Rational/irrational numbers and surd operations | Year 11

01 / LEARN

Understand the idea.

Which roots are irrational?

A rational number can be written as a ratio of integers with nonzero denominator; its decimal terminates or eventually repeats. A square root may be rational, as $\sqrt{49} = 7$ shows, so inspect the value before classifying it.

To simplify a surd, factor out the largest square factor. For nonnegative factors, a square root of a product can split into the product of their square roots. This exposes any whole-number coefficient.

Collect like radicals and rationalise

Simplify radicals before trying to add or subtract them. Only like radicals combine: their coefficients can be collected because they multiply the same exact number. Multiplication follows a different rule and can combine nonnegative radicands.

Rationalise a denominator by multiplying the entire fraction by 1 in a useful form. A conjugate removes a two-term radical denominator through the difference of two squares. Keep the exact expression unless a decimal is requested.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Simplify then collect

Simplify $\sqrt{72} - 2 \sqrt{8}$ .

Worked solution

1
Extract square factors from both radicals.
$\sqrt{72} = \sqrt{36 \cdot 2} = 6 \sqrt{2}, \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$
Each radicand is written as a perfect square times the same remaining factor.
2
Apply the outside coefficient.
$\sqrt{72} - 2 \sqrt{8} = 6 \sqrt{2} - 2 (2 \sqrt{2}) = 6 \sqrt{2} - 4 \sqrt{2}$
The coefficient 2 multiplies the whole simplified value of $\sqrt{8}$ .
3
Collect the like radicals.
$6 \sqrt{2} - 4 \sqrt{2} = (6 - 4) \sqrt{2} = 2 \sqrt{2}$
Both terms contain the same radical, so their coefficients can be subtracted.

Answer

$2 \sqrt{2}$

EXAMPLE 02

Use a conjugate

Rationalise $3 / (\sqrt{5} - 1)$ .

Worked solution

1
Multiply by the conjugate over itself.
$\frac{3}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1}$
The second fraction equals 1, so the value is unchanged.
2
Use the difference of two squares.
$\frac{3 (\sqrt{5} + 1)}{{(\sqrt{5})}^{2} - 1^{2}} = \frac{3 (\sqrt{5} + 1)}{5 - 1}$
The conjugate product cancels the two middle terms.
3
Simplify the rational denominator.
$\frac{3 (\sqrt{5} + 1)}{4}$
The denominator is now a nonzero rational number.

Answer

$3 (\sqrt{5} + 1) / 4$

03 / YOUR TURN

Try it yourself.

Check your understanding

Is $\sqrt{0.04}$ rational?

Show answer & reasoning

Yes: 0.2.

The principal square root is $\sqrt{0.04} = 0.2 = \frac{1}{5}$ . It is rational because it is a ratio of integers, and ${0.2}^{2} = 0.04$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Classify $\sqrt{81}$ and $\sqrt{18}$ .
2 marks
Simplify $\sqrt{50}$ .
2 marks
Simplify $3 \sqrt{12} - \sqrt{27}$ .
2 marks
Expand $(\sqrt{7} + 2) (\sqrt{7} - 2)$ .
2 marks
Rationalise $\frac{5}{\sqrt{3}}$ .
2 marks
Rationalise $\frac{2}{\sqrt{6} - 2}$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Classify $\sqrt{81}$ and $\sqrt{18}$ .
2 marks
$\sqrt{81} = 9$ ; $\sqrt{18} = 3 \sqrt{2}$ .
9 is rational; the square root of 18 is irrational.
Simplify $\sqrt{50}$ .
2 marks
$50 = 25 \times 2$ so extract $\sqrt{25} = 5$ .
$5 \sqrt{2}$
Simplify $3 \sqrt{12} - \sqrt{27}$ .
2 marks
$3 (2 \sqrt{3}) - 3 \sqrt{3} = 3 \sqrt{3}$ .
$3 \sqrt{3}$
Expand $(\sqrt{7} + 2) (\sqrt{7} - 2)$ .
2 marks
Use difference of squares: $7 - 4 = 3$ .
3
Rationalise $\frac{5}{\sqrt{3}}$ .
2 marks
Multiply top and bottom by $\sqrt{3}$ .
$\frac{5 \sqrt{3}}{3}$
Rationalise $\frac{2}{\sqrt{6} - 2}$ .
3 marks
Multiply by the conjugate: $\frac{2 (\sqrt{6} + 2)}{6 - 4}$ .
Cancel the factor 2.
$\sqrt{6} + 2$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

In this lesson

Understand the idea.

Which roots are irrational?

Collect like radicals and rationalise

See the method in action.

Simplify then collect

Extract square factors from both radicals.

Apply the outside coefficient.

Collect the like radicals.

Use a conjugate

Multiply by the conjugate over itself.

Use the difference of two squares.

Simplify the rational denominator.

Try it yourself.

Check your understanding

Keep practising