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Try This

Introducing the Square Root

Why do we call the power of \(4\) to the exponent \(2\) or \(4^2\), “four squared”?

The area of a square is where the term squaring a number comes from.

Now, what if we are given the area of a square and we need to find the side length? 

Solution

Lesson Goals

• Identify the relationship between squaring a number and taking the square root of a number.
• Discuss perfect squares.
• Approximate the square root of a positive integer that is not a perfect square.

Try This!

Joe wants to build a fence around his backyard so that his dog can play outside. His backyard is a square and has an area of about \(52\) m\(^2\).

Use this information to try and approximate the length of the square backyard.

Think about this problem, then move on to the next part of the lesson.

The Square Root of a Perfect Square Versus Non-Perfect Square

Perfect Squares and Square Roots

We can show that

\(4^2=4\times 4 = 16\)

We say that \(16\) is a perfect square because it is the square of the integer \(4\).

The first few perfect squares are:

\(1,~ 4,~ 9,~ 16,~ 25, ...\)

It would be a good idea to take a minute and be sure that you can recall the first twelve perfect squares. And to find these, you're going to take an integer and multiply it by itself.

In addition, we call \(4\) the square root of \(16\), we write \(\sqrt{16}=4\). The radical symbol, \(\sqrt{ \phantom{16} }\), is used to show a square root.

To summarize,

Do a couple more examples on your own to really establish the relationship between squaring a number and finding the square root of a number.

Check Your Understanding 1

Question

Evaluate \(\sqrt{121}\).

Answer

\(11\)

Feedback

We can confirm that \(11^2=121\).  As a result, we can write \(\sqrt{121} = 11\).

Non-Perfect Squares

Now, while we've been picking out specific numbers that are perfect squares, you may have started to notice that most integers are not perfect squares.

For instance, we know that \(25\) and \(36\) are perfect squares of the consecutive integers \(5\) and \(6\). What this means is that all of the integers between \(25\) and \(36\) cannot be perfect squares because there's no integer between \(5\) and \(6\) that we can square to achieve them.

Let's look specifically at the number \(29\). Since \(29\) is in the range from \(25\) to \(36\), we can conclude that \(29\) is not a perfect square. What this means is that the square root of \(29\) is not an integer.

So if \(\sqrt{29}\) is not an integer, what kind of number is it? To be able to answer this question, we need to be able to find a number that squares to \(29\).

Example 1

Solution 

We just made the observation that \(29\) is between the consecutive perfect squares \(25\) and \(36\).

So we know that the side length of a square with area \(25\) is going to be \(5\) because \(5^2\) is equal to \(25\). Similarly, the side length of a square with area \(36\) is \(6\) because \(6^2\) is equal to \(36\). Now, if we have a square with an area of \(29\), then the side lengths of this square should be larger than \(5\), but less than \(6\).

So we've determined that the \(\sqrt{29}\) is going to be \(5\) point something, but it's natural to ask ourselves if we can do better than this. Absolutely, we can.

Now that we have a starting point, we can start testing some numbers between \(5\) and \(6\) in order to find a good approximation.

Let's begin by trying \(5.1\) as well as \(5.9\). We can calculate \(5.1^2\) to be equal to \(26.01\). Furthermore, \(5.9^2\) is equal to \(34.81\).

Now although all these values are closer to \(29\) than our original numbers \(25\) and \(36\) were, we can still get closer and find a better approximation.

As we go through this, keep in mind that we do not need to test all values. Right away, since \(5.1^2\) is equal to \(26.01\), I can estimate that \(5.2^2\) is probably not going to be as close to \(29\) as we can get. So I'm going to skip to \(5.3\) and try that. \(5.3^2\) is equal to \(28.09\). The square of \(5.3\) is even closer to \(29\).

So let's now try \(5.4\) and see if we can do better again. \(5.4^2\) is equal to \(29.16\).

It looks like the square of \(5.4\) is as close to \(29\) as we can get using only one decimal place.

We already showed that the square of \(5.3\) is farther away, and if we square a number larger than \(5.4\), then the results are going to be larger still than \(29.16\), but we haven't actually found an exact answer.

If we increase the number of decimal digits that we can use, and we explore numbers now between \(5.3\) and \(5.4\), would we actually come across an exact answer?

If you continue, you would start to build the following number

\(5.385164...\)

But no matter how many digits you add, you will never find a number that squares to exactly \(29\).

In fact, no rational number squares to \(29\), so \(\sqrt{29}\) is not rational.

Take a moment and think back to what the word rational means. If a number is not rational, then that means it cannot be written as a fraction, meaning that it is not a terminating or a repeating decimal.
\(\sqrt{29}\) is a number that is a non-terminating, non-repeating decimal, and with this in mind, the best we can do is approximate the value of \(\sqrt{29}\).

To finish our question and to wrap up our example, we conclude that

\(\sqrt{29}\approx 5.4 \text{ since } 5.4^2=29.16\)

Before we move on to the next part of our lesson, notice that to show \(\sqrt{29}\) is approximately \(5.4\), we have used a squiggly equal sign, \(\approx\). This is to show that the value is approximate and not equal.

Square Roots of Non-Perfect Squares

Example 2

The square root of a non-perfect square is irrational, meaning that it is a never ending, never repeating decimal number. If we are trying to find the square root of a positive integer that is not a perfect square, then the best we can do is approximate its value. Let's do a couple more examples to practice this calculation.

Solution

Our first step to approximating the value of the \(\sqrt{95}\) is to find perfect squares on either side of \(95\).

We know that \(81\) is less than \(95\), and we also know that \(81\) is a perfect square, since \(9^2\) is equal to \(81\). The next perfect square would be \(10^2\), which is equal to \(100\).

So if we bring all of this information together, since \(95\) is between the consecutive perfect squares \(81\) and \(100\), a square with an area of \(95\) will have a side length that is greater than \(9\) but less than \(10\).

\(9 \lt \sqrt{95} \lt 10\)

Now we ask ourselves, can we do better than this? And just as before, our answer is yes, we can.

Now that we have a starting point, we can start testing some numbers between \(9\) and \(10\). Keep in mind what I said in our last example, we don't have to test every single value. We can use different strategies to narrow in on a good approximation.

Testing:  

So instead of working my way from the ends, working from \(9\) and \(10\) and coming in from the boundaries, I'm going to start in the middle and see if we can narrow in to our value faster.

What this tells us is that \(9.7\) is a good approximation. We can conclude that since \(9.7^2=94.09\), then

\(\sqrt{95} \approx 9.7\)

Try This Problem Revisited

We now have all the tools that we need to solve the try this problem.

Solution

We need to approximate \(\sqrt{52}\) to one decimal place.

Our first step to approximating the value of the \(\sqrt{52}\) is to find perfect squares on either side of \(52\).

We know that \(49\) is less than \(52\), and \(49\) is a perfect square, since \(7^2\) is equal to \(49\). The next perfect square would be \(8^2\), which is equal to \(64\).

\(\underbrace{49}_{=7^2} \lt 52 \lt \underbrace{64}_{= 8^2}\)

A square with an area of \(52\) will have a side length that is larger than \(7\) but less than \(8\).

The next thing I notice is that \(52\) is significantly closer to \(49\) than it is to \(64\). So I'm going to start my approximation with a value that is closer to \(7\) than it is to \(8\).

Testing:

In fact, it turns out that \(7.2\) is a good approximation for the value of the \(\sqrt{52}\) when we're only using one decimal place. Since \(7.2^2=51.84\), then

\(\sqrt{52} \approx 7.2\)

Therefore, Joe's backyard is approximately \(7.2\) m long by \(7.2\) m wide.

Check Your Understanding 2

Question

Approximate \(\sqrt{42}\) to one decimal place, without using a calculator.

Answer

\(6.5\)

Feedback

To approximate the value of \(\sqrt{42}\) we must find the perfect squares on either side of \(38\).  

\(\underbrace{36}_{6^2} \lt 42 \lt \underbrace{49}_{7^2}\)

Next we test values:

\(6.4^2=40.96\)

\(6.5^2=42.25\)

\(6.6^2=43.56\)

We conclude that \(6.5\) is a good approximation. Since \(6.5^2=42.25, \sqrt{42} \approx 6.5\). 

Squaring Negative Numbers

Squaring Negative Numbers

We know that \(4^2=16\).  

Take a moment and try this problem on your own.

Solution

We conclude that both \(4\) and \(-4\) square to \(16\).  

Remember that we write \(\sqrt{16}=4\), so we may also write \(-\sqrt{16}=-4\).

Example 3

Take a moment and try this problem on your own.

Solution

We recognize \(81\) to be a perfect square.

We conclude that \(9\) and \(-9\) are the two numbers that square to \(81\).

Check Your Understanding 3

Question

Find the two integers that square to \(144\).

Answer

\(12\) and \(-12\)

Feedback

We recognize \(144\) to be a perfect square.  Since \(12^2=144\), we have \(\sqrt{144}=12\). We know the second number that will square to \(144\) is \(-\sqrt{144}=-12\).  

We verify:

\( 12^2 = 144\) and \( (-12)^2 = 144 \)

Example 4

Solution

We can conclude that \(3.6\) and \(-3.6\) are good approximations for the two numbers that square to \(13\).

Take It With You

Consider the expressions

\(\sqrt{(9+16)} \quad \text{ and } \quad \sqrt{9} + \sqrt{16}\)

1. Describe the difference between the two expressions in words.
2. Evaluate each expression and compare the answers.