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Indices & Surds

Powers give you a shorthand for repeated multiplication, and surds let you keep roots exact instead of rounding. Master the handful of rules here and a lot of algebra suddenly gets tidier.

MYP 4-5ExtendedNumberCriteria A · C~12 min read

Written by Fayek Jawad Haider · MYP student

The laws of indices

An index (plural indices), also called a power or exponent, tells you how many times to multiply a number by itself. The three core laws all follow from that idea:

\[ a^m \times a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn} \]

Read them in words: multiplying powers of the same base, you add; dividing, you subtract; a power of a power, you multiply. The "same base" condition matters, the laws only apply when the base is identical.

Worked example

Simplify \((2x^2)^3 \times x^{-4}\).

1
Deal with the bracket first, the power hits both the 2 and the \(x^2\): \((2x^2)^3 = 2^3 x^{6} = 8x^6\).
2
Now multiply by \(x^{-4}\), adding the indices: \(8x^{6} \times x^{-4} = 8x^{6+(-4)}\).
\(8x^{2}\)

Zero & negative powers

Two special cases fall straight out of the division law:

\[ a^0 = 1 \qquad\qquad a^{-n} = \frac{1}{a^{n}} \]

Anything (except 0) to the power zero is 1, and a negative power means "one over" the positive power. So \(5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}\). A negative index never makes the answer negative, it makes it a fraction.

Common slip

\(3^{-2}\) is \(\tfrac{1}{9}\), not \(-9\) and not \(-6\). The minus sign moves the term to the denominator, it does not touch the sign of the value.

Fractional powers

A fractional index is a root. The denominator is the root you take; the numerator is the power you raise it to:

\[ a^{\frac{1}{n}} = \sqrt[n]{a} \qquad\qquad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^{m} \]
Worked example

Evaluate \(16^{\frac{3}{4}}\).

1
Take the root on the bottom first, it keeps the numbers small: \(16^{\frac{1}{4}} = \sqrt[4]{16} = 2\).
2
Then raise to the power on top: \(2^{3} = 8\).
\(16^{\frac{3}{4}} = 8\)

Root first, power second

Always take the root before the power. \(\sqrt[4]{16}=2\) then cube gives 8, whereas cubing 16 first gives 4096 and a much harder fourth root. Same answer, far more work.

What a surd is

A surd is a root that cannot be simplified to a whole number, so we leave it in root form to stay exact. \(\sqrt{2}\), \(\sqrt{3}\) and \(\sqrt{10}\) are surds; \(\sqrt{9}=3\) is not. Two rules do most of the work:

\[ \sqrt{a} \times \sqrt{b} = \sqrt{ab} \qquad\qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \]

There is no matching rule for adding surds: \(\sqrt{a}+\sqrt{b}\) is not \(\sqrt{a+b}\). You can only add surds that are already "like", such as \(2\sqrt{3}+4\sqrt{3}=6\sqrt{3}\).

Simplifying surds

To simplify a surd, split the number under the root into a factor pair where one factor is a perfect square, then take that square root out.

Worked example

Simplify \(\sqrt{50}\).

1
Find the largest perfect-square factor of 50: that is 25, since \(50 = 25 \times 2\).
2
Split the root: \(\sqrt{50} = \sqrt{25}\times\sqrt{2}\).
3
\(\sqrt{25}=5\), so it comes outside the root.
\(\sqrt{50} = 5\sqrt{2}\)

Rationalising the denominator

Convention says we do not leave a surd on the bottom of a fraction. Rationalising clears it, by multiplying top and bottom by that same surd (which is really just multiplying by 1).

Worked example

Rationalise \(\dfrac{6}{\sqrt{3}}\).

1
Multiply top and bottom by \(\sqrt{3}\): \(\dfrac{6}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3}\).
2
The denominator is now the whole number 3. Simplify the fraction: \(\dfrac{6}{3}=2\).
\(\dfrac{6}{\sqrt{3}} = 2\sqrt{3}\)

Where this is assessed

Indices and surds sit squarely in Criterion A (applying the right rule) and Criterion C (exact, correctly-notated answers). Exams often insist on an exact surd form, a rounded decimal there loses marks.

Check yourself

1. Simplify \((3x^4)^2\). +

Square both parts: \(3^2 (x^4)^2 = \mathbf{9x^{8}}\).

2. Evaluate \(27^{\frac{2}{3}}\). +

Cube root first: \(\sqrt[3]{27}=3\), then square: \(3^2 = \mathbf{9}\).

3. Simplify \(\sqrt{18}+\sqrt{8}\). +

\(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\). They are now like surds: \(3\sqrt{2}+2\sqrt{2}=\mathbf{5\sqrt{2}}\).

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