1.07 Operations with radical expressions

Multiplying and dividing radicals

So far we know that radicals 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 radicals.

 

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. Using the commutative property, we can also write this expression as $2\times7\times\sqrt{5}\times\sqrt{3}$2×7×√5×√3. We can simplify this by multiplying the numbers together in one group and the radicals together in another.

So this becomes

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

$=$=

$14\times\sqrt{15}$14×√15

$15$15 does not have any perfect square factors other than $1$1, so this can't be simplified any further so the answer is $14\sqrt{15}$14√15.

Worked example

Question 1

Use the grouping concept for division to simplify $6\sqrt{10}\div3\sqrt{2}$6√10÷​3√2.

Think: We can group and then use the property that $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$√a√b​=√ab​ to simplify.

Do:

$6\sqrt{10}\div3\sqrt{2}$6√10÷​3√2	$=$=	$\frac{6}{3}\times\frac{\sqrt{10}}{\sqrt{2}}$63​×√10√2​
	$=$=	$\frac{6}{3}\times\sqrt{\frac{10}{2}}$63​×√102​
	$=$=	$2\times\sqrt{5}$2×√5

Reflect: $5$5 does not have any factors other than $1$1 that are a perfect square, so this can't be simplified any further, our answer is $2\sqrt{5}$2√5.

 

Using simplified radicals in multiplication and division

Although simplifying radicals before starting the question is not necessary with multiplication and division 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.

Worked example

Question 2

Simplify $9\sqrt{28}\div3\sqrt{7}$9√28÷​3√7.

Think: We can simplify by grouping and dividing $\sqrt{28}$√28 by $\sqrt{7}$√7, however, let's try simplifying the radicals first to make the values smaller. 

Do: 

$\sqrt{28}$√28 can be simplified to $\sqrt{4}\times\sqrt{7}=2\sqrt{7}$√4×√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
	$=$=	$\frac{18}{3}\times\frac{\sqrt{7}}{\sqrt{7}}$183​×√7√7​
	$=$=	$6\times1$6×1
	$=$=	$6$6

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

Practice questions

Question 3

Question 4

 

Adding and subtracting radicals

We looked at simplifying radicals which related to multiplication and division, but what about addition and subtraction? Let's explore addition and subtraction of radicals.

Exploration

Let's look at $\sqrt{16}+\sqrt{9}$√16+√9.

$16$16 and $9$9 are perfect squares, so we can simplify $\sqrt{16}+\sqrt{9}$√16+√9 to $4+3=7$4+3=7.

Let's look at whether $\sqrt{16}+\sqrt{9}$√16+√9 is the same as $\sqrt{16+9}$√16+9

$\sqrt{16+9}$√16+9 = $\sqrt{25}$√25 = $5$5

So we can see that, in general, $\sqrt{a}+\sqrt{b}$√a+√b $\ne$≠ $\sqrt{a+b}$√a+b, and similarly $\sqrt{a}-\sqrt{b}$√a−√b $\ne$≠ $\sqrt{a-b}$√a−b.

 

So how can we add and subtract radicals? Well unfortunately, if the radicands, $a$a and $b$b, are different values, we can not simplify the expression $\sqrt{a}+\sqrt{b}$√a+√b. But if they were the same number, then $\sqrt{a}+\sqrt{a}$√a+√a = $2\times\sqrt{a}$2×√a = $2\sqrt{a}$2√a just like collecting like terms in algebra. We can summarize this as:

$c\sqrt{a}+d\sqrt{a}$c√a+d√a = $\left(c+d\right)\sqrt{a}$(c+d)√a

$c\sqrt{a}-d\sqrt{a}$c√a−d√a = $\left(c-d\right)\sqrt{a}$(c−d)√a

 

Using simplified radicals

Sometimes we are asked to add and subtract radicals that have different radicands (arguments). In this case, we can try to simplify one of the radicals so that we have the same radicands.

Worked example

Question 5

Simplify $\sqrt{12}-\sqrt{3}$√12−√3

Think: $\sqrt{12}$√12 and $\sqrt{3}$√3 do not have the same radicand, so can't be subtracted as they are. However, we can simplify $\sqrt{12}$√12 using our technique for simplifying radicals.

Do: Start by simplifying $\sqrt{12}$√12 and then combine like terms.

$\sqrt{12}-\sqrt{3}$√12−√3	$=$=	$\sqrt{4\times3}-\sqrt{3}$√4×3−√3
	$=$=	$\sqrt{4}\sqrt{3}-\sqrt{3}$√4√3−√3
	$=$=	$2\sqrt{3}-\sqrt{3}$2√3−√3
	$=$=	$\sqrt{3}$√3

 

Practice questions

Question 6

Question 7

Question 8