N7 - Calculating with roots and indices | AQA GCSE Mathematics

You will calculate with roots, combine powers efficiently and work out what zero and negative indices mean. The laws come from counting and cancelling equal factors, so you can reconstruct them when memory lets you down. These methods keep large calculations manageable on a non-calculator paper.

Calculating with roots

Which is larger, $2^{10}$ or $4^5$ ? Their indices suggest different answers, but the bases matter too. Keep that puzzle in mind while we build a reliable way to compare powers.

First, roots are numbers, so they can be added, multiplied and divided once you know their values.

You already know

A square root undoes squaring, and a cube root undoes cubing: $\sqrt{81}=9$ and $\sqrt[3]{64}=4$ . Work out roots and powers before multiplication, division, addition and subtraction.

The root bar acts like a bracket. Compare $\sqrt{9+16}=5$ with $\sqrt9+\sqrt{16}=7$ . The first root covers the whole sum; the second expression has two separate roots. The length of the bar changes the question.

Roots do split over a product or a quotient of suitable numbers. For example, $\sqrt{36\times49}=6\times7=42$ , and $\sqrt{\frac{49}{81}}=\frac79$ . Squaring the proposed answer checks each fact.

Roots of products and quotients

For $a\geqslant0$ , $b\geqslant0$ ,

\sqrt{ab}=\sqrt a\times\sqrt b

For $a\geqslant0$ , $b>0$ ,

\sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b}

There is no corresponding rule that splits a square root over addition or subtraction.

Not on the formulae sheet: learn it.

For short calculations, evaluating each root separately makes the order of operations visible.

Worked example 1: adding root values

Question. Work out $\sqrt{81}+\sqrt[3]{64}$ . Non-calculator.

Plan. Evaluate both roots, then add their values.

\begin{aligned} \sqrt{81}&=9 && 9^2=81 \\ \sqrt[3]{64}&=4 && 4^3=64 \\ \sqrt{81}+\sqrt[3]{64}&=9+4=13 && \text{add} \end{aligned}

Why evaluate separately? The plus sign lies outside both root bars, so it joins the two root values.

Answer. $13$ .

Check. The answer is greater than $9$ , as a positive number has been added.

Change only the final operation from addition to subtraction. The root calculations stay the same.

Worked example 2: subtracting root values

Question. Work out $\sqrt{81}-\sqrt[3]{64}$ . Non-calculator.

Plan. Evaluate each root, then subtract in the written order.

\begin{aligned} \sqrt{81}&=9 && \text{square root} \\ \sqrt[3]{64}&=4 && \text{cube root} \\ \sqrt{81}-\sqrt[3]{64}&=9-4=5 && \text{subtract} \end{aligned}

Why not combine the numbers under one root? There is no subtraction law for roots, and these two roots have different indices as well.

Answer. $5$ .

Check. Adding the subtracted $4$ back gives $9$ .

A root operation never “reaches across” an addition sign beyond its bar.

Watch out

Wrong: $\sqrt{9+16}=\sqrt9+\sqrt{16}$ . The left side is $5$ , and the right side is $7$ . Right: follow the root bar exactly; the product rule for square roots does not work for sums.

Now apply the same two-root method with new values.

Multiplying and dividing powers with the same base

When equal bases are multiplied, count all their factors; when they are divided, cancel matching factors.

For instance, $2^3\times2^2$ contains three factors of $2$ , followed by two more. That is five factors, so it equals $2^5$ . Division reverses the count: $\frac{2^5}{2^2}$ leaves three factors after two pairs have cancelled.

Same-base multiplication and division

\begin{aligned} a^m\times a^n&=a^{m+n} \\ a^m\div a^n&=a^{m-n} \end{aligned}

The bases must match. For division, $a\ne0$ . These laws apply to integer indices, including the zero and negative indices introduced later in this lesson.

Not on the formulae sheet: learn it.

The result is often required “as a power”. In that case, stop at the requested index form rather than writing a long ordinary number.

Worked example 3: combining equal factors

Question. Write $3^4\times3^2$ as a single power of $3$ . Non-calculator.

Plan. Keep the base $3$ and add the indices.

\begin{aligned} 3^4\times3^2&=3^{4+2} && \text{count all factors} \\ &=3^6 && \text{six factors of }3 \end{aligned}

Why add the indices? Multiplying joins the four-factor group and the two-factor group; it does not multiply their lengths.

Answer. $3^6$ .

Check. $81\times9=729=3^6$ .

Change only multiplication to division. Matching factors now cancel instead of joining.

Worked example 4: cancelling equal factors

Question. Write $3^4\div3^2$ as a single power of $3$ . Non-calculator.

Plan. Keep the base and subtract the denominator's index.

\begin{aligned} 3^4\div3^2&=3^{4-2} && \text{cancel two factors} \\ &=3^2 && \text{two factors remain} \end{aligned}

Why subtract in this order? The numerator supplies four factors and the denominator cancels two of them.

Answer. $3^2$ .

Check. $81\div9=9=3^2$ .

Neither law works for addition: $2^3+2^2=8+4=12$ , which is not $2^5$ .

Watch out

Wrong: $3^4\times3^2=3^8$ . Multiplying the indices is the power-of-a-power rule, not this rule. Right: $3^4\times3^2=3^{4+2}=3^6$ . Check the operation joining the powers.

The same reasoning works with a larger index; the size of the ordinary number need not enter the calculation.

Raising a power to a power

A power raised to another power contains equal-sized groups of factors.

For $(2^3)^2$ , the outside square means two copies of $2^3$ . Each copy has three factors of $2$ , so there are $3\times2=6$ factors altogether. This is why the indices multiply in this situation.

Power of a power

(a^m)^n=a^{mn}

For integer indices, use non-zero $a$ whenever a zero or negative index occurs. With positive indices the base may also be zero.

Not on the formulae sheet: learn it.

The opening puzzle is already close to solved. Because $4=2^2$ , we have $4^5=(2^2)^5=2^{10}$ . In fact, both equal $1024$ , which is also $32^2$ . Three rather different-looking powers name one number.

Worked example 5: a power of a power in a product

Question. Write $(2^3)^2\times2^4$ as a single power of $2$ . Non-calculator.

Plan. Deal with the bracketed power first, then combine equal bases.

\begin{aligned} (2^3)^2\times2^4&=2^{3\times2}\times2^4 && \text{multiply indices inside power} \\ &=2^6\times2^4 && \text{evaluate index} \\ &=2^{10} && \text{add for multiplication} \end{aligned}

Why use two different rules? The outside square repeats a whole group, while the final multiplication joins two groups.

Answer. $2^{10}$ .

Check. $8^2\times16=64\times16=1024=2^{10}$ .

Change only the operation joining the two powers: divide by $2^4$ instead of multiplying.

Worked example 6: a power of a power in a quotient

Question. Write $(2^3)^2\div2^4$ as a single power of $2$ . Non-calculator.

Plan. Simplify the bracketed power, then subtract indices for division.

\begin{aligned} (2^3)^2\div2^4&=2^{3\times2}\div2^4 && \text{power of a power} \\ &=2^{6-4} && \text{cancel four factors} \\ &=2^2 && \text{simplify} \end{aligned}

Why not subtract before multiplying? The bracketed power is one complete numerator. Its index must first be $3\times2=6$ .

Answer. $2^2$ .

Check. $8^2\div16=64\div16=4=2^2$ .

When the bases differ, look for a shared base before using the laws. For example, $8^2\times2^3=(2^3)^2\times2^3=2^9$ . There is no rule that directly adds indices across unmatched bases.

Watch out

Wrong: $(3^2)^4=3^6$ . Right: four groups of two factors make eight factors, so $(3^2)^4=3^8$ . Add indices when multiplying powers; multiply them when raising a power to a power.

A line showing the bracketed power first makes this distinction easy to mark and easy to check.

Zero and negative indices

Keep the division law working as the index passes through zero, and a new meaning for an index emerges.

Dividing $5^3$ by $5^3$ gives $1$ . The index law writes the same calculation as $5^{3-3}=5^0$ , so $5^0=1$ . Continue down the sequence by dividing by $5$ each time:

\begin{array}{c|rrrrrr} \text{index}&3&2&1&0&-1&-2\\ \hline \text{value}&125&25&5&1&\frac15&\frac1{25} \end{array}

A negative index describes a reciprocal. It does not make the number negative.

Zero and negative indices

For $a\ne0$ and a positive integer $n$ ,

a^0=1\qquad a^{-n}=\frac1{a^n}

In particular, $a^{-1}=\frac1a$ . These rules do not define $0^0$ or negative powers of zero.

Not on the formulae sheet: learn it.

A reciprocal is a number that multiplies with the original to make $1$ . For a non-zero fraction, turn it upside down: the reciprocal of $\frac23$ is $\frac32$ .

Worked example 7: a negative index on a whole number

Question. Work out $5^{-2}$ . Give your answer as a fraction. Non-calculator.

Plan. Use the negative index to take a reciprocal, then square the base.

\begin{aligned} 5^{-2}&=\frac1{5^2} && \text{negative index means reciprocal} \\ &=\frac1{25} && \text{square }5 \end{aligned}

Why not $-25$ ? The minus sign is in the index, so it changes the power into its reciprocal.

Answer. $\frac1{25}$ .

Check. $\frac1{25}\times5^2=1$ .

Change only the base: make it a fraction. Taking the reciprocal now swaps numerator and denominator.

Worked example 8: a negative index on a fraction

Question. Work out $\left(\frac23\right)^{-2}$ . Non-calculator.

Plan. Take the reciprocal of the base, then square.

\begin{aligned} \left(\frac23\right)^{-2}&=\left(\frac32\right)^2 && \text{reciprocal base} \\ &=\frac{3^2}{2^2} && \text{square numerator and denominator} \\ &=\frac94 && \text{evaluate} \end{aligned}

Why turn the whole fraction upside down? The reciprocal is $\frac32$ , not a negative version of $\frac23$ .

Answer. $\frac94$ .

Check. $\frac94\times\left(\frac23\right)^2=\frac94\times\frac49=1$ .

The index laws now cover results such as $2^3\div2^5=2^{-2}=\frac14$ . Expanding the fraction $\frac8{32}$ confirms the same value.

Watch out

Wrong: $7^0=0$ or $7^{-2}=-49$ . Right: $7^0=1$ and $7^{-2}=\frac1{49}$ . A zero index is not a zero base, and a negative index is not a negative sign in front.

That final distinction matters on Paper 1: leave a reciprocal as an exact fraction rather than rounding it.

Quick Recap

Evaluate roots before the operations outside them; the root bar controls what is included.
For equal bases, multiply by adding indices and divide by subtracting indices.
For a power of a power, multiply the indices.
For a non-zero base, a zero index gives $1$ , and a negative index gives a reciprocal.

Neither opening power is larger: $4^5=(2^2)^5=2^{10}=1024$ .

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N7 - Calculating with roots and indices

Calculating with roots

Multiplying and dividing powers with the same base

Raising a power to a power

Zero and negative indices

Quick Recap

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