HN6–HN7 - Fractional indices | Eduqas GCSE Mathematics

A fractional power describes a root followed by a power. Learn to calculate these values exactly, combine them with negative indices and estimate roots between nearby known values. HN7 supplies fractional-index calculations; HN6 adds estimates of powers and roots of positive numbers.

Root powers

Can a power turn $81$ into $27$ ? Multiplying $81$ by itself only makes it larger, but a fractional index lets us take a root first.

Recall that $\sqrt{25}=5$ . We write the same value as $25^{1/2}$ : squaring it gives $25$ . Similarly, $8^{1/3}=\sqrt[3]{8}=2$ , since cubing $2$ gives $8$ .

Fractional index

For $a>0$ and positive integers $m,n$ ,

a^{1/n}=\sqrt[n]{a},\qquad a^{m/n}=\left(\sqrt[n]{a}\right)^m.

The denominator names the root; the numerator names the power. For example, $16^{3/4}=(\sqrt[4]{16})^3=2^3=8$ .

This agrees with the index laws: raising $a^{1/n}$ to power $n$ gives $a^{n/n}=a$ . For a positive base, taking the power before the root gives the same result, but finding a small exact root first usually saves work.

Worked example 1: root before power

Question. Work out $32^{2/5}$ without a calculator.

Plan. Find the fifth root of $32$ , then square it.

\begin{aligned} 32^{2/5}&=(\sqrt[5]{32})^2 &&\text{denominator }5\text{ means fifth root}\\ &=2^2 &&\text{because }2^5=32\\ &=4 \end{aligned}

Answer. $4$ .

Check. Raising the answer to power $5$ gives $4^5=1024$ ; raising the original base to power $2$ gives $32^2=1024$ . The values agree.

Change the numerator of the index from $2$ to $3$ . The root stays the same; the final power changes.

Worked example 2: changing the numerator

Question. Work out $32^{3/5}$ without a calculator.

Plan. Take the same fifth root, then cube it.

\begin{aligned} 32^{3/5}&=(\sqrt[5]{32})^3 &&\text{root first}\\ &=2^3 &&\text{the fifth root is }2\\ &=8 \end{aligned}

Answer. $8$ .

Check. $8^5=32768$ and $32^3=32768$ , confirming the result by a different pair of integer powers.

A square-root sign means the principal, nonnegative root. Thus $25^{1/2}=5$ , not $\pm5$ . The equation $x^2=25$ has two solutions, but the expression $\sqrt{25}$ has one value.

Watch out

$32^{2/5}$ is not $32\times2/5$ . An index acts on the base as a power: find the fifth root and then square it.

Use a different exact fifth root.

Combined indices

A minus sign in a fractional index still means take a reciprocal. Deal with the positive fractional power first, then invert its value.

For example, $9^{-1/2}=1/\sqrt9=1/3$ . The answer is positive: the minus sign belongs to the index, not the base.

Negative fractional powers

For $a>0$ and positive integers $m,n$ ,

a^{-m/n}=\frac1{a^{m/n}}= \frac1{(\sqrt[n]{a})^m}.

The usual index laws apply to positive bases with fractional indices. For example, $a^{1/2}a^{3/2}=a^2$ when $a>0$ .

Not in the formula sheets: learn these rules.

If a base is a positive fraction, take the root of its numerator and denominator. A negative index then swaps the resulting numerator and denominator.

Worked example 3: a negative fractional index

Question. Work out $16^{-3/4}$ .

Plan. Find the fourth root, cube it, then take the reciprocal.

\begin{aligned} 16^{-3/4}&=\frac1{(\sqrt[4]{16})^3} &&\text{the negative index means reciprocal}\\ &=\frac1{2^3} &&\text{because }2^4=16\\ &=\frac18 \end{aligned}

Answer. $\frac18$ .

Check. $16^{3/4}=8$ , and $8\times(1/8)=1$ , as reciprocal values must.

Keep the index but change the base to a fraction smaller than $1$ . Its reciprocal power will be larger than $1$ .

Worked example 4: a fractional base

Question. Work out $\left(\frac{16}{81}\right)^{-3/4}$ .

Plan. Take the fourth root of each part, cube, then invert.

\begin{aligned} \left(\frac{16}{81}\right)^{3/4} &=\left(\frac23\right)^3 &&\text{fourth roots of }16\text{ and }81\\ &=\frac8{27} &&\text{cube numerator and denominator}\\ \left(\frac{16}{81}\right)^{-3/4} &=\frac{27}{8} &&\text{take the reciprocal} \end{aligned}

Answer. $\frac{27}{8}$ .

Check. The reciprocal base is $81/16$ , whose fourth root is $3/2$ . Cubing gives $27/8$ again.

The base matters when deciding whether a real answer exists:

Base and index	Real-number meaning
Positive base	All fractional powers here are defined.
Zero base, positive index	The result is $0$ .
Zero base, negative index	Undefined: it would require division by zero.
Negative base, reduced index with an odd denominator	An odd root is possible: $(-8)^{1/3}=-2$ .
Negative base, reduced index with an even denominator	No real value: $(-8)^{1/2}$ would need a real square root of $-8$ .

For a negative base, reduce the fractional index first, take the indicated odd root, then apply its integer numerator. For example, $(-8)^{2/3}=(-2)^2=4$ . Do not freely rearrange powers of negative bases using fractional power-of-power rules: a principal even root can change a sign. The general rearrangements in the formula box use positive bases.

Watch out

$16^{-3/4}=1/8$ , not $-8$ . A zero or negative base also needs its own domain check; a calculator error message is a reason to inspect the expression.

Try the fractional-base method with another fourth power.

Estimating roots

When a root is not an integer, nearby known powers put it in a useful interval. On positive numbers, increasing the number increases its square, cube and other positive integer powers.

For example, $4^2=16<20<25=5^2$ , so $4<\sqrt{20}<5$ . This locates the root; it does not mean that $\sqrt{20}$ is halfway between $4$ and $5$ . Equal steps in a number do not make equal steps in its square.

Bracketing a root

To bracket a positive root, find a lower value whose corresponding power is too small and an upper value whose power is too large. Narrowing the interval gives a more precise estimate.

Try tenths only after finding the neighbouring integer values. That keeps the search small and provides an immediate sense-check.

Worked example 5: estimating a square root

Question. Show that $7<\sqrt{50}<8$ , then find two consecutive tenths between which $\sqrt{50}$ lies.

Plan. Compare $50$ with nearby squares, first of integers and then of tenths.

\begin{aligned} 7^2=49&<50<64=8^2 &&\text{integer squares}\\ 7&<\sqrt{50}<8 &&\text{take positive square roots}\\ 7.0^2=49&<50<50.41=7.1^2 &&\text{test the next tenth} \end{aligned}

Answer. $7<\sqrt{50}<8$ , with the tighter bounds $7.0<\sqrt{50}<7.1$ .

Check. The midpoint $7.05$ squares to $49.7025<50$ , so the root is above $7.05$ but below $7.1$ , consistent with the bracket.

Change the type of root to a cube root. Use cubes, rather than squares, to test the endpoints.

Worked example 6: estimating a cube root

Question. Find consecutive integer bounds, then consecutive tenth bounds, for $\sqrt[3]{50}$ .

Plan. Locate $50$ between cubes and narrow the interval with tenths.

\begin{aligned} 3^3=27&<50<64=4^3 &&\text{integer cubes}\\ 3&<\sqrt[3]{50}<4\\ 3.6^3=46.656&<50<50.653=3.7^3 &&\text{test nearby tenths} \end{aligned}

Answer. $3<\sqrt[3]{50}<4$ , and $3.6<\sqrt[3]{50}<3.7$ .

Check. $3.65^3=48.627125<50$ , so the root lies in the upper half of the tenth interval. It still has not reached $3.7$ .

The comparison also works forward for powers. Since $3.6<3.65<3.7$ , we know $46.656<3.65^3<50.653$ without multiplying $3.65$ three times. Similarly, $1.4^2<1.45^2<1.5^2$ gives $1.96<1.45^2<2.25$ .

The opening question has an exact answer rather than an estimate: $81^{3/4}=(\sqrt[4]{81})^3=3^3=27$ .

Watch out

$49<50<64$ gives $7<\sqrt{50}<8$ , not $49<\sqrt{50}<64$ . Root the endpoints too, and use only positive endpoints for this square-root comparison.

Use cubes to justify both endpoints of a new bracket.

Quick Recap

The denominator of a fractional index names a root; the numerator names a power.
A negative fractional index also takes a reciprocal; zero cannot have a negative power.
Even roots use principal nonnegative values and cannot take negative inputs in real-number calculations.
Estimate positive roots by comparing nearby powers and tightening the bracket.

A power can turn $81$ into $27$ : use the index $3/4$ .

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HN6–HN7 - Fractional indices

Root powers

Combined indices

Estimating roots

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