# Simplifying Integer Square Roots

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Lesson Priority: VIP Knowledge

Objectives
• Understand the Root Product Property and how it applies to integers
• Learn how to write any integer square root in simplest form
Lesson Description

On the subject of square roots, simplifying the square root of an integer is one of the most common things we are asked to do. Unlike estimating the exact value of a square root, simplifying square roots is a method in which we can employ for any integer square root, and always end up with the same, unique answer.

Practice Problems

Practice problems and worksheet coming soon!

## The Eye of the Beholder

Simplifying square roots is an important skill that we'll use from the roots of Algebra to the depths of Calculus to the obscure branches of advanced math. However, unlike simplifying fractions, which truly does fit its description, simplifying square roots will not always leave you sure that your answers are truly "simpler" in the literal sense of the word. That will require a small amount of faith and patience, as you'll better understand why "fully simplified" square roots will ultimately make our lives easier. In this way, seeing a simplified square roots and appreciating that it is indeed "more simple", is much like beauty - in the eye of the beholder. However unlike beauty, when you simplify a square root, the answer you get is either right or wrong (as is nearly always the case in math!).
You Should Know This lesson will teach us and help us practice simplifying integer square roots into simplest terms, such as turning something like $\sqrt{72}$ into its fully simplified form of $6\sqrt{2}$. If you want to learn about and practice rationalizing square root denominators », we will look at that in its own imminently upcoming lesson!
Some square roots simplify to integers, as we saw with the recent lesson on perfect square integers ». Since we will not always be lucky enough to be asked to take square roots of such integers, this lesson will help us see specifically how to simplify square roots of non-perfect square integers. That way, we will be able to say we can take the square root of anything!
Pro Tip The TI-84 Calculator doesn't simplify square roots, but the TI-89 does. Either way, your teacher probably isn't going to allow you to use a calculator for quizzes and exams on this topic, which is why we need to know the contents of this lesson.

## The Mechanics of Manipulating Roots

Working with and re-tooling square roots will require a thorough understanding of the root product property, now and in future related topics.
Define: The Root Product PropertyThe square root of a product is equal to the product of each factor's square root, and vice versa.In math symbols:$$\sqrt{xy} = \sqrt{x}\sqrt{y}$$and$$\sqrt{x}\sqrt{y} = \sqrt{xy}$$
We'll need to understand this very well.So what makes this definition useful? On the surface it seems like it doesn't get us anywhere, and if used incorrectly, it won't. The way we will leverage it to simplify integer square roots is by factoring the number under the radical into two pieces, one of which is a perfect square integer.
You Should Know Plain and simple - one of the reasons that fully simplified square roots are preferable is that it makes things consistent. No two students can turn in different correct answers, if both students simplified their answers to the fullest extent.
Let's look at the simplifying $\sqrt{72}$, which we said above in passing will turn out to be $6\sqrt{2}$.

Example 1Simplify $\sqrt{72}$.$\blacktriangleright$ The way we get from starting with $\sqrt{72}$ and ending with $6\sqrt{2}$ is to first factor $72$ into two pieces, one of which is a perfect square, and then use the root product property.$$72 = 36 \cdot 2$$Note that $72$ has many factors and therefore there are many ways in which we could have written $72$ as a product, including other ways to write it with other perfect square integers. Why did we pick $36$? We'll discuss how to know which factors to choose in a minute.Since $72 = 36 \cdot 2$, let's use the root product property.$$\begin{array}{rl} \sqrt{72} & = \sqrt{36 \cdot 2} & = \sqrt{36}\sqrt{2} \end{array}$$Finally, having separated the problem into two roots with the root product property, we get the payoff - we picked the factors particularly so that one of them is a perfect square integer, so it's square root is an integer.$$\sqrt{36}\sqrt{2} = 6\sqrt{2}$$That's how it's done!Next, let's look at the difference between partially simplified and fully simplified square roots. Typically, partially simplified square roots are a mistake - not because they are incorrect but because they violate the instructions we are given to simplify fully.

Example 1 (Revisited)Simplify $\sqrt{72}$.$\blacktriangleright$ What if we hadn't though to factor $72$ as $36 \cdot 2$, but rather our first instinct was to factor it as $4 \cdot 18$?$$72 = 4 \cdot 18$$$$\longrightarrow \sqrt{72} = \sqrt{4 \cdot 18}$$$$=\sqrt{4}\sqrt{18}$$$$=2\sqrt{18}$$Because we factored $72$ into a product with one of the factors being a perfect square integer, we were able to proceed with using the root product property and obtain a simplified answer. However, it is not the same answer we got before, and furthermore, this answer ($2\sqrt{18}$) is not fully simplified. Why? Because $\sqrt{18}$ is not fully simplified. On its own, we could simplify $\sqrt{18}$ by breaking it into two factors, one of which is a perfect square integer. If we did, we would get$$2\sqrt{18} = 2 \sqrt{2 \cdot 9}$$$$= 2 \sqrt{2} \sqrt{9}$$$$= 2 \sqrt{2} (3)$$$$=6\sqrt{2}$$This is the same answer we got when we did Example 1 with a single step.
Define: Fully Simplified Integer Square RootsWhen simplifying an integer square root, your answer is fully simplified if one of the two things are true:
• The integer was a perfect square, and thus your answer is itself an integer
• Your simplified answer has a number under the radical, and that number cannot be factored into two numbers such that one of the factors is a perfect square
In the case of $6\sqrt{2}$, we know we have a fully simplified answer because $2$ cannot be factored into a product that involves a perfect square integer. In the case of the equivalent answer $2\sqrt{18}$ that we said was partially simplified, we can prove that it wasn't fully simplified because the number under the radical, $18$, can be written as a product that involves a perfect square integer: $18 = 2 \cdot 9$.
Remember! You have to simplify down completely. Partial simplification doesn't even guarantee partial credit with a lot of teachers. If you have used the root product property to simplify a square root, look again and make sure you can't go further with it. Read on below to see how the "one-half" approach guarantees success without multiple iterations.
The trick to ensuring you get the answer with one step is to choose your factoring to use the biggest possible perfect square factor. Let's look at some tricks to attaining this maximum efficiency approach.

## Amass Mastery

Overall, this is a skill that definitely gets cleaner and more refined as you practice it. Additionally, here are a few things you should do to make your life easier and ultimately to do the least amount of work while getting the correct answer.1. Know Your PerfectsAlthough you'll probably memorize them without trying, you should be able to at least quickly derive the perfect square integers so that you can test out and decide which ones will work when you seek out factors in this root simplifying process. Remember that you can derive them by counting along your integers starting from $1$, and square as you count.$$1, \,\,\,\, 2, \,\,\,\, 3, \,\,\,\, 4....$$becomes$$1^2 = \boxed{1}, \,\,\,\, 2^2 = \boxed{4}, \,\,\,\, 3^2 = \boxed{9}, \,\,\,\, 4^2 = \boxed{16}....$$2. Don't Start SmallTry and use bigger numbers when you factor, if you have a choice. The goal is to avoid coming up with a partially simplified square root that leaves you with another iteration to conduct in your answer. Students mistakenly choose this approach thinking that it's easier because the factoring is easier, but getting the question wrong because you didn't go far enough doesn't help anybody! Here's an example:$$\sqrt{512} = \sqrt{4 \cdot 128}$$$$=2\sqrt{128}$$It seemed like a good idea to take out the "easier" to identify perfect square of $4$ instead of the largest possible perfect square. The result is not wrong, but it's not complete. In fact, if you were to keep along this path and take out another $4$, you still wouldn't be done! Do you really want to play this game three or four times before you arrive at the final answer? If you want to execute this a single time, you should invest the time at the start of the problem to identify the largest perfect square factor, which in this case, would have been $256$:$$\sqrt{512} = \sqrt{2 \cdot 256}$$$$=16\sqrt{2}$$3. Use the "One-half" ApproachThere is a sure-fire way to find the largest perfect square integer factor of any larger number, which as we said, is the way to ensure you simplify integer square roots with minimal effort. The idea is that because $2$ is the smallest integer you can divide by (since dividing by $1$ doesn't actually change you answer), the biggest perfect square integer factor can be at most as large as half the number. For example, if we are trying to simplify $\sqrt{588}$, we don't need to bother trying to see if large perfect squares like $324$ are factors, since $588 \div 2 = 294$, and thus nothing between $294$ and $588$ can be a factor of $588$.At the same token, we don't want to start with checking $4$, because we may then find ourselves having to do the process over again, or be at risk for not fully simplifying. The right way to fix both concerns and get to the answer fastest is to start with half of the number, and work down from there to check each perfect square integer along the way. This works for odd numbers as well, because you can still start at approximately half the number even though you can't exactly divide by $2$.Let's see two examples of this.

Example 2Simplify $\sqrt{588}$.$\blacktriangleright$ Start with half of $588$, which is $294$, and find the first biggest perfect square as you work down from there.$17^2 = 289$, but $289$ is not a factor of $588$. $16^2 = 256$ but that doesn't work either. $15^2 = 225$, neither, but $14^2 = 196$ does!$$\sqrt{588} = \sqrt{196 \cdot 3}$$$$=\sqrt{196} \sqrt{3}$$$$=14 \sqrt{3}$$Again, this method ensures one iteration of the root product property, and even though it doesn't eliminate the need to iteratively search, we load the work up front with an easier process that we're more familiar with (factoring).

Example 3Simplify $\sqrt{675}$.$\blacktriangleright$ Half of $675$ is about $338$, so start there, with $324$, $289$, $256$, and landing on $225$. Since $675 = 225 \cdot 3$, we have$$\sqrt{675} = \sqrt{225 \cdot 3}$$$$=\sqrt{225}\sqrt{3}$$$$=15\sqrt{3}$$
Pro Tip The one-half rule is an efficient way to simplify square roots, but there is no way (aside from just plain using a computational device) to just "know" the answer. You still have to do a little thinking and grunt work. However, you will develop some intuition about what works and what doesn't. In Example 3, you should still start with $324$ per the one-half rule, but you don't have to bother checking $324$ and $256$ because they are even numbers, and $675$ is not, so they aren't going to be factors of $675$.

## Put It To The Test

Fully simplify each square root in the following examples.

Example 4$$\sqrt{200}$$
Show solution
$\blacktriangleright$ Half of the number is $100$, which is itself a perfect square. So we have$$\sqrt{200} = \sqrt{100 \cdot 2}$$$$=\sqrt{100}\sqrt{2}$$$$=10\sqrt{2}$$

Example 5$$\sqrt{1000}$$
Show solution
$\blacktriangleright$ Half of $1000$ is $500$, but this is a great example of when we can layer in some intuition instead of blindly trying every perfect square from there down. $1000$ is only "built" from $10$'s, which means we should be looking for the biggest perfect square that is built from $10$'s, $5$'s, and $2$'s. This leaves numbers like $25$, $100$, $400$, and $625$, but we know nothing that ends in a $5$ will divide into $1000$, and $100$ will be our winner here.$$\sqrt{1000} = \sqrt{100 \cdot 10}$$$$=\sqrt{100} \sqrt{10}$$$$=10\sqrt{10}$$

Example 6$$\sqrt{525}$$
Show solution
$\blacktriangleright$ Half of $525$ is about $250$ (rounding), and, again leveraging some intuition, we should expect to be looking for a perfect square factor that ends in $5$. $225$ doesn't work and the next one down that does is $25$. Note that you should recognize that it's not even worth trying even numbers to divide into $525$, since $525$ is odd.$$\sqrt{525} = \sqrt{25 \cdot 21}$$$$=\sqrt{25}\sqrt{21}$$$$=5\sqrt{21}$$

Example 7$$\sqrt{1014}$$
Show solution
$\blacktriangleright$ As the numbers climb higher, the one-half rule is really the most valuable approach, to make sure we not only get the right answer, but also do the least amount of work!Half of $1014$ is $507$, so start with $22^2 = 484$ (which we'll quickly find, doesn't work) and work down. Eventually, you should figure out that $13^2 = 169$ is the biggest perfect square that $1014$ will divide.$$\sqrt{1014} = \sqrt{169 \cdot 6}$$$$=\sqrt{169} \sqrt{6}$$$$=13\sqrt{6}$$

Lesson Takeaways
• Understand what is meant when you're asked to "simplify" integer square roots
• Be aware that you need to completely simplify, and that partially simplified answers are likely to be marked completely wrong
• Know that factoring using the integer's biggest perfect square factor will give you the correct answer in one step
• Practice to gain speed and familiarity with the types of factors that you're looking for
• Leverage the one-half trick to find the biggest perfect square factor as quickly as possible
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