Multiply numeric radicals

So far we know that surds only like to combine and split when multiplication and division are involved and not addition and subtraction. This gives us a lot of freedom when multiplying and dividing more complicated surds.

 

Grouping

Say we have $2\sqrt{5}\times7\sqrt{3}$2√5×7√3, which we can rewrite as $2\times\sqrt{5}\times7\times\sqrt{3}$2×√5×7×√3. Straight away we can simplify this by multiplying the numbers together in one group and the surds together in another.

So this becomes

$2\times7\times\sqrt{5}\times\sqrt{3}$2×7×√5×√3

$=$=

$14\times\sqrt{15}$14×√15

This can't be simplified any further so the answer is $14\sqrt{15}$14√15.

 

We can also use grouping for division.

For example $6\sqrt{10}\div3\sqrt{2}$6√10÷​3√2 can be written as the fraction

$6\sqrt{10}\div3\sqrt{2}$6√10÷​3√2	$=$=	$\frac{6}{3}\times\frac{\sqrt{10}}{\sqrt{2}}$63​×√10√2​
	$=$=	$2\times\sqrt{5}$2×√5

This can't be simplified either so our answer is $2\sqrt{5}$2√5.

What have you noticed with both these examples? Yes that's right, we group all the numbers together and the surds in another group and work them out separately, then combine them to get our answer.

 

Using Simplified Surds

Although simplifying surds before attempting the question is not necessary here like it is with addition and subtraction, it can still help us get the job done quicker sometimes. Let's have a look at an example that might give you a surprising solution.

$9\sqrt{28}\div3\sqrt{7}$9√28÷​3√7 can be solved by grouping straight away, but let's try and simplify the surds first.

$\sqrt{28}$√28 can be simplified to $2\sqrt{7}$2√7

$\sqrt{7}$√7 cannot be simplified so our question becomes

$9\times2\sqrt{7}\div3\sqrt{7}$9×2√7÷​3√7	$=$=	$18\sqrt{7}\div3\sqrt{7}$18√7÷​3√7
	$=$=	$6\div1$6÷​1
	$=$=	$6$6	as the $\sqrt{7}$√7 term just cancels out!

This is very interesting as we started out with a problem involving surds that ended in an answer without surds! Can you think of any other examples like this?

 

Practice questions

Question 1

Question 2

Question 3

Question 4