Note:  Square roots are pretty tricky to teach and learn because the tendency is to seek answer-getting methods.  Patience at the onset, allowing for full development of conceptual understanding is key.  Do not revert to tricks and quick "gets" when first learning square roots.  Always revert back to the question they ask and how you know if you've answered that question.

square roots part 1

Square Roots

Part 1

Introduction: Square roots are consistently among the most misunderstood topics in developmental math. Similar to exponents, students must possess both procedural fluency but also a solid conceptual foundation and the ability to read and understand what square roots mean, in order to be proficient with them. It is often the case that problems with square roots do not lend themselves to a correct first step, but rather, offer many equally viable methods of approach.

Square Roots Ask a Question: What number squared is equal to the radicand? The radicand is the number inside the square root symbol (radical). This expression asks, what number times itself (squared) is 11?

$\sqrt{11}$

This is a number. It is not 11. It turns out this number is irrational and we can never actually write what it is more accurately than this.

Big Idea: The area of a square is calculated by squaring a side (multiplying it by itself). Since all sides of a square are equal, this about as easy of an area to calculate as possible. A square root is giving us the area of a square and asking us to find out how long a side is.

$\overline{)42}$

For example, this square has an area of forty-two. Instead of writing out the question, “How long is the side of a square whose area is forty-two?” we simply write, $\sqrt{42}.$

The majority of the confusion with square roots comes back to this definition of what a square root is. To make it as clear as possible, please consider the following table.

 English Math How long is the side of a square that has an area of 100? $\sqrt{100}$ How long is the side of a square that has an area of 10? $\sqrt{10}$

These two numbers were chosen because students inevitably write $\sqrt{100}=\sqrt{10}.$

 English Math Answer How long is the side of a square that has an area of 100? $\sqrt{100}$ 10 How long is the side of a square that has an area of 10? $\sqrt{10}$ 3.162277660168379…

Key Knowledge: In order to be proficient with square roots we need to know about perfect squares. A perfect square is a number that is the product of a number squared. Sixteen is a perfect because four times four is sixteen.

The reason you need to know perfect squares is because square roots are asking for numbers squared that equal the radicand. So if the radicand is a perfect square, we have an easy ‘get,’ that is, simplification.

For example, since 42 = 16, and the square root of sixteen $\left(\sqrt{16}\right)$ asks for what number squared is 16, the answer is just four.

Let’s take a look at the first twenty perfect squares and what number has been squared to arrive at the perfect square, which we will call the parent.

 Perfect Square “Parent” 1 1 4 2 9 3 16 4 25 5 36 6 49 7 64 8 81 9 100 10 121 11 144 12 169 13 196 14 225 15 256 16 289 17 324 18 361 19 400 20

You should recognize these numbers as perfect squares as that is a key piece of knowledge required!

Pro-Tip: When dealing with square roots it is wise to have a list of perfect squares handy to help you familiarize yourself with them.

How to Simplify a Square Root: To simplify a square root all you do is answer the question it is asking.

The best way to go about that is to see if the radicand is a perfect square. If so, then just answer the question. For example:

Simplify $\sqrt{256}$

Since this is asking, “What number squared is 256?” and 256 is a perfect square, 162, the answer to the question is just 16.

$\sqrt{256}=16$

What if we had something like this:

$\sqrt{{x}^{2}}$

If you’re confused by this, revert back to the question it is asking. This is asking, “What squared is x $–$ squared?” All you have to do is answer it.

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

What if the radicand was not a perfect square?

If you end up with an ugly square root, like $\sqrt{48},$ all you have to do is factor the radicand to find the largest perfect square.

List all factors, not just the prime factors. In fact, the prime factors are of little use because prime numbers are not perfect squares. And again, we are looking for perfect squares because they help us answer the question posed by the square root.

48

1, 48

2, 24

3, 16

4, 12

6, 8

Pro-Tip: When factoring, do not skip around. Check divisibility by all of the numbers in order until you get a turn around. For example, after 6, check 7. Seven doesn’t divide into 48, but 8 does. Eight times six is forty eight, but you already have that pair. That’s how you know you’re done!

In our list we need to find the largest perfect square. While four is a perfect square, sixteen is larger. So we need to use three and sixteen like shown below.

$\sqrt{48}=\sqrt{16}\cdot \sqrt{3}$

The square root of three is irrational (square roots of prime numbers are all irrational), but the square root of sixteen is four. So rewriting this we get:

$\begin{array}{l}\sqrt{48}=\sqrt{16}\cdot \sqrt{3}\\ \sqrt{48}=4\cdot \sqrt{3}\end{array}$

See Note 1 and Note 2 below for an explanation of why the above works.

Fact:

That means that Let’s see if it is true.

${\left(4\sqrt{3}\right)}^{2}=4\sqrt{3}×4\sqrt{3}$

Because we can change the order in which we multiply, we can rearrange this and multiply the rational numbers together first and the irrational numbers together first.

$4\sqrt{3}×4\sqrt{3}=4×4×\sqrt{3}×\sqrt{3}$

The square root of three times itself is the square root of nine.

$4×4×\sqrt{3}×\sqrt{3}=16×\sqrt{9}$

The square root of nine asks, what squared is nine. The answer to that is three.

$16×\sqrt{9}=16×3$

So,

Note 1: $4\sqrt{3}$ cannot be simplified because the square root of three is irrational. That means we cannot write it more accurately than this. Also, the product of rational number and an irrational number is irrational. So, $4\sqrt{3}$ is just written as “four root three.”

Note 2: We can separate square roots into the product of two different square roots like this:

$\sqrt{75}=\sqrt{25\cdot 3}$ figure a.

or

$\sqrt{75}=\sqrt{25}\cdot \sqrt{3}$ figure b.

If we consider the question being asked, what number squared is seventy five, we can see why this works. What number squared is seventy five is the same as what number squared is twenty five times three,” (figure a). The number squared that is twenty five times the number squared that is three is the same as the number times itself that is twenty five times three.

For example:

But also: $\sqrt{64}=\sqrt{4}\cdot \sqrt{16}$

And this simplifies to:

$\sqrt{64}=2\cdot 4=8$

What we will see in a future section is that square roots are actually exponents, exponents are repeated multiplication and the order in which you multiply does not matter. This allows us to manipulate square root expressions in such a fashion.

Let us work through two examples. Before we do, let us define what simplify means in the context of square roots. Simplify with square roots means that the radicand does not contain a factor that is a perfect square and that all terms are multiplied together.

Simplify: $9\sqrt{8}$

What is the nine doing with the square root of eight? It is multiplying by it. We cannot carry out that operation. However, eight, the radicand, does contain a perfect square, four. Do not allow the fact that 9 is also a perfect square confuse you. This is just 9, as in 1, 2, 3, 4, 5, 6, 7, 8, 9. The square root of eight cannot be counted. It is asking a question, remember?

$9\sqrt{8}=9\sqrt{4}\cdot \sqrt{2}$

Pro-Tip: When rewriting radical expressions (square roots), write the perfect square first as it is easier to manipulate (you won’t mess up as easily).

$\begin{array}{l}9\sqrt{4}\cdot \sqrt{2}\\ 9\cdot 2\cdot \sqrt{2}\\ 18\sqrt{2}\end{array}$

Example 2:

Simplify $8\cdot \sqrt{\frac{32}{{x}^{4}}}$

The eight is multiplying with the radical expression. Just like we could separate the multiplication of square roots, we can also separate the division, provided it is written as multiplication by the reciprocal. So, let’s consider these separately, to break this down into smaller pieces that are easier to manage.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{\sqrt{32}}{\sqrt{{x}^{4}}}$

Let’s factor each square root, looking for a perfect square. Note that x2 times x2 is x4.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{\sqrt{16}×\sqrt{2}}{\sqrt{{x}^{4}}}$

Let’s answer the square root questions we can answer:

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{4×\sqrt{2}}{{x}^{2}}$

Notice that 8 is a fraction 8/1.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=\frac{8}{1}×\frac{4×\sqrt{2}}{{x}^{2}}$

Multiplication of fractions is easy as π.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=\frac{32\sqrt{2}}{{x}^{2}}$

Summary: Square roots ask a question: What number squared is the radicand? This comes from the area of a square. Given the area of a square, how long is the side?

To answer the question you factor the radicand and find the largest perfect square.

Time for some practice problems:

1.7 Square Roots Part 1 Practice Set 1

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{125}=\\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{27}=\\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{162}=\\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{75}=\\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{45}=\end{array}$ $\begin{array}{l}6.\text{\hspace{0.17em}}\sqrt{4{x}^{2}}=\\ 7.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{4{x}^{4}}=\\ 8.\text{\hspace{0.17em}}\sqrt{\frac{4}{25}}=\\ 9.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{98}=\\ 10.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{48}=\end{array}$ $\begin{array}{l}11.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{4}\cdot \sqrt{8}=\\ 12.\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\sqrt{8}=\\ 13.\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\sqrt{9}=\\ 14.\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\sqrt{\frac{1}{4}}=\\ 15.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{300}=\end{array}$

1.7 Square Roots Part 1, Practice 2

Simplify problems 1 through 8.

1.      $\sqrt{24}$ 2. $\sqrt{8{x}^{3}}$

3. $\sqrt{200}$ 4. $\sqrt{27}$

5. $\sqrt{7x}$ 6. $4\sqrt{{a}^{2}b}$

7. $3x\sqrt{98{x}^{2}}$ 8. $\frac{1}{3}\cdot \sqrt{\frac{9}{{x}^{2}}}$

9. Show that ${a}^{2}b=\sqrt{{a}^{4}{b}^{2}}$ 10. Why is finding perfect squares appropriate

when simplifying square roots?