For Higher GCSE and beyond, you need to be able to work with surds.

Surds basics

What is a surd?

A surd is the (positive) square root of a number that isn’t a square number.

For example, $\sqrt{3}$ and $\sqrt{24}$ are surds, because neither 3 nor 24 is a square number…
… but $\sqrt{9}$ isn’t a surd because 9 is a square number and so $\sqrt{9}$ can be simplified to just 3.

A surd is also an irrational number, which means it can’t be written as the ratio of two integers – or as a fraction, which is just a ratio written in a slightly different form.

Squaring and square rooting are inverse operations, i.e. they cancel each other out.
Using a square number to illustrate:
$\sqrt{9}^2 = 3^2 = 9$ and similarly, $\sqrt{9^2} = \sqrt{81} = 9$
– it doesn’t matter whether you square first or take the square root first; you get the same result either way.

This works with surds too:
$\sqrt{5}^2 = \sqrt{5^2} = 5$

Combining surds

You can multiply and divide surds:
(Again, we’re using square numbers to illustrate the principle, but it works with non-square numbers too)

$\sqrt{4} \times \sqrt{9} = 2 \times 3 = 6$
and
$6 = \sqrt{36} = \sqrt{4 \times 9}$
so
$\sqrt{4} \times \sqrt{9} = \sqrt {4 \times 9}$

Or, more generally,
$\sqrt{a} \times \sqrt{b} = \sqrt {ab}$

and similarly,
$\sqrt{a} \div \sqrt{b} = \sqrt {\frac{a}{b}}$

BUT you can’t add/subtract the number under the root sign:

$\sqrt{4} + \sqrt{9} = 2 + 3 = 5$
but
$\sqrt{4+9} = \sqrt{13}$ which clearly is NOT the same as 5.

So
$\sqrt{a} + \sqrt{b} \neq \sqrt {a+b}$
and similarly
$\sqrt{a} - \sqrt{b} \neq \sqrt {a-b}$

If you need to multiply or divide a mixture of surds and rational numbers, deal with each type of number separately, e.g.
$4\sqrt{2} \times 5\sqrt{3} = 4 \times 5 \times \sqrt{2 \times 3} = 20\sqrt{6}$

Your turn 1

Simplify where possible:

$\sqrt{7}^2$
$\sqrt{2} \times\sqrt{8}$
$3\sqrt{2} \times 4\sqrt{5}$
$\sqrt{2} + \sqrt{5}$
$\sqrt{56} \div \sqrt{8}$
$\sqrt{11} - \sqrt{7}$

Click here for the answers

Manipulating surds

Simplifying a surd

This means making the number under the square root sign as small as possible. We do this by taking out a square factor – the biggest one we can.

Examples:
$\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}$
$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}$

Just as with cancelling down fractions, it doesn’t matter if you don’t immediately spot the biggest square factor. For example, with $\sqrt{80}$ you might initially only spot that 4 is a factor, giving
$\sqrt{80} = \sqrt{4 \times 20} = \sqrt{4}\sqrt{20} = 2\sqrt{20}$
but 20 also has a factor of 4 so we can take it further:
$=2 \times \sqrt{4 \times5} = 2\sqrt{4}\sqrt{5} = 2 \times 2 \times \sqrt{5} = 4\sqrt{5}$

If you’d spotted straight away that 16 was a factor then it would have got you to the same result more quickly:
$\sqrt{80} = \sqrt{16 \times 5} = \sqrt{16}\sqrt{5} = 4\sqrt{5}$

Adding and subtracting with surds

You can add and subtract quantities of the same surds – think of it as adding like terms in algebra.

For example, $\sqrt{24} + \sqrt{54}$ can be calculated if we can express both values as multiples of the same surd.
We already know that $\sqrt{24} = 2\sqrt{6}$
and the square number 9 is a factor of 54 so we can say
$\sqrt{54} = \sqrt{9 \times 6} = \sqrt{9} \times \sqrt{6} = 3\sqrt{6}$
Now we have both values as multiples of $\sqrt{6}$
so
$\sqrt{24} + \sqrt{54} = 2\sqrt{6} + 3\sqrt{6} = 5\sqrt{6}$

Expanding brackets with surds

This works just as it does with algebra.

Examples:

$\sqrt{2}(\sqrt{2} + 3)$
$= \sqrt{2}^2 + 3\sqrt{2}$
$= 2 + 3\sqrt{2}$

$(\sqrt{2} + 3)(\sqrt{2} + 4)$
$= \sqrt{2}^2 + 4\sqrt{2} + 3\sqrt{2} + 3 \times 4$
$= 2 + 7\sqrt{2} + 12$
$= 14 + 7\sqrt{2}$

What happens when you expand $(\sqrt{2} + 3)(\sqrt{2} - 3)$ ? Try it, then scroll down to see if you got it right.

Solution:
$(\sqrt{2} + 3)(\sqrt{2} - 3)$
$= \sqrt{2}^2 - 3\sqrt{2} + 3\sqrt{2} - 3 \times 3$
$= 2 - 9$
$= -7$

Notice how the surds cancel out so that you’re left with only rational numbers.
In fact it’s a case of “difference of two squares”, where
$(a+b)(a-b) = a^2 - ab + ab - b^2= a^2 - b^2$ .
(You’ll have encountered this before with quadratic expressions such as $x^2 - 9 = (x + 3)(x - 3)$ .)

In this particular example $a = \sqrt{2}$ and $b = 3$ so we end up with
$\sqrt{2}^2 - 3^2 = 2 - 9 = -7$

What would you need to multiply $(3\sqrt{5} - 2)$ by to get an answer without any surds in it? Try it and see if you were right.
(If you do it right then you should end up with an answer of $(3\sqrt{5})^2 - 2^2 = 41$ .)

This pattern will come in handy when we look at rationalising the denominator, which I’ll cover in my next blog post.

Your turn 2

Part A: Simplify:

$\sqrt{48}$
$\sqrt{27}$
$\sqrt{48} + \sqrt{27}$
$\sqrt{75} - \sqrt{27}$

Part B: Expand and simplify as far as possible:

$\sqrt{5}(2\sqrt{5} - 3)$
$(2\sqrt{3} - 1)(\sqrt{3} + 4)$
$(2\sqrt{7} - 5)(2\sqrt{7} + 5)$
$(\sqrt{2} + 3)(3\sqrt{3} - 4)$

Click here for the answers

Now that you can manipulate surds, in the next post we’ll look at using these techniques to rationalise the denominator of a fraction.

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