YEAR 10 · EXPONENTIAL RELATIONSHIPS · 15 MIN

Operations with surds

Add like surds and multiply or divide surd expressions exactly.

In this lesson

Recognise like surds
Expand and simplify surd products

Before you start

Simplifying surds
Expanding brackets

01 / LEARN

Understand the idea.

Like surds behave like like terms

Like surds have the same simplified root part. Simplify each radical first, then add or subtract its coefficients, just as you combine like algebraic terms. For example, $2 \sqrt{3} + 5 \sqrt{3} = 7 \sqrt{3}$ .

Terms such as $2 \sqrt{3}$ and $5 \sqrt{2}$ have different root parts and cannot be combined into one like term. Leaving their sum in exact form is a finished answer, not an incomplete decimal calculation.

Products use distribution

For nonnegative radicands, $\sqrt{a} \sqrt{b} = \sqrt{a b}$ . Multiply any outside coefficients separately, then simplify the resulting radical. For division, $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ requires $b > 0$ .

Bracket products still obey distribution. Conjugates, $(a + b) (a - b)$ , give $a^{2} - b^{2}$ because the two middle terms cancel. Squaring a bracket such as ${(a + b)}^{2}$ instead includes the middle term $2 a b$ .

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Simplify before adding

Simplify $\sqrt{12} + \sqrt{27}$ .

Worked solution

1
Simplify the first radical.
$\sqrt{12} = \sqrt{4 \times 3} = 2 \sqrt{3}$
The factor $4$ is a perfect square.
2
Simplify the second radical.
$\sqrt{27} = \sqrt{9 \times 3} = 3 \sqrt{3}$
Both terms now have the same root part.
3
Add the coefficients.
$\sqrt{12} + \sqrt{27} = (2 + 3) \sqrt{3} = 5 \sqrt{3}$
The common factor $\sqrt{3}$ is retained.

Answer

$5 \sqrt{3}$

EXAMPLE 02

Expand a product

Expand $(\sqrt{5} + 2) (\sqrt{5} - 2)$ .

Worked solution

1
Recognise conjugate factors.
$(\sqrt{5} + 2) (\sqrt{5} - 2) = {(\sqrt{5})}^{2} - 2^{2}$
The identity $(a + b) (a - b) = a^{2} - b^{2}$ cancels the middle terms.
2
Evaluate each square.
${(\sqrt{5})}^{2} - 2^{2} = 5 - 4$
Squaring the principal square root of $5$ recovers $5$ .
3
Subtract.
$(\sqrt{5} + 2) (\sqrt{5} - 2) = 1$
The two irrational factors have a rational product.

Answer

03 / YOUR TURN

Try it yourself.

Check your understanding

Simplify $\sqrt{6} \sqrt{15}$ .

Show answer & reasoning

$3 \sqrt{10}$

Multiply the radicands, then simplify: $\sqrt{6} \sqrt{15} = \sqrt{90} = \sqrt{9 \times 10} = 3 \sqrt{10}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Simplify $4 \sqrt{2} + 3 \sqrt{2}$ .
2 marks
Simplify $\sqrt{18} - \sqrt{8}$ .
2 marks
Find $\sqrt{3} \sqrt{12}$ .
2 marks
Expand ${(2 + \sqrt{3})}^{2}$ .
3 marks
Simplify $\frac{\sqrt{75}}{\sqrt{3}}$ .
2 marks
A rectangle has sides $2 \sqrt{3}$ cm and $3 \sqrt{6}$ cm. Find its exact area.
3 marks

Read the worked solutions

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

Simplify $4 \sqrt{2} + 3 \sqrt{2}$ .
2 marks
Add coefficients: $4 + 3 = 7$ .
$7 \sqrt{2}$
Simplify $\sqrt{18} - \sqrt{8}$ .
2 marks
$3 \sqrt{2} - 2 \sqrt{2} = \sqrt{2}$ .
$\sqrt{2}$
Find $\sqrt{3} \sqrt{12}$ .
2 marks
$\sqrt{36} = 6$ .
6
Expand ${(2 + \sqrt{3})}^{2}$ .
3 marks
$4 + 4 \sqrt{3} + 3 = 7 + 4 \sqrt{3}$ .
$7 + 4 \sqrt{3}$
Simplify $\frac{\sqrt{75}}{\sqrt{3}}$ .
2 marks
$\sqrt{\frac{75}{3}} = \sqrt{25} = 5$ .
5
A rectangle has sides $2 \sqrt{3}$ cm and $3 \sqrt{6}$ cm. Find its exact area.
3 marks
$2 \sqrt{3} \times 3 \sqrt{6} = 6 \sqrt{18} = 18 \sqrt{2}$ .
$18 \sqrt{2} {cm}^{2}$

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Integer and fractional index laws; connecting fractional powers with roots; simplifying surds; operations with surds; rationalising denominators; exact and approximate values; exponential tables and graphs; growth/decay factors; comparing linear/quadratic/exponential patterns; exponential equations; logarithms as inverses; logarithm laws; interpreting logarithmic scales; graph transformations and parameter experiments; choosing and evaluating models.

Official curriculum source ↗

In this lesson

Understand the idea.

Like surds behave like like terms

Products use distribution

See the method in action.

Simplify before adding

Simplify the first radical.

Simplify the second radical.

Add the coefficients.

Expand a product

Recognise conjugate factors.

Evaluate each square.

Subtract.

Try it yourself.

Check your understanding

Keep practising