Powers, Roots and Indices

Squares, cubes and their roots: the recall list

Cambridge names a short recall list for this topic: squares and their corresponding roots from 1 to 15, and cubes and their corresponding roots of 1, 2, 3, 4, 5 and 10. Recall means a question can say 'Write down the value of $\sqrt{169}$' and expect the exact answer 13 with no working shown. If you are reaching for a calculator here, you are spending time the later questions need.

A square number is an integer multiplied by itself, and a square root undoes that: ${13}^2=169$, so $\sqrt{169}=13$. A cube number is an integer multiplied by itself twice, and a cube root undoes that: $4^3=64$, so $\sqrt[3]{64}=4$. A root sign with no small number in it is a square root; the small 3 in $\sqrt[3]{}$ makes it a cube root.

Watch the signs. $\sqrt{25}$ is 5, not -5: the root sign on its own gives the positive root, and it is the solutions of $x^2=25$ that are $\pm 5$. A negative number has no square root, because no number multiplied by itself gives a negative answer, but cube roots of negative numbers do exist: $\sqrt[3]{-8}=-2$, because $-2\times -2\times -2=-8$.

The check is always the same one. Square your answer to test a square root and cube it to test a cube root: if $8^3=512$ then $\sqrt[3]{512}=8$. Do that check in your head rather than on the calculator, and a slipped digit never survives as far as the answer line.

Powers and roots beyond squares and cubes

A power (also called an index) records how many copies of the base are multiplied together. $2^5$ means $2\times 2\times 2\times 2\times 2=32$: base 2, index 5. $3^4=81$ and ${10}^4=10000$, and the index does not have to be small.

The syllabus asks you to calculate with 'other powers and roots of numbers', so higher roots are examinable too. $\sqrt[4]{81}=3$ because $3^4=81$, and $\sqrt[5]{32}=2$ because $2^5=32$. The small number written inside the root sign tells you which power you are undoing.

Most roots are not whole numbers, and then the skill is knowing where the answer sits. $\sqrt{50}$ lies between 7 and 8, because $7^2=49$ and $8^2=64$. Give 7.07 only when the question asks for a decimal or for 3 significant figures; if it asks for an estimate, 'between 7 and 8' is the answer.

• $2^5=32$, $3^4=81$, ${10}^4=10000$
• $\sqrt[4]{81}=3$ and $\sqrt[5]{32}=2$
• $\sqrt{50}=7.07$ to 3 significant figures, and it sits between 7 and 8
• On a calculator use the $x^y$ button for a power and the root button for a root, then test it on a value you know, such as $\sqrt{49}$

The rules of indices

The index laws only work when the bases are the same. To multiply two powers of the same base, add the indices: $a^m\times a^n=a^{m+n}$. To divide them, subtract the indices: $a^m÷a^n=a^{m-n}$.

The syllabus gives three examples worth knowing by heart: $2^{-3}\times 2^4=2^{-3+4}=2^1=2$, $(2^3{)}^2=2^{3\times 2}=2^6=64$, and $2^3÷2^4=2^{3-4}=2^{-1}=\frac{1}{2}$. Notice that a division can leave you with a negative index, which is why the next block matters.

The reason is only counting. $2^3\times 2^4$ is three 2s multiplied by four 2s, which is seven 2s altogether, so $2^7$. Division cancels: $2^3÷2^4$ leaves one 2 underneath, $\frac{1}{2}$. Nothing mysterious is happening; you are counting how many copies of the base survive.

When the bases are different, no rule applies. $2^3\times 3^2$ has to be worked out one power at a time: $8\times 9=72$. Writing $6^5$ there would be wrong, and it is one of the quickest ways to lose a mark in this topic.

Zero and negative indices

$a^0=1$ for every non-zero number a: $5^0=1$, $(-7{)}^0=1$ and ${100}^0=1$. The division rule explains it, because $2^3÷2^3=2^{3-3}=2^0$, and a number divided by itself is 1.

A negative index means a reciprocal, not a negative answer: $a^{-n}=\frac{1}{a^n}$. So $7^{-2}=\frac{1}{7^2}=\frac{1}{49}$ and $2^{-3}=\frac{1}{8}$. The syllabus names $7^{-2}$ as its example, so expect it as a short 'find the value of' question.

A negative index moves the power to the other side of the fraction line. $\frac{1}{5^{-2}}=5^2=25$, and ${\left(\frac{2}{3}\right)}^{-1}=\frac{3}{2}$, because inverting the fraction is the same as changing the sign of the index. Carry on with the same idea: ${\left(\frac{2}{3}\right)}^{-2}={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$.

The usual slip is writing $7^{-2}=-49$. Squaring 7 gives 49 whatever the sign of the index; the index only decides whether the answer is 49 or $\frac{1}{49}$. If your answer to a negative index question has come out negative, you have misread the sign.

Powers of powers and brackets

When a power is raised to another power, multiply the indices: $(a^m{)}^n=a^{mn}$. So $(2^3{)}^2=2^{3\times 2}=2^6=64$, and $(5^2{)}^3=5^6=15625$.

If the bracket holds a product, every factor is raised to the power: $(2\times 5{)}^3=2^3\times 5^3=8\times 125=1000$. That is the same as ${10}^3$, which is a useful check that the rule is behaving itself.

Addition inside a bracket is not the same thing at all. $(2+5{)}^2=7^2=49$, not $2^2+5^2=29$. Powers do not distribute over a sum, so work out the bracket first and raise the answer to the power.

Keep the two main rules apart in your head: multiplying two powers of the same base adds the indices, while raising a power to a power multiplies them. Nearly every lost mark in an indices question comes from swapping those two.

• $2^3\times 2^4=2^7=128$ (indices added)
• $(2^3{)}^4=2^{12}=4096$ (indices multiplied)
• $2^3÷2^4=2^{-1}=\frac{1}{2}$ (indices subtracted)
• $2^3+2^4=8+16=24$ (no rule here, so work out both powers first)

Fractional indices (Extended only)

A fractional index is a root in disguise: $a^{\frac{1}{n}}=\sqrt[n]{a}$. So ${36}^{\frac{1}{2}}=\sqrt{36}=6$, ${27}^{\frac{1}{3}}=\sqrt[3]{27}=3$, and ${16}^{\frac{1}{4}}=\sqrt[4]{16}=2$. A half power is a square root, a third power is a cube root, and so on.

When the numerator is not 1 the index does two jobs: the denominator is the root and the numerator is the power. In symbols, $a^{\frac{m}{n}}=(\sqrt[n]{a}{)}^m$. So $8^{\frac{2}{3}}=(\sqrt[3]{8}{)}^2=2^2=4$, and ${16}^{\frac{3}{4}}=(\sqrt[4]{16}{)}^3=2^3=8$.

Always take the root first. For $8^{\frac{2}{3}}$ the other order means working out $8^2=64$ and then hunting for $\sqrt[3]{64}$, which is harder and is where the arithmetic goes wrong. Root first keeps the numbers small and the answer exact.

Fractional indices obey the same laws as whole-number indices, and that is the real reason they are written this way: $9^{\frac{1}{2}}\times 9^{\frac{1}{2}}=9^1=9$, and indeed $3\times 3=9$. The rule $a^m\times a^n=a^{m+n}$ would break if a half power meant anything else.

• ${25}^{\frac{1}{2}}=5$, ${125}^{\frac{1}{3}}=5$, ${81}^{\frac{1}{4}}=3$
• ${25}^{\frac{3}{2}}=(\sqrt{25}{)}^3=5^3=125$
• ${64}^{\frac{2}{3}}=(\sqrt[3]{64}{)}^2=4^2=16$
• $8^{-\frac{2}{3}}=\frac{1}{(\sqrt[3]{8}{)}^2}=\frac{1}{4}$, so a fractional index can be negative as well

Surds: exact answers without decimals (Extended only)

A surd is a root that cannot be worked out exactly, left in its root form. $\sqrt{2}=1.41421356\dots$ never ends and never repeats, so writing 1.414 is only an approximation. An exact answer keeps the surd: $3\sqrt{5}$ is exact, while 6.708 is a rounded decimal.

To simplify a surd, look for the largest square factor hiding inside it and take its root outside. The syllabus example is $\sqrt{20}=\sqrt{4\times 5}=\sqrt{4}\times \sqrt{5}=2\sqrt{5}$. For a bigger number, split it into a square factor and what is left over: $\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$, and $\sqrt{200}=\sqrt{100\times 2}=10\sqrt{2}$.

Surds add and multiply like algebra with a letter. Only like surds can be collected: $3\sqrt{2}+5\sqrt{2}=8\sqrt{2}$, in the same way that $3x+5x=8x$. Multiplication combines the numbers in front and the surds separately: $3\sqrt{2}\times 5\sqrt{2}=15\times 2=30$, because $\sqrt{2}\times \sqrt{2}=2$.

Simplify before you add or subtract. The syllabus example is $\sqrt{200}-\sqrt{32}=10\sqrt{2}-4\sqrt{2}=6\sqrt{2}$, which only works once both surds are written in terms of $\sqrt{2}$.

Rationalising the denominator (Extended only)

An answer with a surd on the bottom of a fraction is not in its simplest form. To clear it, multiply the top and the bottom by the same surd, which is the same as multiplying by 1. The syllabus example: $\frac{10}{\sqrt{5}}=\frac{10\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}=\frac{10\sqrt{5}}{5}=2\sqrt{5}$.

When the denominator is a sum or a difference involving a surd, multiply by its conjugate — the same expression with the middle sign flipped. $(\sqrt{3}+1)(\sqrt{3}-1)=3-1=2$: the surd terms cancel and the root squares away. So $\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{\sqrt{3}-1}{2}$.

The conjugate works because of the difference of two squares, $(a+b)(a-b)=a^2-b^2$, and squaring a surd gives a whole number. That is the only reason the denominator comes out rational, so it is worth knowing why rather than just remembering to flip the sign.

Rationalising never changes the value, only the appearance. $\frac{10}{\sqrt{5}}$ and $2\sqrt{5}$ are the same number, about 4.47. A question that says 'rationalise the denominator', or gives you a target form such as $k\sqrt{5}$, wants the version with the whole number underneath.

• $\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}$
• $\frac{10}{\sqrt{5}}=\frac{10\sqrt{5}}{5}=2\sqrt{5}$
• $\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}$
• $\frac{4}{\sqrt{5}-1}=\frac{4(\sqrt{5}+1)}{4}=\sqrt{5}+1$