# Arithmetic with n-th Roots

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Objectives
• Add and subtract root expressions if possible
• When it is not possible to add or subtract, understand why
• Learn to multiply n-th root expressions together and obtain the simplest final answer
Lesson Description

Now that we understand a bit about what n-th roots are and how to simplify them, we now turn to how to work with them when it comes to performing arithmetic. We'll multiply, add, and subtract various root expressions in this lesson.

Practice Problems

Practice problems and worksheet coming soon!

## Remembering "Like Terms"

For a while now we've been used to the idea of adding like terms in Algebra. "Like terms" generally refers to objects that have the same variable and exponent, but different coefficients, and they can be combined together with addition or subtraction. For example,$$5x^2 + 3x^2 = 8x^2$$When it comes to $n$-th roots, terms will be like (and therefore combinable with addition or subtraction) if the two terms have the same variable expression underneath the root, and the same $n$ degree of root.For example, $\displaystyle 4\sqrt{xy^2}$ and $\displaystyle 2\sqrt{xy^2}$ can be added together to make $\displaystyle 6\sqrt{xy^2}$, but the following two examples cannot be added as like terms:$$4\sqrt{xy^2} + 2\sqrt{xy^2}$$(because the degrees of each root do not match)$$4\sqrt{xy} + 2\sqrt{xy^2}$$(because the variable expressions under each root do not match)

Adding and subtracting like terms isn't new to us, and the mechanics will look exactly as you expect them to.
Define: $n$-th Root CombinationFor two $n$-th root terms of the same degree, and with the same variable expression under the root (including coefficient), combine the terms with addition or subtraction by adding or subtracting their coefficients, respectively.
To recap, you could not combine expressions like$$2\sqrt{t^5} - \sqrt{t^5}$$$$-6\sqrt{2r^6} + 4\sqrt{2r^4}$$$$5\sqrt{2x} + 3\sqrt{3x}$$but you could combine expressions such as$$2\sqrt{t^5} - \sqrt{t^5} = \sqrt{t^5}$$$$-6\sqrt{2r^6} + 4\sqrt{2r^6} = -2\sqrt{2r^6}$$$$5\sqrt{3x} + 3\sqrt{3x} = 8\sqrt{3x}$$
Remember! Combining $n$-th root expressions with addition or subtraction is only possible when the root degrees match (e.g. both roots are square roots, both roots are cube roots, etc.). Then, we simplify each term if possible, because we also need to have the same expressions under the radical in order to combine them. If this is not the case, then the problem cannot be further simplified.

Example 1Simplify $\displaystyle 4\sqrt{3x^2} + 7\sqrt{3x^2}$.$\blacktriangleright$ These two terms meet the definition of like terms for $n$-th roots, and so we can combine them:$$4\sqrt{3x^2} + 7\sqrt{3x^2} = 11\sqrt{3x^2} \;\; \blacksquare$$

The concept is very familiar and intuitive, so it's not surprising that we're really not going to see quizzes and test ask us to compute the addition or subtraction of simple like terms. Instead, with $n$-th roots, the challenge is often that we're given unsimplified terms that cannot initially be combined as like terms, but will be able to be combined once simplified.Therefore, being tested on this knowledge is essentially a review of something we learned very recently - simplifying n-th roots ».

The best way to deal with this issue is to always simplify any root before adding or subtracting. This is something we probably became accustomed to with square roots in early algebra, and being consistent with simplifying will ultimately make for less work on your part. Therefore we must practice and simply get in the habit of simplifying everything we can to lowest terms.Here's a refresher example from Algebra One involving square roots.

Example 2Simplify $\displaystyle \sqrt{27} + \sqrt{18}$.$\blacktriangleright$ Simplify each root first to see if they can be combined.$$3\sqrt{3} + 2\sqrt{3}$$These are now like terms, and can be combined as such.$$= 5 \sqrt{3} \;\; \blacksquare$$

Pro Tip Make sure you've thoroughly mastered the recent lesson on simplifying n-th roots ». If you are having trouble with it, I recommend going back to early algebra and re-mastering simplifying square roots », which may help you better understand what's similar and different when it comes to $n$-th roots.
Now let's try a problem with $n$-th root expressions.

Example 3Simplify $\displaystyle \sqrt{32} + \sqrt{162}$.$\blacktriangleright$ While these are both $4$-th roots, they are not yet combinable since they do not have the same expressions under the root. However, if we simplify, we'll have terms that we can add together.$$\sqrt{32} = \sqrt{16 \cdot 2} = 2\sqrt{2}$$$$\sqrt{162} = \sqrt{81 \cdot 2} = 3\sqrt{2}$$Therefore,$$\sqrt{32} + \sqrt{162}$$$$= 2\sqrt{2} + 3\sqrt{2}$$$$= 5\sqrt{2} \;\; \blacksquare$$It's important to note also that teachers will always expect simplified answers, if possible - yet another reason we need to already be masterful at the prerequisite skill of simplifying $n$-th root expressions.

Example 4$$\sqrt{16a^5 b^4 c^2} - \sqrt{54a^5 b^4 c^2}$$$\blacktriangleright$ Even though the variables and exponents under the root match, the coefficients underneath the roots do not - but they will once we simplify.$$\sqrt{16a^5 b^4 c^2} = 2\sqrt{2a^5 b^4 c^2}$$$$\sqrt{54a^5 b^4 c^2} = 3\sqrt{2a^5 b^4 c^2}$$However, the exponents under the root symbol are larger than the root we're taking, so we would be expected to simplify. If you turn in the answer $\displaystyle -\sqrt{2a^5 b^4 c^2}$ you will most likely lose points. Let's repeat the simplification step again but this time take care of the variables as well.$$\sqrt{16a^5 b^4 c^2} = 2ab \sqrt{2a^2 b c^2}$$$$\sqrt{54a^5 b^4 c^2} = 3ab\sqrt{2a^2 b c^2}$$Overall,$$\sqrt{16a^5 b^4 c^2} - \sqrt{54a^5 b^4 c^2}$$$$=2ab\sqrt{2a^2 b c^2} - 3ab\sqrt{2a^2 b c^2}$$$$= -ab\sqrt{a^2 b c^2} \;\;\; \blacksquare$$
Warning! Remember that, as we proved in Algebra One, $\sqrt{x+y}$ is not equal to $\sqrt{x} + \sqrt{y}$. The caution about this very common mistake extends to $n$-th roots as well, not just square roots. The root of a sum is not equal to the sum of each root!$$\sqrt[n]{x+y} \ne \sqrt[n]{x} + \sqrt[n]{y}$$

## Multiplying n-th Roots

Perhaps against your intuition, multiplying is a little easier than combination when it comes to $n$-th roots, because it doesn't have any restrictions. In fact, the two expressions don't even need to have the same degree of root, although in this lesson we will stick to ones that do until we learn about fraction exponents » in the very near future, because problems with mismatched root degrees are not common, and are much easier to handle with fraction exponents. So for now, we will only be looking at multiplying matching degrees of roots.As we recently learned with our updated root product property » for $n$-th roots, we can multiply objects of the same root degree together by simply multiplying the expressions under the roots, and putting it under one big radical symbol.

Example 5$$\sqrt{w^4y^2} \cdot \sqrt{wy^2}$$$$\blacktriangleright \;\;\; \sqrt{\left( w^4 y^2 \right) \cdot\left( wy^2 \right)}$$$$= \sqrt{w^5 y^4}$$
Pro Tip If there are any coefficients already outside of the roots, keep them outside of the roots. The coefficients will still need to get multiplied together, but since multiplication is commutative, we can keep them all in front and uninvolved with the "root stuff".e.g.$$2\sqrt{x^2} \cdot 5\sqrt{y^2} = 10 \sqrt{x^2 y^2}$$
As with combining $n$-th roots (i.e. addition and subtraction operations), the basic mechanic is fairly simple, so the major skill that will be tested on quizzes and exams will be correctly simplifying our answers, which we already practiced in the prior lesson ».Here's an example that uses square roots, which we practiced in Algebra One ».

Example 6$\displaystyle \sqrt{5x^3} \cdot \sqrt{10 x^5}$$\blacktriangleright According to the root product property, we can re-write this product as a single radical expression:$$\sqrt{5x^3} \cdot \sqrt{10 x^5} = \sqrt{5x^3 \cdot 10x^5}$$Multiply, then simplify the radical.$$ = \sqrt{50x^8} = \sqrt{25x^8 \cdot 2} = 5x^4 \sqrt{2} \;\; \blacksquare$$You Should Know If you're wondering if you should try and do any simplifying immediately when multiplying root expressions, the answer is no. It's not wrong to do, but it ultimately creates more work.Radicals can be frustrating for that reason, because in many other algebra topics, simplifying as the first step is often easier or even necessary. Now let's work the process with a similar problem that uses a third root. Example 7\displaystyle \sqrt{20a^4 b^2} \cdot \sqrt{6a^2 b^6}$$\blacktriangleright$ Let's use the root product property to put all of this under one big root.$$\sqrt{20a^4 b^2} \cdot \sqrt{6a^2 b^6}$$$$= \sqrt {120a^6 b^8}$$Now, remember what we recently learned about simplifying n-th roots » we must find simplify by factoring out the largest perfect cubes we can find, for the coefficient as well as the variables.$$\sqrt{120a^6 b^8} = \sqrt{8 \cdot 12 \cdot a^6 \cdot b^6 \cdot b^2}$$$$= 2a^2 b^2 \sqrt{12b^2}$$Many teachers give little or no credit for turning in the answer without simplifying. $\blacksquare$
You Should Know Shortly after this lesson, we'll be learning about fraction exponents » which will give us an easier way to handle both multiplying and simplifying n-th root expressions.

## Multiplying Mixed Root Expressions

If you're asked to multiply mixed expressions that involve sums of numbers and roots, you'll need to use the FOIL method, or generally speaking, the distributive property. This isn't something we'll be asked to do often but it's still something we'd do well to firmly understand.

## Put It To The Test

Example 10$\displaystyle \sqrt{40} + \sqrt{160}$
Show solution
$\blacktriangleright$ Simplify each square root and see if we end up with like terms.$$\sqrt{40} = \sqrt{4 \cdot 10}$$$$= 2 \sqrt{10}$$$$\sqrt{160} = \sqrt{16 \cdot 10}$$$$= 4 \sqrt{10}$$We can add these together.$$\sqrt{40} + \sqrt{160}$$$$= 2\sqrt{10} + 4\sqrt{10}$$$$=6\sqrt{10} \;\;\; \blacksquare$$

Example 11$\displaystyle 3\sqrt{5b^2} + 6\sqrt{5b^2} - 2\sqrt{5b^2}$
Show solution
$\blacktriangleright$ All terms are like terms here, since the expression under the radical is the same. All we are doing is "counting" how many $\sqrt{5b^2}$'s we have.$$\longrightarrow 7\sqrt{5b^2} \;\;\; \blacksquare$$

Example 12$\displaystyle 2\sqrt{16} - 3\sqrt{54}$
Show solution
$\blacktriangleright$ Simplify each and see if we end up with like terms. We are looking for the biggest perfect third power factor when we simplify third roots.$$\sqrt{16} = \sqrt{8 \cdot 2}$$$$= 2 \sqrt{2}$$$$\sqrt{54} = \sqrt{27 \cdot 2}$$$$= 3 \sqrt{2}$$Therefore,$$2\sqrt{16} - 3\sqrt{54}$$$$= 2\cdot 2 \sqrt{2} - 3 \cdot 3 \sqrt{2}$$$$= 4 \sqrt{2} - 9\sqrt{2}$$$$= -5 \sqrt{2} \;\;\; \blacksquare$$

Example 13$\displaystyle \sqrt{432x} - \sqrt{2x} + \sqrt{686x} - \sqrt{512x}$
Show solution
$\blacktriangleright$ While $n$-th root problems tend to give us annoyingly large integers to work with, remember that you're seeking out cubes and fourth power numbers, so there aren't a lot of choices for reducing (even $10^3$ is $1000$ so you only have $9$ or so options to think about when working with cube roots under $1000$).Reduce each root, then combine like terms, keeping in mind that third roots and fourth roots cannot combine together.$$\sqrt{432x} = \sqrt{216 \cdot 2x} = 6\sqrt{2x}$$$$\sqrt{2x} = \sqrt{2x}$$(no reducing is possible)$$\sqrt{686x} = \sqrt{343 \cdot 2x} = 7\sqrt{2x}$$$$\sqrt{512x} = \sqrt{256 \cdot 2x} = 4\sqrt{512x}$$Therefore, we can re-write our problem as$$6\sqrt{2x} - \sqrt{2x} + 7\sqrt{2x} - 4\sqrt{2x}$$and, combining like terms, get our final answer:$$-\sqrt{2x} - 5\sqrt{2x} \;\;\; \blacksquare$$

Example 14$\displaystyle \sqrt{125y^3} - \sqrt{8y} + \sqrt{27y^3}$
Show solution
$\blacktriangleright$ Simplify each root, remembering the rules for $n$-th roots of variable expressions.$$\sqrt{125y^3} = 5y$$$$\sqrt{8y} = 2\sqrt{y}$$$$\sqrt{27y^3} = 3y$$Only the first and last terms are like, so the middle term will stand alone and not be able to be combined.$$\sqrt{125y^3} - \sqrt{8y} + \sqrt{27y^3}$$$$= 5y - 2\sqrt{y} + 3y$$$$= 8y - 2\sqrt{y} \;\;\; \blacksquare$$

Example 15$\displaystyle \sqrt{24a^3 b} \cdot \sqrt{3a^2 b^4}$
Show solution
$\blacktriangleright$ This one is a square root, like we may have already seen in Algebra One. However, the process is the same for any degree of root - combine and simplify.$$\sqrt{72 a^5 b^5}$$$$\sqrt{36a^4 b^4 \cdot 2ab}$$$$6a^2 b^2 \sqrt{2ab} \;\; \blacksquare$$

Example 16$\displaystyle \sqrt{40x^3 y} \cdot \sqrt{8 x^2 y^2}$
Show solution
$\blacktriangleright$ Once again, this is a square root problem, but the process is the same.$$\sqrt{40x^3 y} \cdot \sqrt{8 x^2 y^2}$$$$= \sqrt{320x^5 y^3}$$For the coefficient, the largest perfect square factor of $320$ is $64$. For the variables, remember that the biggest factor we can remove is a multiple of the root degree - in this case, $2$.$$\sqrt{320x^5 y^3} = \sqrt{64 \cdot 5 \cdot x^4 \cdot x \cdot y^2 \cdot y}$$$$= 8x^2 y \sqrt{5xy} \;\;\; \blacksquare$$

Example 17$\displaystyle \left( 3\sqrt{6} \right) \left( 4\sqrt{6} \right)$
Show solution
$\blacktriangleright$ Multiply the coefficients, and also the expressions under the root.$$\left( 3\sqrt{6} \right) \left( 4\sqrt{6} \right)$$$$= 12 \sqrt{36}$$There is no perfect cube factor of $36$ other than $1$, so this answer is fully simplified. $\blacksquare$

Example 18$\displaystyle \sqrt{10} \cdot \sqrt{6}$
Show solution
$\blacktriangleright$ Multiply the expressions under the roots:$$\sqrt{10} \cdot \sqrt{6}$$$$= \sqrt{60}$$To simplify, we seek out the biggest perfect cube factor of $60$. However, the only two options are $8$ and $27$ (since the next one is $64$ and a number larger than $60$ can't be a factor of $60$) and neither of these numbers is a factor. Therefore, the results is as simplified as it can be. $\blacksquare$

Example 19$\displaystyle \sqrt{54x^2 y^3} \cdot \sqrt{-2x^3 y^4}$
Show solution
$\blacktriangleright$ Let's start by multiplying everything together:$$\sqrt{54x^2 y^3} \cdot \sqrt{-2x^3 y^4}$$$$= \sqrt{-108x^5 y^7}$$For the coefficient, the biggest perfect cube factor is $27$. For the exponents, remember that perfect cube factors have exponents that are multiples of $3$.$$= \sqrt{-27 \cdot 4 \cdot x^3 \cdot x^2 \cdot y^6 \cdot y}$$$$= -3xy^2 \sqrt{4x^2 y} \;\;\; \blacksquare$$
Remember! You can take the root of a negative number when working with odd root degrees!

Example 20$\displaystyle \sqrt{4x^7 y^3} \cdot \sqrt{48x^2 y^3}$
Show solution
$$\blacktriangleright \;\; \sqrt{192 x^9 y^6}$$$$\sqrt{16x^8 y^4 \cdot 12 x y^2}$$$$2x^2 y \sqrt{12xy^2} \;\; \blacksquare$$

Example 21$\displaystyle 3\sqrt{15a^6 b} \cdot \sqrt{-3a^3b^4}$
Show solution
$$\blacktriangleright \;\; \sqrt{-45 a^9 b^5}$$$$\sqrt{-a^9 b^3 \cdot 45 b^2}$$$$-a^3 b \sqrt{45b^2} \;\; \blacksquare$$

Lesson Takeaways
• Know the principles of adding and subtracting $n$-th root terms
• Extend your view of "like terms" to include varying degrees of roots
• Use the idea of like terms to understand when you can and can't combine root expressions
• Be able to efficiently multiply n-root terms and obtain fully simplified answers
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