In this section, we'll look at two things. First, what *is* a **square root** (and at the end of the page, what is a root at all)? Second, how can we simplify square roots so that we can express $\sqrt{20}$ in its simpler form, $2 \sqrt{5}.$ That's a skill you'll need as you move along in your math studies.

You can always use a calculator to say that

$$\sqrt{1000} = 31.623,$$

but that's just an approximation. Though it might be good enough for whatever your purpose is, in mathematics, we like to be exact, so we'll want to express $\sqrt{1000}$ as $10 \sqrt{10}.$ We'll see how to do that later.

The square root sign, $\sqrt{\phantom{000}}$, is called a **radical**. When written as

$$\sqrt{x},$$

it asks the question, what number, when multiplied by itself (i.e. when squared) gives x? Some roots are easy to figure out. For example, $\sqrt{4} = 2$ because $2^2 = 2 \cdot 2 = 4.$

We need to be careful with that example, however, because every number has precisely two square roots. Notice that while $(2)^2 = 4, \; (-2)^2 = 4$ as well. So when we find the square root of 4, we should write

$$\sqrt{4} = ± 2,$$

using the **plus-or-minus** sign ( ± ). Here are more easy examples:

$$ \begin{matrix} \sqrt{1} = ± 1 && \sqrt{36} = ± 6 \\[4pt] \sqrt{4} = ± 5 && \sqrt{49} = ± 7 \\[4pt] \sqrt{9} = ± 3 && \sqrt{64} = ± 8 \\[4pt] \sqrt{16} = ± 4 && \sqrt{81} = ± 9 \\[4pt] \sqrt{25} = ± 5 && \sqrt{100} = ± 10 \end{matrix}$$

We call squaring and the square-root **inverse** operations because they are, in a sense, opposites. One operation on a certain number undoes the operation of another. For example, we know that

$$\sqrt{16} = ±4$$

But it's also true that

$$(±2)^2 = 4.$$

Now think of one operation inside of the other:

$$ \begin{align} (\sqrt{4})^2 &= (±2)^2 = 4 \; \; \color{#E90F89}{\text{ and}} \\[4pt] \sqrt{4^2} &= \sqrt{16} = ±4 \end{align}$$

In general, we have:

$$(\sqrt{x})^2 = \sqrt{x^2} = x$$

If we successively square, then take the square root, or square a root, it's like we really did nothing at all, because these operations "undo" one another.

*

Now we have a bunch of easy square roots like $\sqrt{16}, \; \sqrt{25},$ and so on, but what about something like $\sqrt{12}$ ? We'd like to be able to simplify any square-root expression as much as possible. We'll cover that below, but first we should look into the properties of square roots and we'll need to take a detour to prime numbers.

Proof: Let $x = a\cdot a$ and $y = b\cdot b$, then

$$ \begin{align} \sqrt{x\cdot y} &= \sqrt{aa \cdot bb} \\[4pt] &= \sqrt{ab \cdot ab} \\[4pt] &= \sqrt{(ab)^2} \\[4pt] &= ab, \; \text{ which} \\[4pt] &= \sqrt{a^2}\sqrt{b^2} = \sqrt{x}\sqrt{y} \end{align}$$

This can be proved in a way similar to the product property above.

Here's a contradiction:

$$ \begin{align} \sqrt{16 + 9} &\overset{?}{=} \sqrt{16} + \sqrt{9}\\[4pt] \sqrt{25} &\overset{?}{=} 4 + 3 \\[4pt] 5 &\ne 7 \end{align}$$

Negative numbers are not in the domain of the square root function, so we won't worry about them here. Later you'll learn how to deal with roots of negative numbers using imaginary numbers (see complex numbers).

The first thing we need to do is to find a list of small prime numbers that will mostly be our factors in simplifying square roots. To do this, we reproduce the **seive of Eratosthenes**. Eratosthenes (about 300 BCE) was former head of the Library of Alexandria, a Greek mathematician and philosopher. He is famous for being first to calculate the diameter of Earth using shadows measured at the same time from distant points. He also worked with prime numbers.

The idea of Eratoshenes' "seive" is to write the first 100 integers, like in the table below. Numbers 1, 2 and 3 are circled because they are prime. That is, they're divisible only by themselves and 1; they have no other factors.

Now all even numbers are divisible by 2, so no even number except for 2 can be prime. We'll cross out the other evens:

On to 3. All numbers except for 3 that are multiples of 3 (6, 9, 12, ...) can't be prime because they have a factor of 3. We'll cross out all of the multiples of 3:

Now all multipls of 4 are also multiples of 2, so we've eliminated those already. Next is 5. Let's cross out all multiples of 5, which lie in the 5th and last columns.

Now crossing out the multiples of 7 gets rid of 49 and 91. In one more step, eliminating multiples of 11 takes out 77, and that's actually it. You can look for multiples of other primes like 13, 17, but they're all covered up.

So we're left with 25 prime numbers in the first 100 integers. They are

These prime numbers will be helpful for learning how to reduce roots to their simplest forms. We'll be looking to find all prime factors of the number under the radical.

A theorem called the **fundamental theorem of arithmetic** says that

**any number is either prime or has a unique set of factors, all of which are prime.**

It's that unique set of prime factors we'll be looking for in simplifying roots below.

Now let's simplify some square roots. There are two methods for doing this, but the difference is only cosmetic. Choose the one you're most comfortable with or might have already seen. Let's begin by simplifying $\sqrt{320}.$ Our goal will be to pick 320 apart, bit-by-bit, by finding prime factors. For example, 320 is even, so 2 is obviously a factor: $320 = 2 \times 160.$ The prime number 2 is also a factor of 160: $160 = 2 \times 180,$ and so on.

One way to organize this process, with which we aren't done yet, is the **factor tree**. Here is the factor tree for 320:

Notice that each line contains a prime factor and the remaining factor of the number above. Once we get to the bottom of the tree, all that's left are prime numbers, so we have

$$320 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$$

Now notice that there are three pairs of 2 there,

So we can also write

$$320 = 4 \times 4 \times 4 \times 5$$

Using the properties of square roots, we also see that

$$ \begin{align} \sqrt{320} &= \sqrt{4\cdot 4\cdot 4\cdot 5} \\[4pt] &= \sqrt{4} \, \sqrt{4} \, \sqrt{4} \, \sqrt{5} \\[4pt] &= 2 \cdot 2 \cdot 2 \cdot \sqrt{5} \\[4pt] &= 8 \sqrt{5} \end{align}$$

This is the simplest way we can represent $\sqrt{320}$ using integers. It is exact, not a decimal approximation.

The second method is very similar, it's just a matter of doing repeated divisions by known factors. The same full process for the prime factors of 320 would look like this:

The idea is very similar. We divide 320 by 2 (an obvious prime factor) with the result of 160. Divide by 2 again to get 180. Keep dividing by 2 until we get 5, a prime number, and then we're done. We collect the pairs of prime factors as above and reconstruct our root.

Now let's do a couple more examples using both methods.

Our factorization of the root, and the result is

$$ \begin{align} \sqrt{162} &= \sqrt{2 \cdot 3 \cdot 3 \cdot 3 \cdot 3} \\[4pt] &= \sqrt{2 \cdot 9\cdot 9} \\[4pt] &= \sqrt{2} \cdot \sqrt{9} \cdot \sqrt{9} \\[4pt] &= 3\cdot 3 \sqrt{2} = \bf 9 \sqrt{2} \end{align}$$

You might also have noticed that 162 factors to 81 × 2, which would lead to the same result.

Our factorization of the root, and the result is

$$ \begin{align} \sqrt{1000} &= \sqrt{2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5} \\[4pt] &= \sqrt{2 \cdot 5 \cdot 4 \cdot 25} \\[4pt] &= \sqrt{10} \cdot \sqrt{4} \cdot \sqrt{25} \\[4pt] &= \bf 10 \sqrt{10} \end{align}$$

You can use this problem generator to generate square roots for you to simplify. Note that you won't be able to simplify them all. Some, for example, will be roots of prime numbers.

Practice as much as you can. Get your confidence up. You'll be simplifying roots throughout your studies in math and science. It's an important skill to have.

Hit answer to view solution.

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