Sec 3 · Additional Mathematics

Surds

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1. Simplifying surds

Rule	Example
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$	$\sqrt{3} \times \sqrt{5} = \sqrt{15}$
$\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$	$\dfrac{\sqrt{12}}{\sqrt{3}} = 2$
$\sqrt{a} \times \sqrt{a} = a$	$\sqrt{7} \times \sqrt{7} = 7$
like surds add	$2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$

Take out the largest square factor

\displaystyle \sqrt{75}

e.g.

\displaystyle \sqrt{75}

\displaystyle = \sqrt{25 \times 3}

\displaystyle = 5\sqrt{3}

Simplify each, then collect

\displaystyle \sqrt{12} + \sqrt{75}

e.g.

\displaystyle \sqrt{12} + \sqrt{75}

\displaystyle = 2\sqrt{3} + 5\sqrt{3}

\displaystyle = 7\sqrt{3}

📖 Worked example

Simplify $2\sqrt{12} - \sqrt{27} + \sqrt{3}$ , giving your answer in the form $a\sqrt{b}$ .

Step 1 of 2

Simplify each surd: take out the biggest square factor, keeping the number in front

2\sqrt{12} = 2 \times 2\sqrt{3}

= 4\sqrt{3}

-\sqrt{27} = -3\sqrt{3}

\sqrt{3}

is already in simplest form

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2. Expanding surds

Each term × each term

\displaystyle (1 - \sqrt{3})(4 + 2\sqrt{3})

e.g.

\displaystyle (1 - \sqrt{3})(4 + 2\sqrt{3})

\displaystyle = 4 + 2\sqrt{3} - 4\sqrt{3} - 6

\displaystyle = -2 - 2\sqrt{3}

Conjugates: a difference of two squares

\displaystyle (a + \sqrt{b})(a - \sqrt{b}) = a^{2} - b

e.g.

\displaystyle (3 + \sqrt{5})(3 - \sqrt{5})

\displaystyle = 9 - 5

\displaystyle = 4

📖 Worked example

Expand and simplify $(4 + \sqrt{3})^2$ .

Step 1 of 2

(a + b)² = a² + 2ab + b²: or multiply every term by every term

First:

4 \times 4 = 16

Outer:

4 \times \sqrt{3} = 4\sqrt{3}

Inner:

\sqrt{3} \times 4 = 4\sqrt{3}

Last:

\sqrt{3} \times \sqrt{3} = 3

Practise this →

3. Rationalising the denominator

Bottom	Multiply top and bottom by
$\sqrt{b}$	$\sqrt{b}$
$a + \sqrt{b}$	$a - \sqrt{b}$
$a - \sqrt{b}$	$a + \sqrt{b}$

A root on the bottom

\displaystyle \dfrac{6}{\sqrt{3}}

e.g.

\displaystyle \dfrac{6}{\sqrt{3}}

\displaystyle = \dfrac{6\sqrt{3}}{3}

\displaystyle = 2\sqrt{3}

Use the conjugate

\displaystyle \dfrac{4}{3 - \sqrt{5}}

e.g.

\displaystyle \dfrac{4}{3 - \sqrt{5}}

\displaystyle = \dfrac{4(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}

\displaystyle = \dfrac{4(3 + \sqrt{5})}{4}

\displaystyle = 3 + \sqrt{5}

📖 Worked example

Express $\dfrac{3 - \sqrt{3}}{2 + \sqrt{3}}$ in the form $\dfrac{a + b\sqrt{3}}{c}$ , where $a$ , $b$ and $c$ are integers.

Step 1 of 4

Multiply the top and the bottom by the conjugate of the bottom (change the sign between its terms)

\dfrac{3 - \sqrt{3}}{2 + \sqrt{3}} = \dfrac{(3 - \sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}

Practise this →

4. Surd equations

Square both sides, solve, then check

\displaystyle \sqrt{x + 7} = x + 1

e.g.

\displaystyle x + 7 = (x + 1)^{2}

\displaystyle x^{2} + x - 6 = 0

\displaystyle x = -3 \text{ or } x = 2

check $x = -3$ : $\sqrt{4} = 2$ but $-3 + 1 = -2$ , so reject it
answer: $x = 2$

📖 Worked example

Solve $x\sqrt{3} - x = 6$ , giving your answer in the form $a + b\sqrt{3}$ , where $a$ and $b$ are integers.

Step 1 of 4

Factorise out x, then divide

x(\sqrt{3} - 1) = 6

x = \dfrac{6}{\sqrt{3} - 1}

Practise this →

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Sec 3 · Additional Mathematics

Surds

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Start free Worksheet (PDF)

1. Simplifying surds

Rule	Example
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$	$\sqrt{3} \times \sqrt{5} = \sqrt{15}$
$\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$	$\dfrac{\sqrt{12}}{\sqrt{3}} = 2$
$\sqrt{a} \times \sqrt{a} = a$	$\sqrt{7} \times \sqrt{7} = 7$
like surds add	$2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$

Take out the largest square factor

\displaystyle \sqrt{75}

e.g.

\displaystyle \sqrt{75}

\displaystyle = \sqrt{25 \times 3}

\displaystyle = 5\sqrt{3}

Simplify each, then collect

\displaystyle \sqrt{12} + \sqrt{75}

e.g.

\displaystyle \sqrt{12} + \sqrt{75}

\displaystyle = 2\sqrt{3} + 5\sqrt{3}

\displaystyle = 7\sqrt{3}

📖 Worked example

Simplify $2\sqrt{12} - \sqrt{27} + \sqrt{3}$ , giving your answer in the form $a\sqrt{b}$ .

Step 1 of 2

Simplify each surd: take out the biggest square factor, keeping the number in front

2\sqrt{12} = 2 \times 2\sqrt{3}

= 4\sqrt{3}

-\sqrt{27} = -3\sqrt{3}

\sqrt{3}

is already in simplest form

Practise this →

2. Expanding surds

Each term × each term

\displaystyle (1 - \sqrt{3})(4 + 2\sqrt{3})

e.g.

\displaystyle (1 - \sqrt{3})(4 + 2\sqrt{3})

\displaystyle = 4 + 2\sqrt{3} - 4\sqrt{3} - 6

\displaystyle = -2 - 2\sqrt{3}

Conjugates: a difference of two squares

\displaystyle (a + \sqrt{b})(a - \sqrt{b}) = a^{2} - b

e.g.

\displaystyle (3 + \sqrt{5})(3 - \sqrt{5})

\displaystyle = 9 - 5

\displaystyle = 4

📖 Worked example

Expand and simplify $(4 + \sqrt{3})^2$ .

Step 1 of 2

(a + b)² = a² + 2ab + b²: or multiply every term by every term

First:

4 \times 4 = 16

Outer:

4 \times \sqrt{3} = 4\sqrt{3}

Inner:

\sqrt{3} \times 4 = 4\sqrt{3}

Last:

\sqrt{3} \times \sqrt{3} = 3

Practise this →

3. Rationalising the denominator

Bottom	Multiply top and bottom by
$\sqrt{b}$	$\sqrt{b}$
$a + \sqrt{b}$	$a - \sqrt{b}$
$a - \sqrt{b}$	$a + \sqrt{b}$

A root on the bottom

\displaystyle \dfrac{6}{\sqrt{3}}

e.g.

\displaystyle \dfrac{6}{\sqrt{3}}

\displaystyle = \dfrac{6\sqrt{3}}{3}

\displaystyle = 2\sqrt{3}

Use the conjugate

\displaystyle \dfrac{4}{3 - \sqrt{5}}

e.g.

\displaystyle \dfrac{4}{3 - \sqrt{5}}

\displaystyle = \dfrac{4(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}

\displaystyle = \dfrac{4(3 + \sqrt{5})}{4}

\displaystyle = 3 + \sqrt{5}

📖 Worked example

Express $\dfrac{3 - \sqrt{3}}{2 + \sqrt{3}}$ in the form $\dfrac{a + b\sqrt{3}}{c}$ , where $a$ , $b$ and $c$ are integers.

Step 1 of 4

Multiply the top and the bottom by the conjugate of the bottom (change the sign between its terms)

\dfrac{3 - \sqrt{3}}{2 + \sqrt{3}} = \dfrac{(3 - \sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}

Practise this →

4. Surd equations

Square both sides, solve, then check

\displaystyle \sqrt{x + 7} = x + 1

e.g.

\displaystyle x + 7 = (x + 1)^{2}

\displaystyle x^{2} + x - 6 = 0

\displaystyle x = -3 \text{ or } x = 2

check $x = -3$ : $\sqrt{4} = 2$ but $-3 + 1 = -2$ , so reject it
answer: $x = 2$

📖 Worked example

Solve $x\sqrt{3} - x = 6$ , giving your answer in the form $a + b\sqrt{3}$ , where $a$ and $b$ are integers.

Step 1 of 4

Factorise out x, then divide

x(\sqrt{3} - 1) = 6

x = \dfrac{6}{\sqrt{3} - 1}

Practise this →

Try a question: Surds

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Practise surds with step-by-step help

Endless questions from basic to exam level. Get one wrong and see where it went wrong. Free account.

Start free