# Integer n-th Roots

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Objectives
• Define the n-th root of a number, where $n$ is an integer
• Learn how to take the n-th root of whole numbers, when $n$ is an integer
• Learn how to simplify or reduce an n-th root of an integer to its simplest form, similar to how we learned to reduce square roots
• Know how to solve simple n-th root equations involving integers
• Correctly identify positive and negative answers depending on whether you are simplifying an expression or solving an equation
Lesson Description

Although, we're already familiar with square roots and how to work with them, we need to be able to take other roots - like third roots, fourth roots, or n-th roots, where n could be anything. We'll introduce these types of roots, understand how they work, and then practice simplifying them for roots with only integers underneath (we'll look at n-th roots of variables in the next lesson).

Practice Problems

Practice problems and worksheet coming soon!

## Higher Roots

When we are first introduced to roots, we always called them "square" roots. Perhaps you've never thought twice about that label, but it's important that we specify the degree of roots from here on out. While it's certainly true that a vast majority of roots and root operations in Algebra and Calculus are square roots, square roots are not the only type.The main reason square roots are named the way they are named has to do with the exponent $2$, to which they are closely related. When we take the square root of a number, we are looking for the number that, when squared, gives us back what we started with. Squaring the answer is the shortcut word for raising it to the power of $2$.For example, $\displaystyle \sqrt{16} = 4$ because $\displaystyle 4^2 = 16$. We can define roots of higher degrees by simply doing the same thing with an exponent other than $2$.
n-th RootsThe $n$-th root of a number $x$ is the number such that, when raised to the power $n$, is equal to $x$.$$\sqrt[n]{x} = a \longrightarrow a^n = x$$
When we work with roots other than the degree $2$ square roots we're used to, we can make generalized rules that will help us see how the things we've known about square roots are just a consequence of a more general rule.For example, you should know by now that we can't take the square root of a negative number. Here's why.
Theorem: Even and Odd RootsIf $n$ is an even integer, then $\sqrt[n]{x}$ is only a real number if $x$ is positive. Even roots of negative numbers do not exist. If $n$ is an odd integer, then $\sqrt[n]{x}$ has a real number result regardless of what $x$ is. The result will be positive if $x$ is positive. The result will be negative if $x$ is negative.
This theorem touches on something we learned in early Algebra - why square roots ($n=2$) do not yield real answers when we try to take the root of a negative number. The reason this works for odd roots is that we are then able to have an odd number of negative signs if we write the equation in exponent form, which gives a negative answer. See Example 2 below for further clarification of this.

Example 1Find $\displaystyle \sqrt{81}$.$\blacktriangleright$ Another way to ask this question is, "what number gives $81$ when raised to the fourth power?" Some quick guess and check will confirm that the correct number is $3$.$$3^4 = 81 \longrightarrow \sqrt{81} = 3 \;\;\; blacksquare$$

Example 2Find $\displaystyle \sqrt{-64}$.$\blacktriangleright$ Unlike even roots, odd degree roots of negative numbers can be found. Again, to solve this we can instead think, "what number gives me $-64$ when raised to the third power?" That number is $-4$. This reinforces the idea that we can take odd roots of negatives, because when we write this fact using multiplication, there is an odd number of negative signs.$$(-4)^3 = (-4) \cdot (-4) \cdot (-4)$$$$= -64$$Therefore, $\displaystyle \sqrt{-64} = -4$. $\blacksquare$
You Should Know Note that up until now, we have been notating the square root symbol as $\sqrt{\,\,\,\,}$ and not as $\sqrt{\,\,\,\,}$. Because square roots are so much more common than third roots, fourth roots, fifth roots, etc., mathematicians have collectively decided that writing the symbol without a number will imply that the root is a second root.

## Expanded Root Product Property

The product of two root expressions is equal to the root of the product of the expressions, if the degree of each root is the same. We've seen this before with square roots, and we called it the root product property ». This is still the case, but now we'll expand the root product property to all degrees of roots, and so what we learned in the past for square roots is just a subset of this generic rule.
Expanded Root Product PropertyFor any two $n$-th root expressions of the same degree, obtain their product by multiplying the expressions under the root together.$$\sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{xy}$$

Example 3Write the following product of roots as a single root.$$\sqrt{9} \cdot \sqrt{8}$$$\blacktriangleright$ As the root product property says, we can re-write this as a single root by multiplying the expressions under the radicals.$$\sqrt{9} \cdot \sqrt{8} = \sqrt{72} \;\;\; \blacksquare$$

Similarly, if need be, we can this equation applies in reverse as a way to separate a product into separate roots, which we'll use to our advantage to simplify roots, again very similar to what we've done in the past with square roots.

Example 4Write the following root as the product of two roots.$$\sqrt{24}$$$\blacktriangleright$ According to the root product property, we can separate this by factoring $24$ under the radical. We can choose any factoring we like, so there are multiple ways to answer this question correctly.$$\sqrt{24} = \sqrt{8 \cdot 3}$$$$= \sqrt{8} \cdot \sqrt{3}$$And, since $8$ is a perfect cube, it happens that we picked a factoring that results in some simplifying.$$\sqrt{24} = \sqrt{8} \cdot \sqrt{3}$$$$= 2 \sqrt{3} \;\;\; \blacksquare$$

In fact, this is precisely how we'll routinely simplify $n$-th roots - by factoring in a clever way. Let's see how to approach simplifying generally using this idea.

## Simplifying n-th Roots

When we wanted to simplify square roots, our strategy was to find the biggest "perfect square" factor when we did our factoring. For an $n$-th root, we'll want to factor out the biggest "perfect $n$". Once we have, the root product property will let us separate the product into two roots, one of which will reduce to an integer.Let's learn by doing.

Example 5Simplify the following roots.a) $\displaystyle \sqrt{40}$b) $\displaystyle \sqrt{40}$c) $\displaystyle \sqrt{72}$d) $\displaystyle \sqrt{32}$e) $\displaystyle \sqrt{729}$$\blacktriangleright Let's take out the biggest perfect "n" from each n-th root, remembering that n is the degree of root.a) \displaystyle \sqrt{40}This is a square root, which is the same as an n-th root where n=2. Look for the biggest perfect square that is a factor of 40, which is 4. Factor 40 and then use the root product property to separate the roots.$$\sqrt{40} = \sqrt{4 \cdot 10} = \sqrt{4} \cdot \sqrt{10} = 2\sqrt{10}$$Because we found the largest perfect square factor of 40, this answer is fully simplified. \blacksquareb) \displaystyle \sqrt{40}Here n=3, so seek out the biggest perfect third. If you don't remember what those numbers are, simply take the integers sequentially starting with 1 and raise them to the power 3 to get the list.$$1^3 = 12^3 = 83^3 = 274^3 = 645^3 = 125$$and so on. It looks like the biggest perfect third power integer that is a factor of 40 is 8.$$\sqrt{40} = \sqrt{8 \cdot 5}= \sqrt{8} \cdot \sqrt{5}= 2 \sqrt{5} \;\;\; \blacksquare$$c) \displaystyle \sqrt{72}Again, we need the biggest perfect third power integer, and again for this example, that will be 8.$$\sqrt{72} = \sqrt{8 \cdot 9}= \sqrt{8} \cdot \sqrt{9} = 2 \sqrt{9} \;\;\; \blacksquare$$d) \displaystyle \sqrt{32}This time, we need the largest perfect fourth power integer. For this problem, 2^4 = 16 is the largest perfect fourth factor.$$\sqrt{32} = \sqrt{16 \cdot 2}\sqrt{16} \cdot \sqrt{2} = 2 \sqrt{2} \;\;\; \blacksquare$$e) \displaystyle \sqrt{729}Perfect fifth integers (and higher ones too for that matter) are actually a little easier in my opinion, because we quickly run out of options. Remember that the factor we seek couldn't possibly be bigger than the number itself, so as soon as we list out perfect fifth integers that surpass 729, we know it's useless to keep listing.$$1^5 = 12^5 = 323^5 = 2434^5 = 1024$$Here, 243 is indeed a factor of 729.$$\sqrt{729} = \sqrt{243 \cdot 3} = \sqrt{243} \cdot \sqrt{3}= 3\sqrt{3} \;\;\; \blacksquare$$This is all there is to simplifying integer n-th roots. We'll tackle variables in one of the next lessons », but that's actually a little easier in that it ultimately follows some simple arithmetic using the exponents. Remember! If you cannot find any perfect factors when simplifying roots of integers, then the expression cannot be further simplified. ## Basic n-th Root Equations There are two types of basic equations that involve n-th roots.Type 1 - Power EquationsKnowing how to take and simplify n-th roots of integers means you can solve super simple n-th power equations, which require taking n-th roots to solve.The principle is based on the definition of n-th roots. n-th Roots of xnThe n-th root of x^n is equal to x.$$\sqrt[n]{x^n} = x$$You Should Know If we are solving an equation and n is even, the answer should include both positive and negative solutions. When simplifying an expression, we should report the positive or negative answer only.For example:Solve \displaystyle x^4 = 16:$$\longrightarrow \sqrt{x^4} = \sqrt{16}x = \pm 2$$compared toSimplify \displaystyle \sqrt{16}:$$\longrightarrow \sqrt{16} = 2$$When you are solving, you are trying to find every solution that makes the original equation true, unlike simplifying which follows preset standards. Example 6Solve.$$2x^5 - 60 = 4$$\blacktriangleright To solve this, first isolate the variable:$$\longrightarrow 2x^5 = 64x^5 = 32$$Finish the problem by taking the fifth root of both sides.$$\sqrt{x^5} = \sqrt{32}x = \sqrt{32} = 2$$This was an odd root, so we do not need a \pm symbol with our answer. \blacksquare Example 7Solve.$$x^4 + 10 = 106$$\blacktriangleright Isolate x^4 and take the fourth root.$$x^4 = 96\sqrt{x^4} = \sqrt{96}x = \pm \sqrt{96}$$And, using our simplification technique and the root product property as discussed early, we can simplify the right side give us a final answer of$$x = \pm \sqrt{16 \cdot 6}x = \pm 2 \sqrt{6}$$Again, we need the \pm symbol because we're solving an equation with an even root, but if we were only simplifying a standalone expression without an equation, we would only report the positive answer. \blacksquare Type 2 - Basic n-th Root EquationsIf an equation appears with an n-th root of x that you can isolate, we'll simply raise both sides to the n power to solve. These are typically less common on tests because they are fairly easy. Example 8Solve.$$\sqrt{x} = 10$$\blacktriangleright Raise both sides to the power 5.$$\left( \sqrt{x} \right)^5 = \left( 10 \right)^5$$By the definition of n-th roots, \displaystyle \left( \sqrt{x} \right)^5 is equal to x. The root and the exponent "undo" one another.$$x = 10,000 \;\;\; \blacksquare$$Example 9Solve.$$\sqrt{x^3} - 6 = 2$$\blacktriangleright This problem combines the last two examples. First let's isolate the variable.$$\sqrt{x^3} = 8$$Then we'll raise each side to the power 4:$$x^3 = 4096$$Finally, take the third degree root of both sides.$$\sqrt{x^3} = \sqrt{4096}x = 16$$Because this was an odd root, we do not use a \pm sign for the answer. \blacksquare ## Put It To The Test Example 10\sqrt{216} Show solution \blacktriangleright 216 is a perfect third power integer.$$\sqrt{216} = 6 \;\;\; \blacksquare$$Example 11\sqrt{6561} Show solution \blacktriangleright 6561 is a perfect fourth power integer.$$\sqrt{6561} = 9 \;\;\; \blacksquare$$Example 12\sqrt{64} Show solution \blacktriangleright We want the largest perfect fifth integer that is a factor of 64. There's only one option:$$\sqrt{64} = \sqrt{32 \cdot 2}= \sqrt{32} \cdot \sqrt{2}= 2\sqrt{2} \;\;\; \blacksquare$$Example 13\sqrt{729} Show solution \blacktriangleright 729 is a perfect third power integer.$$\sqrt{729} = 9 \;\;\; \blacksquare$$Example 14\sqrt{180} Show solution \blacktriangleright List out the perfect third power integers to seek out the largest one that is a factor of 180.$$1^3 = 12^3 = 83^3 = 274^3 = 645^3 = 1256^3 = 216$$None of the perfect cube integers is a factor of 180, so this expression is already simplified as written. \blacksquare Example 15\sqrt{1024} Show solution \blacktriangleright The largest perfect third power integer that 1024 divides is 512, or 8^3.$$\sqrt{1024} = \sqrt{512 \cdot 2}= \sqrt{512} \cdot \sqrt{2} = 8 \sqrt{2} \;\;\; \blacksquare$$Example 16Solve \displaystyle 2x^4 - 15 = 147. Show solution \blacktriangleright Isolate the variable:$$2x^4 = 162x^4 = 81$$Take the fourth root of each side.$$\sqrt{x^4} = \sqrt{81}x = \pm 3$$Note that the \pm side is required, since the root was an even root, and we're solving an equation and not merely simplifying an expression. \blacksquare Example 17Solve \displaystyle 3\sqrt{x} + 25 = 19 Show solution \blacktriangleright Isolate the root expression:$$3\sqrt{x} = -6\sqrt{x} = -2$$Raise both sides to the sixth power.$$\left( \sqrt{x} \right)^6 = (-2)^6x = 64 \;\;\; \blacksquare$$Lesson Takeaways • Understand what$n$-th roots are, by definition and conceptually • Make connections to what you already knew about square roots • Know how to fully simplify$n$-th roots of integers • Solve basic equations that use the definition of$n\$-th roots
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