Rationalizing and Roots of Numerical Fractions

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Lesson Priority: VIP Knowledge

 
Objectives
  • Know the rule regarding how to take the square root of a fraction, both forward and backwards
  • Learn methods for working with fractions where the numerator and/or the denominator has a root in it
  • Understand why we must ultimately simplify to an answer that does not have a square root in the denominator
Lesson Description

Most importantly in this lesson, we want to understand how to proceed when we're asked to take the root of a fraction, or when part of a fraction has a root in it. This lesson shows us how to deal with any situation of this nature.

 
Practice Problems

Practice problems and worksheet coming soon!

 
You Should Know
This lesson covers integer square roots only.For rationalizing square roots of variable expressions, jump ahead a few lessons to this one on rationalizing » with variables.If you are looking for lessons on advanced roots of higher degrees, such as cube roots, you will find them in Algebra Two ».

The Division Problem

Rules about how square roots work on fractions are closely related to rules about dividing by square roots, since fractions are merely a different way to express division. Learn the content of this lesson with that in mind.The goal is a general one that we see in many Algebra topics - to simplify. The two types of "non-simplified" expressions that we will learn how to remedy in this lesson are unsimplified fractions and square roots in the denominator, or a combination of the two. In Algebra, we are not allowed to leave any root expression in the denominator of a fraction.

The Square Root Fraction Rule

The master rule that we'll use when it comes to square roots of entire fractions is a fairly simple one that is a direct result of the root product property ».
Define: Square Roots of FractionsFor any fraction,$$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$
 
Example 1Simplify $\displaystyle \sqrt{\frac{16}{25}}$$\blacktriangleright$ According to the rule, the square root of this fraction is equal to the square root of the numerator over the square root of the denominator.$$\sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}}$$$$=\frac{4}{5}$$
 
We will often need to use this rule backwards as well, meaning if we start with a fraction that has a square root in each its numerator and denominator, we may end up writing$$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$
 
Example 2Simplify$$\frac{\sqrt{32}}{\sqrt{50}}$$$\blacktriangleright$ $32$ and $50$ have a common factor of $2$, but before we can properly show how their ratio reduces, we need this to be written as a ratio of integers, not a ratio of square roots.We'll use the rule backwards:$$\frac{\sqrt{32}}{\sqrt{50}} = \sqrt{\frac{32}{50}}$$Now, underneath the root symbol we have a fraction that can be simplified.$$\sqrt{\frac{32}{50}} = \sqrt{\frac{16}{25}}$$$$=\frac{4}{5}$$
 
While Example 2 was just a small modification of Example 1, notice how the mechanics we used to answer Example 1 wouldn't have exactly worked for Example 2. We will categorize the possible problems we'll need to know how to simplify into three major cases. First, however, we need to know how to rationalize fractions that don't happen to contain perfect square integers.

Rationalizing a Fraction

When a fraction does not have an integer denominator, we are not allowed to say that the fraction is fully simplified. Since we nearly always are asked to turn in answers that are simplified, we need to know how to manipulate the fraction into a form we like, without changing its value. As we've saw recently when working with imaginary numbers », the way to do that is to multiply by a form of $1$.
 
Example 3Simplify$$\frac{4}{\sqrt{7}}$$$\blacktriangleright$ The numerator and denominator are each simplified on their own, since $4$ is an integer and $\sqrt{7}$ cannot reduce as a simpler square root. However, the fraction is not considered simplified because we aren't allowed to leave roots in the denominator.Remedy this by multiplying by $\sqrt{7} / \sqrt{7}$, which is a form of $1$.$$\frac{4}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}}$$$$=\frac{4\sqrt{7}}{7}$$This is the simplified answer. Square roots are present in the numerator but that's ok - it's the denominator that we aren't allowed to leave roots in.
I Used To Know That!
When multiplying fractions together, multiply across. The numerators are multiplied together to get the final numerator, and the denominators are multiplied together to get the final denominator.
You Should Know
Rationalizing a fraction is a skill you'll need to use often enough at random times in other future topics, so make sure you not only understand how to do it, but remember what it is for future use!

Simplifying Generally with Fractions and Roots

Simplifying fractions with root stuff is challenging because we are constantly using the rule backward and forward, and without guidance and a little practice, it can be difficult to understand when we need each. As you practice you may feel like there are a whole bunch of problem varieties to master, but in reality there are three types of problems and therefore three approaches to simplifying based on the situation we encounter. Let's take a look at the three cases with accompanying examples for each.
Pro Tip
No matter what type of problem you work on, make sure you simplify fractions when you can before you take square roots. Then, once you take square roots, you should also try to simplify roots » on their own if possible (e.g. $\sqrt{12} = 2\sqrt{3}$).
Warning!
When asked to simplify, you must check that square roots are also fully reduced on their own. The answer$$\frac{\sqrt{60}}{3}$$is rationalized, but is not fully simplified because $\sqrt{60}$ can be reduced to a smaller root. $\sqrt{60} = 2\sqrt{15}$, and so the simplified answer would be$$\frac{2\sqrt{15}}{3}$$
Case 1 - Root of the Entire FractionSimilar to Example 1, when presented with the root of an entire fraction, we usually end up "separating" the root by applying our fraction root rule and taking the square root of each top and bottom on their own. However, you absolutely must check to see if you can simplify the fraction before separating. If you can and you don't, you're going to either get the problem wrong or create double the work for yourself.In short, when presented with the root of a whole fraction, first try to simplify, and then separate by rooting the top and bottom each. We'll often have to quickly rationalize the denominator at the end. Rationalizing is not required only when the denominator turned into an integer (because it was a perfect square).
 
Example 4$$\sqrt{\frac{9}{18}}$$$\blacktriangleright$ It is tempting to square root each top and bottom right away but that will make the problem worse later - always simplify whole fractions if possible.$$\sqrt{\frac{9}{18}} = \sqrt{\frac{1}{2}}$$$$=\frac{1}{\sqrt{2}}$$Finally, rationalize the denominator.$$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$$$$= \frac{\sqrt{2}}{2}$$
 
Example 5Simplify.$$\sqrt{\frac{3}{4}}$$$\blacktriangleright$ $3$ and $4$ do not reduce with one another, so let's now separate the roots on top and bottom:$$\frac{\sqrt{3}}{\sqrt{4}}$$$$=\frac{\sqrt{3}}{2}$$This problem is actually finished - no more steps are required because the denominator was a perfect square integer. Integer denominators do not need to be rationalized.
 
Example 6Simplify.$$\sqrt{\frac{11}{20}}$$$\blacktriangleright$ These numbers do not have a common factor to cancel, so proceed with separating the square roots.$$\sqrt{\frac{11}{20}} = \frac{\sqrt{11}}{\sqrt{20}}$$Before rationalizing the denominator, it will be easier if we recognize that $\sqrt{20}$ on its own simplifies to a smaller root. Let's re-write:$$\frac{\sqrt{11}}{2\sqrt{5}}$$Now rationalize.$$\frac{\sqrt{11}}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$$$$= \frac{\sqrt{55}}{10}$$This is the final answer because $\sqrt{55}$ does not simplify to a smaller radical, and the denominator is now an integer.
 
Case 2 - Rationalize the DenominatorIf you are given a fraction in the form$$\frac{a}{\sqrt{b}}$$you should immediately rationalize the denominator. Be prepared for further simplification, if $a$ and $b$ have any common factors. However, if $\sqrt{b}$ can be simplified on its own (e.g. $\sqrt{12} = 2\sqrt{3}$), you should do that before rationalizing.
 
Example 7Simplify.$$\frac{9}{\sqrt{6}}$$$\blacktriangleright$ $\sqrt{6}$ cannot be simplified to a smaller radical, so we'll go right to work rationalizing the denominator.$$\frac{9}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}}$$$$=\frac{9\sqrt{6}}{6}$$The $9$ and $6$ coefficients have a common factor and therefore can reduce, giving us our final answer:$$=\frac{3\sqrt{6}}{2}$$
 
Example 8Simplify.$$\frac{10}{\sqrt{24}}$$$\blacktriangleright$ Before we rationalize, let's reduce $\sqrt{24}$:$$\frac{10}{2\sqrt{6}}$$Before we proceed, we can do some simplifying with the coefficients $10$ and $2$.$$=\frac{5}{\sqrt{6}}$$Now rationalize.$$\frac{5}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}}$$$$=\frac{5\sqrt{6}}{6}$$Since the remaining coefficients $5$ and $6$ do not have any common factors, the result is fully simplified.
 
Example 9Simplify.$$\frac{40}{\sqrt{20}}$$$\blacktriangleright$ Once again let's simplify the denominator before rationalizing:$$\frac{40}{\sqrt{20}} = \frac{40}{2\sqrt{5}} = \frac{20}{\sqrt{5}}$$Now rationalize:$$\frac{20}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$$$$=\frac{20\sqrt{5}}{5}$$Since $20$ and $5$ reduce with one another, let's cancel their common factor to finish the problem.$$= \frac{4 \sqrt{5}}{1} = 4 \sqrt{5}$$
 
Case 3 - A Ratio of RootsIf you are given a fraction with two separate roots, which will look something like$$\frac{\sqrt{a}}{\sqrt{b}}$$then the first thing you should do is examine $a$ and $b$. If they have any common factor, you should re-write it as one big square root around the fraction, effectively turning it into Case 1. We can do this by applying the fraction root rule backward:$$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$Then, you can cancel the common factor and proceed as you do in Case 1.If $a$ and $b$ do not have anything in common, leave them separate and rationalize the denominator.
 
Example 10Simplify.$$\frac{\sqrt{3}}{\sqrt{10}}$$$\blacktriangleright$ These numbers have nothing in common, so proceed with rationalizing the denominator.$$\frac{\sqrt{3}}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}$$$$=\frac{\sqrt{30}}{10}$$Nothing else reduces, so this is the final answer.
 
Example 11Simplify.$$\frac{\sqrt{8}}{\sqrt{15}}$$$\blacktriangleright$ $8$ and $15$ do not have a common factor, so let's leave them as separate roots. We can still recognize that $\sqrt{8}$ reduces on its own:$$\frac{2\sqrt{2}}{\sqrt{15}}$$Now rationalize the denominator.$$\frac{2\sqrt{2}}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}}$$$$= \frac{2\sqrt{30}}{15}$$The coefficients $2$ and $15$ have nothing in common with one another, so we're done.
 
Example 12Simplify.$$\frac{\sqrt{6}}{\sqrt{18}}$$$\blacktriangleright$ Since $6$ and $18$ have a common factor, we should re-write this as a single fraction and reduce.$$\sqrt{\frac{6}{18}}$$$$=\sqrt{\frac{1}{3}}$$Now separate these into separate roots in each the numerator and denominator, rationalize, and finish.$$\frac{\sqrt{1}}{\sqrt{3}}$$$$\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$$$$=\frac{\sqrt{3}}{3}$$
 

Mr. Math Makes It Mean

While the methods above will always work, on occasion, these fail-safe three case procedures are not efficient. Particularly, if the denominator is the square root of a perfect square, it is actually more work to procede as normal. In this case, you should separate the root and take the square root.Again - this is just about being efficient. The three methods will always work.
 
Example 13Simplify$$\sqrt{\frac{10}{16}}$$$\blacktriangleright$ Because $\sqrt{16}$ is $4$, this is a problem where we should immediately separate and take roots:$$\sqrt{\frac{10}{16}} = \frac{\sqrt{10}}{\sqrt{16}}$$$$=\frac{\sqrt{10}}{4}$$This is the final answer in simplest form.
 
Let's quickly re-do Example 13 using the method that we would have used in the usual three-case method. It will give us the same answer but you'll see why we end up with more work.
 
Example 13 (Re-done)Simplify$$\sqrt{\frac{10}{16}}$$$\blacktriangleright$ According to the cases, this is case 1, so we would reduce the fraction before rooting each piece:$$\sqrt{\frac{10}{16}} = \sqrt{\frac{5}{8}}$$Then we separate the roots:$$\frac{\sqrt{5}}{\sqrt{8}}$$Before we rationalize, simplify $\sqrt{8}$:$$\frac{\sqrt{5}}{2\sqrt{2}}$$Finally, we need to rationalize the denominator, since we aren't allowed to leave $\sqrt{2}$ in the denominator.$$\frac{\sqrt{5}}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$$$$= \frac{\sqrt{10}}{4}$$This is the same answer we got with the shortcut. The extra steps happen because the factor that we used to originally cancel $10/16$ it will be ultimately needed for the denominator to be rationalized, so we ended up having to put it back later.
 
Remember!
Most questions are not so tedious, and this is an optional shortcut that is really about efficiency. You can always follow the usual three methods, even for these annoying problems. Both methods are completely correct and result in the same answer.

Put It To The Test

This is a skill that takes practice overall - don't be discouraged if you need to hammer out a few handfuls of these before you feel like you're ready for a quiz! It will also be a huge help to have this mastered before moving on to the next lesson on taking roots of variable expressions ».Simplify the expression in each problem completely.
 
Example 14$$\sqrt{\frac{25}{81}}$$
Show solution
$\blacktriangleright$ These numbers have nothing in common - apply the root to each numerator and denominator.$$\frac{\sqrt{25}}{\sqrt{81}} = \frac{5}{9}$$Since both integers were perfect squares, no rationalization was needed and no roots ended up in the answer.
 
Example 15$$\frac{\sqrt{21}}{\sqrt{7}}$$
Show solution
$\blacktriangleright$ These numbers do have a common factor, so let's use the fraction root rule backward and make this one fraction under one square root, and then reduce.$$\frac{\sqrt{21}}{\sqrt{7}} = \sqrt{\frac{21}{7}}$$$$=\sqrt{3}$$
 
Example 16$$\sqrt{\frac{80}{25}}$$
Show solution
$\blacktriangleright$ Reduce the common factor of $5$ before proceeding.$$\sqrt{\frac{80}{25}} = \sqrt{\frac{16}{5}}$$With nothing in common, apply the root to each numerator and denominator, and then rationalize.$$\sqrt{\frac{16}{5}} = \frac{\sqrt{16}}{\sqrt{5}}$$$$\frac{4}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$$$$\frac{4\sqrt{5}}{5}$$Note that in accordance with our discussion on efficiency, we could have also taken roots before reducing, since $25$ is a perfect square denominator. Both processes are correct.
 
Example 17$$\sqrt{\frac{36}{15}}$$
Show solution
$\blacktriangleright$ A common factor is present, so reduce, and then root each piece and rationalize.$$\sqrt{\frac{36}{15}} = \sqrt{\frac{12}{5}} = \frac{\sqrt{12}}{\sqrt{5}}$$$$\frac{2\sqrt{3}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{15}}{5}$$
 
Example 18$$\frac{\sqrt{72}}{\sqrt{2}}$$
Show solution
$\blacktriangleright$$$\sqrt{\frac{72}{2}} = \sqrt{36} = 6$$
 
Example 19$$\frac{18}{\sqrt{20}}$$
Show solution
$\blacktriangleright$$$\frac{18}{\sqrt{20}} = \frac{18}{2\sqrt{5}} = \frac{9}{\sqrt{5}}$$$$\frac{9}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{9\sqrt{5}}{5}$$
 
Example 20$$\sqrt{\frac{45}{6}}$$
Show solution
$\blacktriangleright$$$\sqrt{\frac{45}{6}} = \sqrt{\frac{15}{2}} = \frac{\sqrt{15}}{\sqrt{2}}$$$$\frac{\sqrt{15}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$$$$=\frac{\sqrt{30}}{2}$$
 
Example 21$$\frac{\sqrt{300}}{\sqrt{4}}$$
Show solution
$\blacktriangleright$$$\sqrt{\frac{300}{4}} = \sqrt{75}$$$$=5\sqrt{3}$$
 
Lesson Takeaways
  • Understand what the word simplified means for fractions that involve roots
  • Apply the idea of rationalization (having an integer denominator only) to square roots
  • Know the way to deal with the handful of ways that we'll be presented with numeric fraction root expressions to simplify

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