Use and Manipulation of Surds

Use and manipulation of surds

Simplifying expressions that contain surds, including rationalising the denominator. For example: simplifying $\frac{\sqrt{5}}{3+2\sqrt{5}}$ and $\frac{3}{\sqrt{7}-2\sqrt{3}}$.

What is a surd? Here we consider a surd to be an expression that has a root in it [usually a square root] that cannot be simplified to a rational expression. For instance, $3 + 5\sqrt{2}$ is a surd as we cannot simplify it to a rational expression; whereas $3 + 5\sqrt{64}$ is not a surd as we can simplify it to 43. Surds are a mathematical way of expressing numbers exactly and they help get around the impossibility of expressing certain irrational numbers precisely using decimal expansion – that is, for instance, it is impossible to express $\sqrt{2}$ exactly as a decimal as the decimal bit of the number $\sqrt{2}$ is never ending.

Before we look at the sorts of things we might expect you to know for the TMUA/ESAT, we note that there is a convention with square root signs that we adopt in the TMUA/ESAT [it is a standard maths convention] and that is that $\sqrt{a}$ is always positive. So $\sqrt{64}$ is 8 and not -8. If we want to have both 8 and -8, we write $\pm\sqrt{64}$.

What sort of things do we expect you to be able to do with surds in the TMUA/ESAT? Let's look at a few:

Simplifying roots

We expect you to be able to simplify expression such as $\sqrt{50}$ or $\sqrt{40}$ in various ways. You should be comfortable with how the following, and similar, examples work:

$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$

$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10} = 2\sqrt{2 \times 5} = 2\sqrt{2}\sqrt{5}$

Multiplying out expressions that involve surds

The best way to multiply out expression with surds in them is to treat the square roots like $x$ and then simplify at the very final stage. Here is an example:

$(2 + 3\sqrt{5})^2 = 2^2 + 2 \times 2 \times 3\sqrt{5} + 3^2\sqrt{5}\sqrt{5} = 4 + 12\sqrt{5} + 45 = 49 + 12\sqrt{5}$

Compared with

$(2 + 3x)^2 = 2^2 + 2 \times 2 \times 3x + 3^2xx = 4 + 12x + 9x^2$

Factorise expression with surds in them

From our earlier discussion, we can see that going from $(2 + 3\sqrt{5})^2$ to $49 + 12\sqrt{5}$ is quite easy. But it is less easy to start with $49 + 12\sqrt{5}$ and factorise it to get $(2 + 3\sqrt{5})^2$. Nevertheless, you should be able to factorise expressions with surds in when it might be useful. For instance, if you were asked to find the exact value of $\sqrt{49 + 12\sqrt{5}}$ you would need to spot that $49 + 12\sqrt{5} = (2 + 3\sqrt{5})^2$ to get to the answer.

How might you go about factorising expression with surds in them? There are a few "tricks" that might help sometimes – although it is best to practise this yourself and devise your own "tricks" and way of approaching the factorising.

Let's look at $49 + 12\sqrt{5}$ and compare it with $4 + 12x + 9x^2$ from above:

The first thing we notice is that the $12\sqrt{5}$ is the middle term [and this is usually where we start with this sort of question], so this suggests that our original expression must be $(a + b\sqrt{5})^2$. We then notice that the middle term is also written as $2ab\sqrt{5}$. This means we must have $ab = 6$ and $a^2 + 5b^2 = 49$. Now all we need to do is substitute pairs of numbers that multiply to give 6 into the equation $a^2 + 5b^2 = 49$ until we find some that work. It doesn't take long to find $a = 2$ and $b = 3$ [you can start with $b$ and work though $b = 1, 2, 3, 6$ until you find which one works].

Exercise

Start with some random surd expression and square them and simplify them. Then look at the expressions, maybe after some days, and see if you can factorise them back to their original form.

Exercise

What is $\sqrt{49 - 12\sqrt{5}}$? Be very careful – you might get the wrong answer! [The answer is not $2 - 3\sqrt{5}$. Why not?]

Rationalising the denominator

This is probably the most common thing you will meet with surds in standard exams. It is about "moving" the surd expression in a fraction from the bottom of the fraction the top. That is, we want to find another expression that has the same value as the original, but which has a surd in the numerator [top] of the fraction and no surds in the denominator [bottom]. [As you will be aware, the top and the bottom of fractions have different roles. The bottom of a fraction [denominator] tells us [denominates] what sort of fraction it is [is it halves, or thirds, or quarters, etc]. The top of a fraction [numerator] enumerates [i.e. tells us how much] of the fraction we have. This seems obvious but it is often misunderstood when fractions are first met – for instance, it is common to see $\frac{1}{2} + \frac{2}{3} = \frac{1+2}{2+3}$. Have a think about how you perform fraction division: usually the rule used is "flip the second fraction and multiply". Can you explain why this method works [hint, make the denominators the same then think about what the denominator tells you]? And is the following true [and, if so, why]: $\frac{a}{b} + \frac{c}{d} = \frac{a-c}{b+d}$?]

In simple cases, this is straightforward. For instance, how do we rationalise the denominator for $\frac{1}{\sqrt{5}}$? The answer is we multiply the expression by 1 but we write 1 in an unusual manner: we use $1 = \frac{\sqrt{5}}{\sqrt{5}}$ and this gives:

$\frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \times 1 = \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{\sqrt{5}}{5}$

In practice, you should be able to convert from $\frac{1}{\sqrt{5}}$ to $\frac{\sqrt{5}}{5}$ almost without thinking about it: $\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a}$.

What about more complicated expressions such as $\frac{3}{2+\sqrt{5}}$? We use the same idea – that is, multiplying by 1, but by writing 1 in a specific way. We also use the standard difference of two squares formula – this is a formula that crops up everywhere and so you should always be on the lookout for it just in case it might be useful, but also be aware that using it can sometimes lead you down the wrong path [doing mathematics is sometimes a bit like playing chess: when you have an idea, you also need to anticipate what the consequences of applying the idea will be before devoting time to pursuing it. Sometimes, ideas that seem perfect at the start will lead you down the wrong path. Mathematics is also a bit like chess in that the more you practise your mathematics, the better you will become at it].

Recall the difference of two square formula: $a^2 - b^2 = (a + b)(a - b)$.

Let's look at using all this to rewrite $\frac{3}{2+\sqrt{5}}$. We need to multiply by 1 written in an appropriate manner. We start by using the difference of two squares and noticing:

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

And also

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

Looking carefully at both of these, we decide to use 1 written as $\frac{\sqrt{5}-2}{\sqrt{5}-2}$ [we could use $\frac{2-\sqrt{5}}{2-\sqrt{5}}$ but it gives us a negative value and that involves a teeny bit more work]. So, we now rewrite our expression $\frac{3}{2+\sqrt{5}}$ as follows:

$\frac{3}{2+\sqrt{5}} = \frac{3}{2+\sqrt{5}} \times 1 = \frac{3}{2+\sqrt{5}} \times \frac{\sqrt{5}-2}{\sqrt{5}-2} = \frac{3\sqrt{5}-6}{5-4} = \frac{3\sqrt{5}-6}{1} = 3\sqrt{5}-6$

Of course, you should be able to rationalise denominators much more quickly than we have done above – we just took our time to explain each step carefully.

As an aside: it is very easy when rationalising denominators to be careless and write $\sqrt{5} \times \sqrt{5} = 25$. Make sure you don't!

Exercise

What is the factorisation of $a^3 - b^3$ and of $a^3 + b^3$? [these factorisations are both useful and you ought to know them]

Can you use these to help you rationalise the denominators in:

$\frac{7}{\sqrt[3]{36} + \sqrt[3]{6} + 1}$

$\frac{x - 27}{\sqrt[3]{x} - 3}$