

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