# Rationalize the denominator - Radicals

## How to rationalize the denominator

When you're working with fractions, you may run into situations where the denominator is messy. What exactly does messy mean? It's when your denominator isn't a whole number and cannot be cancelled off. When we've got, say, a radical in the denominator, you're not done answering the question yet. Your final answer has to have a clean denominator. What can we do to fix this? We can rationalize the denominator.

Do you remember how you first worked with fractions with unequal denominators? Take for example the following:

$\frac{1}{7} + \frac{3}{9}$

You would then go multiply each individual fraction's numerator and denominator with the number the denominator needed. In this case, you'd do:

$(\frac{1}{7} \bullet \frac{9}{9})+(\frac{3}{9} \bullet \frac{7}{7})$

$= \frac{9}{63} + \frac{21}{63}$

$= \frac{30}{63}$

$= \frac{10}{21}$

When we multiply $\frac{1}{7}$ by $\frac{9}{9}$, we're actually multiplying the whole thing by $1$, since $\frac{9}{9}$ equals $1$. This essentially means we haven't changed the original $\frac{1}{7}$, but instead have made it so that it's possible to add it together with $\frac{3}{9}$. This is the concept we'll use when we start rationalizing denominators.

So now that we recall how we added or subtracted fractions that didn't have the same denominators, we will use a similar method to change the denominators when we're rationalizing the denominator. When you've got a radical in your denominator, such as in the case of $\frac{1}{\sqrt{3}}$, we can rationalize this by multiplying the numerator and denominator by $\sqrt{3}$. This means, again, that we're really not changing the original fraction since we're multiplying both the top and bottom by $\sqrt{3}$, essentially meaning we multiply the original fraction by $1$. A radical multiplied by itself gets rid of the radical sign. Meaning our final fraction will become:

$\frac{1*\sqrt{3}}{\sqrt{3}*\sqrt{3}}$

$= \frac{\sqrt{3}}{3}$

You can see how we've got rid of the radical in the denominator. Having a radical in the numerator is acceptable for a final answer, so you've got the final answer here. Make sure that if you can simplify the fraction further, make sure you rationalize the denominator and simplify it in order to get the correct answer. Now you've learned how to rationalize denominators in fractions.

## Example problems

Question 1:

Simplify and rationalize the denominator

$\frac{14\sqrt{20}}{3\sqrt{16}}$

Solution:

$= \frac{14}{3} \sqrt{\frac{20}{16}}$

$= \frac{14}{3} \sqrt{\frac{5}{4}}$

$= \frac{14}{3}\bullet\frac{\sqrt{5}}{\sqrt{4}}$

$= \frac{14}{3}\bullet\frac{\sqrt{5}}{2}$

$= \frac{14\sqrt{5}}{6}$

$= \frac{7\sqrt{5}}{3}$

Question 2:

Simplify and rationalize the denominator

$\frac{\sqrt{7} - 5}{\sqrt{3}}$

Solution:

$= \frac{\sqrt{7} - 5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$= \frac{\sqrt{21} - 5\sqrt{3}}{\sqrt{9}}$

$= \frac{\sqrt{21} - 5\sqrt{3}}{3}$

Question 3:

Simplify and rationalize denominator

$\frac{3\sqrt{6}}{5} + \frac{5\sqrt{3}}{\sqrt{7}}$

Solution:

Find the lowest common denominator of the $2$ fractions

$= \frac{(\sqrt{7})3\sqrt{6}}{(\sqrt{7})5} + \frac{(5)5\sqrt{3}}{(5)\sqrt{7}}$

$= \frac{3\sqrt{42}+25\sqrt{3}}{5\sqrt{7}}$

$= \frac{3\sqrt{42}+25\sqrt{3}}{5\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}$

$= \frac{3\sqrt{294}+25\sqrt{21}}{5(\sqrt{7})(\sqrt{7})}$

$= \frac{3\sqrt{49\bullet 6}+25\sqrt{21}}{35}$

$= \frac{21\sqrt{6}+25\sqrt{21}}{35}$

### Rationalize the denominator

When there is a radical in the denominator, the fraction is not in its simplest form. Therefore, we need to rationalize the denominator to move the root from the denominator/bottom of the fraction to the numerator/ top. In this lesson, we will learn how to simplify radicals by rationalizing the denominator.