Surds and rational exponents (VCE Maths Methods Unit 1)

What this dot point is asking

VCAA wants you to simplify and operate on surds (including rationalising denominators) and to apply the index laws to rational and negative exponents.

Surd basics

A surd is an irrational root that cannot be simplified to a rational number.

Surds are simplified by removing perfect-square factors. $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt 2$.

Adding and subtracting surds

Like terms only: $3\sqrt 2 + 5\sqrt 2 = 8\sqrt 2$. Unlike surds do not combine: $\sqrt 2 + \sqrt 3$ stays as is.

Rationalising denominators

Eliminate surds from a denominator.

Monomial denominator. Multiply top and bottom by the surd: $\dfrac{1}{\sqrt 3} = \dfrac{\sqrt 3}{3}$.

Binomial denominator. Multiply by the conjugate: $\dfrac{1}{\sqrt 5 - 1} = \dfrac{\sqrt 5 + 1}{(\sqrt 5 - 1)(\sqrt 5 + 1)} = \dfrac{\sqrt 5 + 1}{4}$.

Rational exponents

Example: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$.

Negative exponents

Example: $5^{-2} = 1/25$.

All index laws apply

The seven index laws ($a^m a^n = a^{m+n}$, etc.) apply to rational and negative exponents:

$x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6}$.

$(x^{2/3})^3 = x^2$.

The conjugate and the difference of squares

Rationalising a binomial surd denominator relies on the identity $(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = a - b$, which clears both surds at once. The pair $\sqrt{a} - \sqrt{b}$ and $\sqrt{a} + \sqrt{b}$ are conjugates. Multiplying the fraction by the conjugate over itself does not change its value but removes the surd from the denominator, leaving a rational denominator. This is the standard exam method whenever a surd appears in a denominator with two terms.

The seven index laws and their fractional forms

The index laws apply unchanged to rational and negative exponents. They are: $a^m a^n = a^{m+n}$; $\dfrac{a^m}{a^n} = a^{m-n}$; $(a^m)^n = a^{mn}$; $(ab)^n = a^n b^n$; $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$; $a^0 = 1$ (for $a \ne 0$); and $a^{-n} = \dfrac{1}{a^n}$. Combined with the definition $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$, these let you rewrite any surd as a power and simplify expressions that mix roots, reciprocals and products. Converting a surd to index form (for example $\sqrt[3]{x^2} = x^{2/3}$) is often the cleanest route through a simplification.

In one sentence

Surds are simplified by extracting perfect-square factors ($\sqrt{72} = 6\sqrt 2$), like surds add and subtract (unlike surds do not), denominators are rationalised by multiplying by the surd or its conjugate, and the index laws extend to rational ($a^{m/n} = \sqrt[n]{a^m}$) and negative ($a^{-n} = 1/a^n$) exponents.

Examples in context

Example 1. Exact side length from area. A square paving tile has area $48 \, \text{cm}^2$, so its side is $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt 3 \, \text{cm}$, an exact surd value (about $6.93$ cm). The perimeter is $4 \times 4\sqrt 3 = 16\sqrt 3 \, \text{cm}$, kept exact rather than rounded so later calculations stay precise.

Example 2. Rational exponents in a scaling law. A model states that an object's strength scales as $S = 8 m^{2/3}$ for mass $m$ kilograms. For $m = 27$: $S = 8 \times 27^{2/3} = 8 \times (\sqrt[3]{27})^2 = 8 \times 3^2 = 8 \times 9 = 72$. The fractional exponent $\tfrac23$ means "cube root, then square."

Try this

Q1. Simplify $\sqrt{75} + \sqrt{12}$. [2 marks]

• Cue. $5\sqrt 3 + 2\sqrt 3 = 7\sqrt 3$.

Q2. Rationalise the denominator of $\dfrac{6}{\sqrt 3}$ and of $\dfrac{2}{\sqrt 5 + 1}$. [2+2 marks]

• Cue. $\dfrac{6}{\sqrt 3} = 2\sqrt 3$; $\dfrac{2}{\sqrt 5 + 1} = \dfrac{2(\sqrt 5 - 1)}{4} = \dfrac{\sqrt 5 - 1}{2}$.

Q3. Evaluate $16^{3/4}$ and $27^{-2/3}$. [2+2 marks]

• Cue. $16^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8$; $27^{-2/3} = \dfrac{1}{(\sqrt[3]{27})^2} = \dfrac{1}{9}$.