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Surds and exponents (QCE Math Methods Unit 1)

Syllabus dot point

“Simplify expressions involving surds and apply the laws of indices to rational and negative exponents”

7 min read

Quick answer

Simplify surds by pulling out perfect-square factors (e.g. 72=62\sqrt{72}=6\sqrt2), and rationalise denominators by multiplying by the surd or conjugate. The index laws extend to negative powers (a−n=1/ana^{-n}=1/a^n) and rational powers (am/n=amna^{m/n}=\sqrt[n]{a^m}).

Jump to a section
  1. What this dot point is asking
  2. Surd basics
  3. Adding and subtracting surds
  4. Rationalising denominators
  5. Index laws
  6. Rational and negative exponents
  7. Why rationalise with the conjugate
  8. In one sentence
  9. Exam-style questions

What this dot point is asking

QCAA wants you to simplify expressions involving surds (including rationalising denominators) and apply the laws of indices to rational and negative exponents. These skills give exact answers, which Methods prefers over decimal approximations, and they recur whenever a calculation produces an irrational result or a fractional power.

Surd basics

A surd is an irrational root, such as 2\sqrt 2, that cannot be reduced to a rational number. Surds obey the multiplication and division rules

a⋅b=ab,ab=ab,\sqrt a \cdot \sqrt b = \sqrt{ab}, \qquad \frac{\sqrt a}{\sqrt b} = \sqrt{\frac{a}{b}},

valid for a,b≥0a, b \geq 0. To simplify a surd, extract the largest perfect-square factor: 72=36⋅2=62\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt 2. There is no corresponding rule for sums, so a+b\sqrt{a + b} does not split into a+b\sqrt a + \sqrt b, a distinction worth stating explicitly because it is a frequent error.

Adding and subtracting surds

Surds add and subtract only when they are "like", that is share the same radicand after simplifying. So 32+52=823\sqrt 2 + 5\sqrt 2 = 8\sqrt 2, but 2+3\sqrt 2 + \sqrt 3 cannot be combined. Often two surds look unlike but become like after simplification, for example 8+18=22+32=52\sqrt{8} + \sqrt{18} = 2\sqrt 2 + 3\sqrt 2 = 5\sqrt 2, so always simplify each surd before deciding whether terms combine.

Rationalising denominators

Eliminate surds from the denominator.

Monomial: 13=33\dfrac{1}{\sqrt 3} = \dfrac{\sqrt 3}{3}.

Binomial: multiply by conjugate. 15−1=5+14\dfrac{1}{\sqrt 5 - 1} = \dfrac{\sqrt 5 + 1}{4}.

Index laws

The index laws combine powers that share a base, and they are the backbone of all exponent simplification:

aman=am+n,aman=am−n,(am)n=amn,a^m a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn},

(ab)n=anbn,(ab)n=anbn,a0=1 (a≠0),a−n=1an.(ab)^n = a^n b^n, \qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}, \qquad a^0 = 1\ (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}.

The first move in almost any simplification is to express every quantity as a power of a common base (for example writing 88 as 232^3), because the laws only act on matching bases. Surds fit in through a=a1/2\sqrt a = a^{1/2}, so a surd expression can always be recast as an index expression and simplified with the same laws.

Rational and negative exponents

Fractional and negative powers extend the index laws so that roots and reciprocals fit the same framework:

a1/n=an,am/n=(an)m=amn,a−n=1an.a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}, \qquad a^{-n} = \frac{1}{a^n}.

The denominator of a fractional index is the root and the numerator is the power, so 82/3=(83)2=22=48^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4 and 163/4=(164)3=23=816^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8, while 5−2=1255^{-2} = \tfrac{1}{25}. This unification is what lets a single set of laws simplify any expression mixing roots, reciprocals and powers, which is the recurring task in this dot point.

Why rationalise with the conjugate

For a binomial surd denominator such as 5−2\sqrt 5 - \sqrt 2, multiplying by the conjugate 5+2\sqrt 5 + \sqrt 2 exploits the difference of squares: (5−2)(5+2)=5−2=3(\sqrt 5 - \sqrt 2)(\sqrt 5 + \sqrt 2) = 5 - 2 = 3, a rational number. The conjugate is the same two terms with the middle sign flipped, and this is the only multiplier that clears both surds at once. The same idea evaluates expressions of the form (a+b)(a−b)=a2−b(a + \sqrt b)(a - \sqrt b) = a^2 - b.

Simplifying with index laws

Expand the power

Simplify (2x2)3⋅x−14x3/2\dfrac{(2x^2)^3 \cdot x^{-1}}{4x^{3/2}}. Apply (2x2)3=23x6=8x6(2x^2)^3 = 2^3 x^6 = 8x^6.

Combine the numerator

8x6⋅x−1=8x6−1=8x5.8x^6 \cdot x^{-1} = 8x^{6 - 1} = 8x^5.

Divide

8x54x3/2=2 x5−3/2=2x7/2.\frac{8x^5}{4x^{3/2}} = 2\,x^{5 - 3/2} = 2x^{7/2}.

Markers reward expanding the bracketed power, combining powers of xx in the numerator, and subtracting indices on division.

Common traps
Adding unlike surds
2+3\sqrt 2 + \sqrt 3 does not equal 5\sqrt 5; only like surds combine.
Sign of −a2-a^2
−32=−9-3^2 = -9 (the power binds tighter than the minus), whereas (−3)2=9(-3)^2 = 9.
Leaving a surd in the denominator
QCAA expects a rational denominator, so rationalise before stating the final form.
Misapplying a fractional index
am/na^{m/n} means the nnth root raised to the mmth power; the denominator of the index is the root.

In one sentence

Surds simplify by extracting perfect-square factors and rationalising denominators (monomial: multiply by the surd; binomial: multiply by the conjugate); the index laws extend to rational (am/n=amna^{m/n} = \sqrt[n]{a^m}) and negative (a−n=1/ana^{-n} = 1/a^n) exponents.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks
Paper 1 (technique). Simplify 75+123\dfrac{\sqrt{75} + \sqrt{12}}{\sqrt 3}, leaving the answer in simplest form.
Show worked answer →

75=53\sqrt{75} = 5\sqrt 3 and 12=23\sqrt{12} = 2\sqrt 3, so the numerator is 53+23=735\sqrt 3 + 2\sqrt 3 = 7\sqrt 3.

Divide: 733=7\dfrac{7\sqrt 3}{\sqrt 3} = 7.

Markers reward simplifying each surd, collecting like surds, and the cancellation.

Practice4 marks
Paper 1 (technique). (a) Rationalise the denominator of 65+2\dfrac{6}{\sqrt 5 + \sqrt 2}. (b) Evaluate 27−2/327^{-2/3} as an exact fraction.
Show worked answer →

(a) Multiply by the conjugate 5−25−2\dfrac{\sqrt 5 - \sqrt 2}{\sqrt 5 - \sqrt 2}: denominator (5)2−(2)2=5−2=3(\sqrt 5)^2 - (\sqrt 2)^2 = 5 - 2 = 3, so 6(5−2)3=2(5−2)\dfrac{6(\sqrt 5 - \sqrt 2)}{3} = 2(\sqrt 5 - \sqrt 2).

(b) 27−2/3=1272/3=1(273)2=132=19.27^{-2/3} = \dfrac{1}{27^{2/3}} = \dfrac{1}{(\sqrt[3]{27})^2} = \dfrac{1}{3^2} = \dfrac{1}{9}.

Markers reward the conjugate, the difference-of-squares denominator, and applying the negative and fractional index laws.

Practise this

Sources & how we know this

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