SurdsAQA A-Level Maths: Revision notes

Section 1

What a surd is

A surd is an irrational root that cannot be written exactly as a fraction, such as $\sqrt2$ , $\sqrt{5}$ or $\sqrt[3]{7}$ . A surd is left in root form in an exact answer; a decimal is only an approximation. $\sqrt9=3$ is not a surd, because it is rational. Useful results for any $a,b>0$ : $\sqrt{ab}=\sqrt a\times\sqrt b,\qquad \sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b},\qquad \sqrt a\times\sqrt a=a.$ Note that $\sqrt{a+b}\neq\sqrt a+\sqrt b$ .

Key termssurdirrational

Common mistake

Writing $\sqrt{9+16}=\sqrt9+\sqrt{16}=7$ . In fact $\sqrt{25}=5$ .

Section 2

Simplifying surds

To simplify a surd, take out the largest square factor: $\sqrt{50}=\sqrt{25\times2}=5\sqrt2,\qquad\sqrt{8}=\sqrt{4\times2}=2\sqrt2.$ A surd is fully simplified when the number under the root has no square factor. To add or subtract, first simplify, then combine like surds (the same root) just like like terms: $\sqrt{50}+\sqrt8=5\sqrt2+2\sqrt2=7\sqrt2$ . To multiply, $3\sqrt2\times4\sqrt5=12\sqrt{10}$ . To divide, $\frac{5\sqrt2}{2\sqrt2}=\frac52$ .

Key termslike surds

Common mistake

Adding the numbers under the roots: $\sqrt{50}+\sqrt8\neq\sqrt{58}$ .

Exam tip

Know the square numbers to 15 $^2$ so you can spot the largest square factor quickly.

Section 3

Expanding brackets with surds

Expand surd brackets exactly as you would algebraic brackets, using $\sqrt a\times\sqrt a=a$ : $(3+\sqrt5)^2=9+6\sqrt5+5=14+6\sqrt5,$ $(3+\sqrt5)(3-\sqrt5)=9-5=4.$ The second case is the difference of two squares, $(a+b)(a-b)=a^2-b^2$ . The surd terms cancel, so the answer is rational. This is the key idea behind rationalising.

Key termsdifference of two squares

Common mistake

Writing $(3+\sqrt5)^2=9+5$ . The cross term $2\times3\times\sqrt5$ is missing.

Section 4

Rationalising the denominator

To rationalise the denominator means to rewrite a fraction so that the denominator contains no surd.

For a single surd, multiply top and bottom by that surd: $\dfrac{6}{\sqrt3}=\dfrac{6\sqrt3}{3}=2\sqrt3$ .
For a denominator $a+\sqrt b$ , multiply top and bottom by the conjugate $a-\sqrt b$ (change the sign), which gives $a^2-b$ on the bottom. Worked example. $\dfrac{2}{3-\sqrt5}=\dfrac{2(3+\sqrt5)}{(3-\sqrt5)(3+\sqrt5)}=\dfrac{2(3+\sqrt5)}{4}=\dfrac{3+\sqrt5}{2}$ .

Key termsrationaliseconjugate

Common mistake

Multiplying only the denominator. Whatever you multiply the bottom by, you must multiply the top by as well.

Exam tip

Simplify the answer fully at the end: $\frac{22+10\sqrt7}{6}$ becomes $\frac{11+5\sqrt7}{3}$ .

Section 5

Using surds in problems

Exact answers appear in geometry (areas, diagonals) and in algebra. Keep surds until the end, and use identities to avoid decimals. For example, given $x=\frac{3+\sqrt5}{2}$ and $y=\frac{3-\sqrt5}{2}$ , you can find $x+y=3$ and $xy=1$ , then $x^2+y^2=(x+y)^2-2xy=7$ . A diagonal of a rectangle with sides $(3+\sqrt5)$ and $(3-\sqrt5)$ has $d^2=28$ , so $d=\sqrt{28}=2\sqrt7$ . If a question says exact, or gives the form $\dfrac{a+b\sqrt c}{d}$ , do not use a decimal.

Key termsexact value

Exam tip

Check with a calculator: if the surd form and the decimal of the original do not agree, one step is wrong.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Surds

Let $p=\sqrt{50}$ and $q=\sqrt{8}$ .
Express $\dfrac{p}{q}$ as a fraction in its simplest form.2 marks
A rectangle has length $(3+\sqrt5)$ cm and width $(3-\sqrt5)$ cm.
Express length divided by width in the form $a+b\sqrt5$ , where $a$ and $b$ are rational.2 marks
A rectangle has area $(8+2\sqrt7)$ cm $^2$ and length $(\sqrt7-1)$ cm.
Find the width of the rectangle, giving your answer in the form $\dfrac{a+b\sqrt7}{c}$ where $a$ , $b$ and $c$ are integers.3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).