Fractions and Rationalizing with n-th Roots

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Objectives
• Learn how to take the n-th root of a fraction
• Learn how to rationalize simple n-th root denominators
• Get tips for special complex n-th root denominators
Lesson Description

This lesson focuses on applying n-th roots to fractions and rationalizing when the denominator is an n-th root, by using similar mechanics and ideas as the ones we saw in our study of applying square roots to fractions in Algebra One.

Practice Problems

Practice problems and worksheet coming soon!

Applying n-th Roots to Fractions

Just like we saw with square roots », it makes sense to discuss fractions and rationalizing together, since we think of rationalizing as something that must be "done" to fractions.Taking the $n$-th root of a fraction is not different than square roots - you must apply the root to both the numerator and the denominator.

Example 1Simplify $\displaystyle \sqrt[3]{\frac{27}{64}}$.$\blacktriangleright$ Apply the operation of "taking the cube root" to each the numerator and denominator.$$\longrightarrow \frac{\sqrt[3]{27}}{\sqrt[3]{64}}$$$$= \frac{3}{4} \;\;\; \blacksquare$$

However, when things don't work out to integers, that's when we have to understand rationalizing.
I Used To Know That!
This lesson focuses on $n$-th roots, building off of what we learned in the past about rationalizing square root denominators. »If you forget what rationalizing is all about, go back and ground yourself with the core concept before we add on to it here.

Next Level Rationalizing

The core idea behind rationalizing involves multiplying a radical denominator by another radical such that we ultimately get an integer denominator. For example, as we saw in Algebra One with square roots:$$\frac{2}{\sqrt{5}} \longrightarrow \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \longrightarrow \frac{2\sqrt{5}}{5}$$The idea is the same with n-th roots, but we must pay attention to the degree of the root, as it will tell us exactly which radical to multiply by, and how many.Remember from the definition of $n$-th roots that it takes $n$ identical $n$-th roots multiplied together to obtain a result without radicals. For example, it takes two square roots as we saw above. Here are examples with cube roots (a.k.a. third roots) and fourth roots:$$\sqrt[3]{2} \cdot \sqrt[3]{2} \cdot \sqrt[3]{2} = 2$$$$\sqrt[4]{7} \cdot \sqrt[4]{7}$$$$\cdot \sqrt[4]{7} \cdot \sqrt[4]{7}= 7$$Another way to look at this is, when you have the $n$ degree root of an object to the $n$ power, your expression simplifies to itself because they are inverse operations.This means, in order to rationalize a standalone $n$-th root expression denominator, repetitively multiply by the $n$-th root of the object enough times until you have $n$ of them.

Example 2Simplify $\displaystyle \frac{6}{\sqrt[3]{4}}$$\blacktriangleright To rationalize a third root, we need three total. One root is present in the denominator, so we need two more.$$\frac{6}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}}= \frac{6 \sqrt[3]{4^2}}{\sqrt[3]{4^3}}= \frac{6 \sqrt[3]{16}}{4}$$Note that by what we know about n-th root arithmetic and the root product property » we are able to multiply "under" the radical. The denominator is 4 because the third root of 4 to the third power is 4 (recall that roots and exponents are inverse operations).Finally, be sure to check for any final simplification in your answer. In this case there is a GCF of 2 in the result.$$\frac{6 \sqrt[3]{16}}{4} = frac{3 \sqrt[3]{16}}{2} \;\;\; \blacksquare$$Because roots are multiplicative underneath the radial sign, another way to say that we need n number of n-th roots is to say, we need the object we are working with to be raised to the n-th power. This can be helpful when the problem starts with something that is already raised to a power. Example 3Simplify \displaystyle \frac{1}{\sqrt[5]{x^2}}$$\blacktriangleright Remember that in order to take the fifth root of a number or variable, we need to have an object that is raised to the fifth power (or any multiple of$5$). Since$x^2$is the object that we want to take the root of, and it is already raised to the second power, we only need three more.$$\frac{1}{\sqrt[5]{x^2}} \cdot \frac{\sqrt[5]{x^3}}{\sqrt[5]{x^3}}$$$$= \frac{\sqrt[5]{x^3}}{\sqrt[5]{x^5}}$$$$= \frac{\sqrt[5]{x^3}}{x}$$ Multiple Variable Expressions Rationalizing expressions that have multiple variables means that you have to pay attention to what each variable needs. The idea is no different - we want to get to a place where each variable under the$n$-th root is raised to the$n$-th power. Example 4Simplify$\displaystyle \frac{2ab^2}{\sqrt[4]{a^3 b c^2}}$$\blacktriangleright In the denominator, we need to get to a fourth root expression that has all fourth powers in it. This will guide our choice for what to multiply by.$$\frac{2ab^2}{\sqrt[4]{a^3 b c^2}} \cdot \frac{\sqrt[4]{ab^3 c^2}}{\sqrt[4]{ab^3 c^2}} = \frac{2ab^2 \sqrt[4]{ab^3 c^2}}{\sqrt[4]{a^4 b^4 c^4}} = \frac{2ab^2 \sqrt[4]{ab^3 c^2}}{abc} \;\;\; \blacksquare$$Mr. Math Makes It Mean Beside the ever-present need to check your answers for further simplification, teachers can make this intricate topic even more challenging by introducing mixed root degrees.Mixed n-th Root DegreesMixed root degrees in a problem can only be properly simplified by using fraction exponents ». Some teachers don't bother trying to make problems that involve mixed roots, not only because of the need to use fraction exponents but also because it's just not commonly useful in algebra. But if you encounter this on your homework, it's probably fair game for a test. Example 5\displaystyle \frac{\sqrt[3]{4}}{\sqrt[4]{5}}$$\blacktriangleright$All we really care about is rationalizing the denominator, so we proceed with the multiplier we need.$$\frac{\sqrt[3]{4}}{\sqrt[4]{5}} \cdot \frac{\sqrt[4]{5^3}}{\sqrt[4]{5^3}}$$$$= \frac{\sqrt[3]{4} \cdot \sqrt[4]{5^3}}{\sqrt[4]{5^4}}$$$$= \frac{\sqrt[3]{4} \cdot \sqrt[4]{5^3}}{5}$$In order to finish simplifying properly, we need to combine the expressions in the numerator into one root. Re-write this with fraction exponents.$$\longrightarrow \frac{4^{1/3} \cdot 5^{3/4}}{5}$$Now re-write the exponents so that they have the same denominator:$$\frac{4^{4/12} \cdot 5^{9/12}}{5}$$$$\frac{\sqrt[12]{4^4 \cdot 5^9}}{5} \;\;\; \blacksquare$$This is not commonly asked for because, as you can see, you usually get funky numbers. Unusual Suspects Complicated n-th Root DenominatorsWhile it is extremely uncommon,$n$-th degree mixed expressions can be rationalized if they happen to be in a familiar form, such as the difference of squares or sum/difference of cubes.Many teachers won't get into this kind of problem because it's complicated and not usually useful, but others use it as a bonus or challenge question in advanced level courses. Example 6$\displaystyle \frac{1}{1 + \sqrt[3]{a}}$$\blacktriangleright The denominator is part of a sum of cubes. Recall that$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$Therefore:$$\frac{1}{1 + \sqrt[3]{a}} \cdot \frac{1 - \sqrt[3]{a} + \sqrt[3]{a^2}}{1 - \sqrt[3]{a} + \sqrt[3]{a^2}} = \frac{1 - \sqrt[3]{a} + \sqrt[3]{a^2}}{1 + a}$$This might not look like it's "better" than what we started with, but it is rationalized. Again, not a common test question except for intentionally difficult questions. Example 7\displaystyle \frac{1}{\sqrt[4]{a} - \sqrt[4]{b}}$$\blacktriangleright$We can create a difference of squares by multiplying by the conjugate. Recall that, generally,$$(x+y)(x-y) = x^2 - y^2$$Therefore:$$\frac{1}{\sqrt[4]{a} - \sqrt[4]{b}} \cdot \frac{\sqrt[4]{a} + \sqrt[4]{b}}{\sqrt[4]{a} + \sqrt[4]{b}}$$$$= \frac{\sqrt[4]{a} + \sqrt[4]{b}}{\sqrt[4]{a^2} - \sqrt[4]{b^2}$$Now, we can use this trick a second time to end up with a denominator without roots.$$= \frac{\sqrt[4]{a} + \sqrt[4]{b}}{\sqrt[4]{a^2} - \sqrt[4]{b^2} \cdot \frac{\sqrt[4]{a^2} + \sqrt[4]{b^2}}{\sqrt[4]{a^2} + \sqrt[4]{b^2}}$$$$= \frac{\left(\sqrt[4]{a} + \sqrt[4]{b}\right) \left( \sqrt[4]{a^2} + \sqrt[4]{b^2} \right)}{a-b}$$The denominator no longer contains a root, so the objective is achieved. The numerator is unpleasant but the fraction is rationalized.$\blacksquare$Put It To The Test Simplify each expression in the following examples. Example 8$\displaystyle \frac{3}{\sqrt[4]{2}}$Show solution$\blacktriangleright$Multiply by enough "extra" fourth roots of two, so that we end up with four fourth roots.$$\frac{3}{\sqrt[4]{2}} \cdot \frac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}$$$$= \frac{3\sqrt[4]{2^3}}{\sqrt[4]{2} \, \sqrt[4]{2^3}}$$$$= \frac{3\sqrt[4]{2^3}}{\sqrt[4]{2^4}}$$$$= \frac{3\sqrt[4]{8}}{2} \;\;\; \blacksquare$$ Example 9$\displaystyle \frac{\sqrt[3]{3}}{\sqrt[3]{9}}$Show solution$\blacktriangleright$You could proceed with rationalizing immediately but you'll have to do more work later to fully simplify. This one could be simplified a bit first, since$3$and$9$have a common factor of$3$.$$\frac{\sqrt[3]{3}}{\sqrt[3]{9}} = \sqrt[3]{\frac{3}{9}}$$$$= \sqrt[3]{\frac{1}{3}}$$$$= \frac{\sqrt[3]{1}}{\sqrt[3]{3}}$$$$= \frac{1}{\sqrt[3]{3}}$$Now, proceed with rationalization.$$\frac{1}{\sqrt[3]{3}} \cdot \frac{\sqrt[3]{3^2}}{\sqrt[3]{3^2}}$$$$= \frac{\sqrt[3]{9}}{\sqrt[3]{3^3}}$$$$= \frac{\sqrt[3]{9}}{3} \;\;\; \blacksquare$$ Example 10$\displaystyle \frac{\sqrt[3]{a}}{\sqrt[3]{5b c^2}}$Show solution$\blacktriangleright$There are no common factors, so we rationalizing right away using the most efficient factor.$$\frac{\sqrt[3]{a}}{\sqrt[3]{5b c^2}} \cdot \frac{\sqrt[3]{25 b^2 c}}{\sqrt[3]{25 b^2 c}}$$$$= \frac{\sqrt[3]{5 a b^2 c}}{\sqrt[3]{125 b^3 c^3}}$$$$= \frac{\sqrt[3]{5 a b^2 c}}{5bc} \;\;\; \blacksquare$$ Example 11$\displaystyle \frac{\sqrt[4]{a^2}}{\sqrt[4]{a^3 c^4}}$Show solution$\blacktriangleright$Before rationalizing, we should consider common factors that we can cancel, because this is a ratio of roots which can be re-written as a single root.$$\frac{\sqrt[4]{a^2}}{\sqrt[4]{a^3 c^4}} = \sqrt[4]{\frac{a^2}{a^3 c^4}}$$$$= \sqrt{\frac{1}{a c^4}}$$Now we can separate it back to two separate roots:$$= \frac{\sqrt[4]{1}}{\sqrt[4]{ac^4}}$$$$= \frac{1}{\sqrt[4]{ac^4}}$$Before we rationalize, there is one more important time-saver we should recognize: The variable$c$is already raised to the fourth power, so it already has a perfect fourth root:$$= \frac{1}{c \sqrt[4]{a}}$$Now, all we have to do for rationalization is multiply by enough fourth roots of$a$- three of them.$$\frac{1}{c \sqrt[4]{a}} \cdot \frac{\sqrt[4]{a^3}}{\sqrt[4]{a^3}}$$$$= \frac{\sqrt[4]{a^3}}{c \sqrt[4]{a^4}}$$$$= \frac{\sqrt[4]{a^3}}{ac} \;\;\; \blacksquare$$ Example 12$\displaystyle \frac{y z^5}{x^2 \sqrt[5]{x^2 y^3 z}}$Show solution$\blacktriangleright$Having only a radical expression in the denominator which cannot be simplified in any way on its own, let's go right for the rationalization.$$\frac{y z^5}{x^2 \sqrt[5]{x^2 y^3 z}} \cdot \frac{\sqrt[5]{x^3 y^2 z^4}}{\sqrt[5]{x^3 y^2 z^4}}$$$$= \frac{y z^5 \sqrt[5]{x^3 y^2 z^4}}{\sqrt[5]{x^5 y^ z^5}}$$$$= \frac{y z^5 \sqrt[5]{x^3 y^2 z^4}}{xyz}$$$$= \frac{z^4 \sqrt[5]{x^3 y^2 z^4}}{x} \;\;\; \blacksquare$$ Lesson Takeaways • Recall and reinforce what we already knew about rationalizing from square roots in Algebra One • Understand the concept of multiplying by the "needed" amount of additional$n$-th roots • Practice working with both numeric and variable$n$-th root simplifying problems • Gain exposure to advanced questions that could be used as "challenge" questions on exams, such as complex$n\$-th root denominator expressions
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