Simplifying surds · Surds

A surd is a square root that isn’t a whole number, such as $\sqrt2$ or $\sqrt{12}$ . Its decimal never ends, so questions ask you to leave answers “in surd form” and keep them exact.

The one rule behind everything:

\sqrt{ab} = \sqrt a \times \sqrt b

Simplifying: take out the largest square

Simplifying surds: split the squarePick a number

72area of the big square36 × 2= 6² × 2: 36 squares of area 2√72 = 6√2side = 6 small sides of √2

The big square has area 72, so its side is √72. Split it into 6 × 6 = 36 small squares: each has area 72 ÷ 36 = 2, so each side is √2. The big side is 6 of those: √72 = 6√2. Splitting by a smaller square such as 9 gives 3√8, which can still be simplified: always use the largest square factor.

$\sqrt{72}$ is the side of a square of area 72. Since $72 = 36 \times 2$ , the square splits into $6 \times 6$ small squares of area 2, so its side is 6 lots of $\sqrt2$ : $\sqrt{72} = 6\sqrt2$ .

In symbols: $\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt2 = 6\sqrt2$ .

√72 = 6√236 small squares of area 2: the side is 6 lots of √2

Adding and subtracting: only like surds

$2\sqrt3 + 5\sqrt3 = 7\sqrt3$ , just like $2x + 5x = 7x$ . But $\sqrt2 + \sqrt3$ can’t be combined, and $\sqrt2 + \sqrt3 \ne \sqrt5$ . So simplify every surd first; often they turn out to be multiples of the same one.

Like surdsCount the √3s, as you would count x's

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q2 (a)

Simplify $3\sqrt{75} - \sqrt{12} + \sqrt{108}$ , leaving the answer in surd form (radicals).

Simplify each surd
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt3$ , $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt3$ and $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt3$ .
Think first. Which square divides 75? 12? 108?
Collect like surds
$\begin{aligned} &3 \times 5\sqrt3 - 2\sqrt3 + 6\sqrt3 \\ &= 15\sqrt3 - 2\sqrt3 + 6\sqrt3 = 19\sqrt3 \end{aligned}$

More: simplifying and collecting surds

Multiplying

Multiply the numbers outside together and the numbers inside together: $2\sqrt3 \times 5\sqrt6 = 10\sqrt{18} = 10 \times 3\sqrt2 = 30\sqrt2$ . And $\sqrt a \times \sqrt a = a$ .

Brackets expand as usual. A useful pair is the difference of two squares:

(a + \sqrt b)(a - \sqrt b) = a^2 - b

The surds cancel and a whole number is left.

The difference of two squares(a + √b)(a − √b) = a² − b

More: multiplying and substituting surds

The square root of a surd expression

Squaring a difference of surds gives a whole number plus a surd: $(\sqrt5 - \sqrt3)^2 = 5 + 3 - 2\sqrt{15} = 8 - 2\sqrt{15}$ . So to find $\sqrt{8 - 2\sqrt{15}}$ , look for two numbers that add to 8 and multiply to 15: they are 5 and 3, and the root is $\sqrt5 - \sqrt3$ .

More: square roots of surd expressions

Surds in other topics

Surds turn up whenever an answer must be exact: in trigonometry with exact values such as $\sin 60^\circ = \frac{\sqrt3}{2}$ , in Pythagoras, and in equations whose solution isn’t a whole number. The same rules apply.

More: surds inside other topics

Your turn

WAEC 2019 · Paper 2 · Q1 (b)

(b)
Simplify $\dfrac{15}{\sqrt{75}} + \sqrt{108} + \sqrt{432}$ , leaving the answer in the form $a\sqrt b$ , where $a$ and $b$ are positive integers.

Worked solution (try it first)

(b)

Rationalise and simplify each term: $\frac{15}{\sqrt{75}} = \frac{15}{5\sqrt3}$
$= \frac{3}{\sqrt3}$
$= \sqrt3$ .
$\sqrt{108} = 6\sqrt3$ .
$\sqrt{432} = \sqrt{144 \times 3}$
$= 12\sqrt3$ .
Total: $\sqrt3 + 6\sqrt3 + 12\sqrt3 = 19\sqrt3$ .

Report a problem with this question