Simplifying n-th Root Variable Expressions

Lesson Features »

Lesson Priority: High

  • Simplify monomial variable expressions involving n-th roots
  • Learn when to apply absolute value bars in your answers depending on the instructions in the problem
Lesson Description

Taking n-th roots of variable expressions requires that we deal with two types of objects: the coefficient and its variables. While we will practice doing both at once, the focus of this lesson is on n-th roots of variables, since the n-th root of the coefficient is something we can already do using what we learned in the lesson immediately prior. Variable expression n-th roots have a very basic simplifying process that we'll want to master before the upcoming lesson on multiplying these types of expressions.

Practice Problems

Practice problems and worksheet coming soon!


A Root By Any Other Name

Recently » we defined what an $n$-th root is as a concept, and how to understand them in a similar way to how we already understood square roots. We also learned how to simplify number-based $n$-th root expressions. Now, it's time to learn how to fully simplify $n$-th root variable expressions.
Pro Tip
The study of roots and radicals is closely tied to that of polynomials. In fact, the reason we use the word root when talking about $x$ intercepts of polynomials is because of the inverse operation of exponents when solving a polynomial equation. When $n$ is an integer and $c$ is any real number, the equation$$x^n - c = 0$$has $n$ distinct solutions, or as we often call them, roots (though they might not all be real number solutions). The real number roots show up graphically as $x$-intercepts. For example, $\displaystyle x^2 - 4 = 0$ has two solutions, $2$ and $-2$, and if you graph the function $\displaystyle x^2-4$ you will find that the parabola crosses the $x$ axis at $2$ and $-2$. We call the solutions roots because we use a root operation to solve that type of equation.
When we need to simplify $n$-th root variable expressions that have both a coefficient and a group of variables under the radical, we'll take a two-step approach. First we'll deal with simplifying the coefficient, which, since it will virtually always be an integer, is exactly the same as what we just finished practicing in the last lesson ». Then, we'll deal with each variable, and use division with remainders to get to the answer quickly.
I Used To Know That!
Once upon a time, you learned how to perform long division on integers with remainders, like$$500 \div 6 = 83 \; \mathrm{R} \; 2$$We'll need to do this frequently while working on this topic, but usually with relatively small integers like $9 \div 2$ or $12 \div 5$.
Recall the new-and-improved root product property from the last lesson, which states:$$\sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{xy}$$We use this property backward to break up complex products into separate pieces for the purposes of simplifying the root. In the prior lesson we discussed looking for the biggest "perfect $n$" integer for simplifying integer $n$-th roots. For variables, simplifying is much, well, simpler - we just divide the exponent by the root degree.
n-th Roots of a Variable MonomialThe result of$$\sqrt[n]{x^a}$$is related to the result of $a \div n$.If $n > a$ then the root is simplified already.If $n \le a$ then we will end up with $x$ coming outside of the root, raised to the $a \div n$ whole number power, and an $x$ term still "left over" under the root raised to the power of the remainder of $a \div n$.
This is an instance where it is absolutely harder to say the rule in words than it is to show with an example, and the root product property will guide our steps, meaning you don't have to memorize this definition verbatim. In fact after just a couple of problems, you'll be an expert.We'll also gain further perspective and reinforcement of this idea in the upcoming lesson on fraction exponents ».
Example 1Simplify $\displaystyle \sqrt[3]{x^{14}}$.$\blacktriangleright$ $14 \div 3$ is $4$ R $2$. Blindly following the definition above tells us that$$\sqrt[3]{x^{14}} = x^4 \sqrt[3]{x^2}$$However, we can use the root product property to better understand why this happens. Keep in mind that$$\sqrt[3]{x^3} = x$$This is where the division / reminder idea comes into play. We want to break $x^{14}$ into the product of the largest perfect third power we can, and whatever is left over. Using the above fact and the root product property, we can see that every perfect third power of $x$ will have a perfect third root:$$\sqrt[3]{x^6} = x^2$$$$\sqrt[3]{x^9} = x^3$$$$\sqrt[3]{x^{12}} = x^4$$and so on. Now let's try the example again.$$\sqrt[3]{x^{14}} = \sqrt[3]{x^{12} \cdot x^2}$$This is a similar perspective to the one we use for simplifying $n$-th roots of integers - find the biggest "perfect $n$".$$\sqrt[3]{x^{12} \cdot x^2} = \sqrt[3]{x^{12}} \cdot \sqrt[3]{x^2}$$$$= x^4 \sqrt[3]{x^2} \;\;\; \blacksquare$$
This will quickly become a very intuitive skill after some exposure. Students tend to pick up on it very quickly with a little independent practice. Remember to focus on the biggest "perfect $n$" perspective, which is just another way of thinking about division and remainder.Let's try another one.
Example 2Simplify $\displaystyle \sqrt[5]{w^8}$$\blacktriangleright$ The biggest perfect fifth factor of $w^8$ is $w^5$. Using the root product property,$$\sqrt[5]{w^8} = \sqrt[5]{w^5 \cdot w^3}$$$$= \sqrt[5]{w^5} \cdot \sqrt[5]{w^3}$$$$= w \sqrt[5]{w^3}$$The most efficient way to get the answer is with division and remainder, noting that $8 \div 5$ is $1$ R $3$, so we have $w^1$ outside the root, and $w^3$ left over inside.Finally, depending on your teacher, you may be asked to report$$|w| \sqrt[5]{w^3}$$as your answer, because we started the problem with the positive quantity $w^8$, and because we only report positive answers when we simplify root expressions. Many teachers and texts will tell you to assume all variables are positive, eliminating the need to report your answer with absolute value bars. $\blacksquare$
Pro Tip
When we simplify $n$-th root stand-alone expressions, we will give a positive answer. A lot of the time, the instructions will tell us to assume all variables are positive anyway. If that isn't stated, check with your teacher - they may require you to put absolute value bars in your answer as needed. This is often considered an unimportant nuance but if you're going for the A+ it's something to keep track of.For example, unless we are told not to bother, we should say that$$\sqrt[3]{a^4} = |a| \sqrt[3]{a}$$but if we are told that all variable values are positive, we can just say$$\sqrt[3]{a^4} = a \sqrt[3]{a}$$
Example 3Simplify $\displaystyle \sqrt[3]{x^2}$.$\blacktriangleright$ This expression is already simplified, because the exponent of $2$ is smaller than the degree of the root. Although we said this in the above definition, we can use the division / remainder idea to reinforce this. $2 \div 3$ is $0$ R $2$, so following our rules we would have $x^0$ outside the root, and $x^2$ inside the root - which is the same as what we started with. $\blacksquare$

Simplifying n-th Root Variable Expressions

Now that we know how to deal with the $n$-th root of a variable, we can tackle expressions with coefficients and multiple variables. All we need to do is process each piece, one at a time.
Example 4Simplify $\displaystyle \sqrt[4]{64 a^5 b^8 c^3}$.$\blacktriangleright$ We'll simplify this one piece at a time, starting with the coefficient before looking at each of the three variables.Looking at $\displaystyle \sqrt[4]{64}$,$$\sqrt[4]{64} = \sqrt[4]{16 \cdot 4}$$$$= 2\sqrt[4]{4}$$So in our answer, we'll have a factor of $2$ outside the root and a factor of $4$ remaining inside the root.Next, for $a$:$$\sqrt[4]{a^5} = \sqrt[4]{a^4 \cdot a}$$$$ = a \sqrt[4]{a}$$This results tells us that we'll have a factor of $a$ outside of the root and a factor of $a$ inside the root.Moving on to $b$:$$\sqrt[4]{b^8} = b^2$$because $\displaystyle b^8$ is already a perfect fourth factor. The factor $b^2$ will be present outside the root and no $b$ factor will remain inside.Finally, for $c$:$$\sqrt[4]{c^3}$$is already simplified because $3$ is smaller than the root degree of $4$.Putting all of this together, looking at which factors have reduced completely and which factors remain under the root:$$\sqrt[4]{64 a^5 b^8 c^3} = 2ab^2 \sqrt[4]{4ac^3} \;\;\; \blacksquare$$
If you are asked to simplify an expression like $\sqrt[4]{162}$, then you should report just the positive answer $3\sqrt[4]{2}$. However, if you are asked to solve the equation$$x^4 = 162$$then you need to report both a positive and negative answer ($x = \pm 3\sqrt[4]{2}$) because both answers would make the original equation true, meaning that both answers are solutions to the equation.
It's possible that your teacher will also jump right into using this technique when doing things like multiplying » or dividing » $n$-th root expressions. Each of those has its own lesson but it's worth practicing all three of these lessons together when preparing for an exam.

Mr. Math Makes It Mean

Roots of RootsWhile it might not sound very fun, it is possible to take higher roots of higher root expressions. Ultimately, this task will be easier to digest once you learn how to use fraction exponents », but some teachers make you learn this before that point. In short, just multiply the root degrees together and pretend that's the root you are taking.For example,$$\sqrt[3]{\sqrt[4]{x^{15}}}$$is the same as$$\sqrt[12]{x^{15}}$$for the purposes of simplifying.

Put It To The Test

Example 5$$\sqrt[4]{112 x^5 y z^{10}}$$
Show solution
$\blacktriangleright$ The largest perfect fourth that is a factor of $112$ is $16$, meaning that$$\sqrt[4]{112} = \sqrt[4]{16 \cdot 7} = 2 \sqrt[4]{7}$$For the variables, use the division-remainder concept, or factor into the largest fourth powers possible.$$\sqrt[4]{x^5} = \sqrt[4]{x^4 \cdot x} = x\sqrt[4]{x}$$$$\sqrt[4]{y} = \sqrt[4]{y}$$$$\sqrt[4]{z^{10}} = \sqrt[4]{z^8 \cdot z^2} = z^2 \sqrt[4]{z^2}$$Putting these pieces together:$$\sqrt[4]{112 x^5 y z^{10}} = 2xz^2 \sqrt[4]{7xyz^2} \;\; \blacksquare$$
Example 6$$\sqrt[3]{-40a^5 b^8}$$
Show solution
$\blacktriangleright$$$\sqrt[3]{-40} = -2\sqrt[3]{5}$$$$\sqrt[3]{a^5} = a \sqrt[3]{a^2}$$$$\sqrt[3]{b^8} = b^2 \sqrt[3]{b^2}$$Therefore,$$\sqrt[3]{-40a^5 b^8} = -2ab^2 \sqrt[3]{5a^2 b^2} \;\; \blacksquare$$
Example 7$$\sqrt[5]{96}$$
Show solution
$$\blacktriangleright \;\; 2\sqrt[5]{3} \;\; \blacksquare$$
Example 8$$\sqrt[3]{135 r^4 s^6 t^{13}}$$
Show solution
$$\blacktriangleright \;\; 3 r t^4 \sqrt[3]{5rt} \;\; \blacksquare$$
Example 9$$\sqrt[3]{125m^2 n^3 p^4}$$
Show solution
$$\blacktriangleright 5np \sqrt[3]{m^2 p } \;\; \blacksquare$$
Example 10$$\sqrt[5]{\sqrt[2]{4096x^{16} y^{20}}}$$
Show solution
$\blacktriangleright$ Recall that the $m$-th root of an $n$-th root expression is equivalent to taking the $m \cdot n$ root. Here,$$\sqrt[5]{\sqrt[2]{4096x^{16} y^{20}}}$$$$ = \sqrt[10]{4096x^{16} y^{20}}$$And so:$$\sqrt[10]{4096} = \sqrt[10]{1024 \cdot 4} = 2\sqrt[10]{4}$$$$\sqrt[10]{x^{16}} = x \sqrt[10]{x^6}$$$$\sqrt[10]{y^{20}} = y^2$$Leaving us with a final answer of$$\sqrt[5]{\sqrt[2]{4096x^{16} y^{20}}}$$$$ = \sqrt[10]{4096x^{16} y^{20}}$$$$ = 2xy^2 \sqrt[10]{4x^6} \;\; \blacksquare$$
Lesson Takeaways
  • Apply what we recently learned about $n$-th roots to variable expressions
  • Manage your work demonstration by tackling the coefficient first and then each variable one at a time
  • Recognize and use the division and remainder pattern for simplifying $n$-th roots of variables
  • Be able to simplify roots of roots by multiplying the root degrees

Lesson Metrics

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