EXT: Rationalise the denominator

Surds on Top and Surds on Bottom

What happens when you have a fraction with surds in it? Well it's a little difficult to work with, especially if there are surds in the denominator.

$\frac{2\sqrt{3}}{5}$2√35​ can be rewritten as $\frac{2}{5}\sqrt{3}$25​√3, so it just becomes a simple radical with a fraction in front of it.

But we're in trouble when we have something like $\frac{8}{3\sqrt{7}}$83√7​, how do we work with that?? 

You've probably realised by now that mathematicians are quite lazy and like to simplify things as much as they can, so they've devised a way to rewrite these complicated fractions without the surds on the bottom.

This is called rationalising the denominator, and it does not change the value of the fraction. 

 

Using Perfect Squares

We know that when we square a radical, the answer is always going to be rational and without surds! This is the key we need to rationalise fractions such as $\frac{8}{3\sqrt{7}}$83√7​.

What do you think will happen if we multiply the denominator here by $\sqrt{7}$√7? Well, the square root sign will disappear right? However, we need the fraction to still have the same value, so let's try multiplying the fraction by $\frac{\sqrt{7}}{\sqrt{7}}=1$√7√7​=1, which will not change the fraction at all!

$\frac{8}{3\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{8\times\sqrt{7}}{3\times\sqrt{7}\times\sqrt{7}}$83√7​×√7√7​=8×√73×√7×√7​

which simplifies down to

$\frac{8\sqrt{7}}{3\left(\sqrt{7}\right)^2}=\frac{8\sqrt{7}}{3\times7}$8√73(√7)2​=8√73×7​

This answer can again be finally simplified down to $\frac{8\sqrt{7}}{21}$8√721​. 

The fraction now looks completely different! But try putting it in your calculator, do you get the same value as $\frac{8}{3\sqrt{7}}$83√7​?

So now we know one way of rationalising the denominator is to multiply top and bottom by the radical in the denominator.

 

Using differences of two squares

The above technique is great for when there is only one term in the denominator, but it doesn't really work when we have more complicated expressions.

Let's take a look at an example: $\frac{5}{\sqrt{6}-1}$5√6−1​.

If we use the squares technique and multiply the bottom by $\sqrt{6}$√6, it'll become $6-\sqrt{6}$6−√6, which has not gotten rid of the radical at all!

Let's see what happens when we multiply$\sqrt{6}-1$√6−1 by $\sqrt{6}+1$√6+1.

Does the expression $\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)$(√6−1)(√6+1) look similar to you? Of course, it's a difference of two squares!

Then the answer is a simple case of $\left(\sqrt{6}\right)^2-1^2$(√6)2−12 = $6-1=5$6−1=5.

Well what do you know, this answer doesn't have any surds!

Let's use this to transform the above fraction.

$\frac{5}{\sqrt{6}-1}\times\frac{\sqrt{6}+1}{\sqrt{6}+1}$5√6−1​×√6+1√6+1​	$=$=	$\frac{5\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}$5(√6+1)(√6−1)(√6+1)​
	$=$=	$\frac{5\sqrt{6}+5}{5}$5√6+55​

We can go even further and divide top and bottom by $5$5, which gives us the final simplified answer $\frac{\sqrt{6}+1}{1}=\sqrt{6}+1$√6+11​=√6+1.

 

Let's have a look at these worked examples.

Question 1

Question 2

Question 3

Question 4