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Section8.2Square Root Properties

In this chapter, we will learn how to do operations with square roots.

Figure8.2.1Alternative Video Lesson

Subsection8.2.1The Definition of Square Roots

Recall that \(\sqrt{16}=4\) because \(4^2=16\text{.}\) You can review the square root basics in Section 1.3.

Definition8.2.2

If \(y^2=x\) for a positive number \(y\text{,}\) then \(\sqrt{x}=y\text{.}\) \(\sqrt{x}\) is called the square root of \(x\text{.}\) We call expressions with a root symbol radical expressions. The number inside the radical is called the radicand

For example, \(\sqrt{2}\) and \(3\sqrt{2}\) are both radical expressions. In both expressions, the number \(2\) is the radicand.

Subsection8.2.2Multiplication and Divison Properties of Square Roots

Here is an example using perfect squares to show how we can multiply square roots:

\begin{gather*} \sqrt{9\cdot16}=\sqrt{144}=12\\ \end{gather*}

or

\begin{gather*} \sqrt{9\cdot16}=\sqrt{9}\cdot\sqrt{16}=3\cdot4=12 \end{gather*}

Whether we multiply the radicands first or take the square roots first, we get the same result. This tells us that in multiplication with radicals, we can combine factors into a single radical or separate them as needed.

Here is another example:

\begin{gather*} \sqrt{4\cdot25}=\sqrt{100}=10\\ \end{gather*}

or

\begin{gather*} \sqrt{4\cdot25}=\sqrt{4}\cdot\sqrt{25}=2\cdot5=10 \end{gather*}

Now let's look at division. When we learned how to find the square root of a fraction, we saw that the numerators and denominators could be simplified separately. This is because of the way fractions are multiplied. We multiply the numerators and denominators independently. Here is an example of two different ways to simplify a fraction in a square root:

\begin{gather*} \sqrt{\frac{4}{9}}=\sqrt{\frac{2}{3}\cdot\frac{2}{3}}=\frac{2}{3}\\ \end{gather*}

or

\begin{gather*} \sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}=\frac{2}{3} \end{gather*}

Just like in multiplication, we can separate a radical expression or combine them as needed. Let's summarize what we have learned in a theorem.

Example8.2.4

Evaluate the following expressions:

\begin{align*} a. \sqrt{2}\cdot\sqrt{32}\amp=\\ b. \sqrt{27}\cdot\sqrt{3}\amp=\\ c. \sqrt{6}\cdot\sqrt{24}\amp= \end{align*}
Solution

The radicands are not perfect squares so we need to combine them before we can evaluate the square roots:

\begin{align*} a. \sqrt{2}\cdot\sqrt{32}\amp=\sqrt{64}\\ \amp=8\\ b. \sqrt{27}\cdot\sqrt{3}\amp=\sqrt{81}\\ \amp=9\\ c. \sqrt{6}\cdot\sqrt{24}\amp=\sqrt{144}\\ \amp=12 \end{align*}
Example8.2.5

Evaluate the following expressions:

\begin{align*} a. \sqrt{\frac{1}{16}}\amp=\\ b. \sqrt{\frac{9}{16}}\amp=\\ c. \frac{\sqrt{50}}{\sqrt{2}}\amp= \end{align*}
Solution

For the first two problems, we use the property \(\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\text{:}\)

\begin{align*} a. \sqrt{\frac{1}{16}}\amp=\frac{\sqrt{1}}{\sqrt{16}}\\ \amp=\frac{1}{4}\\ b. \sqrt{\frac{9}{16}}\amp=\frac{\sqrt{9}}{\sqrt{16}}\\ \amp=\frac{3}{4} \end{align*}

For the third problem, we use the same property backward: \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\text{:}\)

\begin{align*} c. \frac{\sqrt{50}}{\sqrt{2}}\amp=\sqrt{\frac{50}{2}}\\ \amp=\sqrt{25}\\ \amp=5 \end{align*}

Subsection8.2.3Simplifying Square Roots

When we learned the square root properties we used perfect squares to make it easier. Now we can learn how to simplify a radical that does not contain a perfect square. Simplifying radicals is similar to simplifying fractions because we want the radicand to be as small as possible.

Let's use a calculator to compare \(\sqrt{12}\) and \(2\sqrt{3}\text{.}\)

\begin{equation*} \sqrt{12}=2\sqrt{3}=3.4641... \end{equation*}

These are equivalent expressions so let's see how we can simplify \(\sqrt{12}\) to \(2\sqrt{3}\text{.}\)

First, we will make a table of factor pairs for the number \(12\text{,}\) as we did in Section 7.3.

\begin{align*} 1\amp\cdot12\\ 2\amp\cdot6\\ 3\amp\cdot4 \end{align*}

We can stop after the factor \(3\) because the next factor would be \(4\) which is already in our table. Now that we have the table, we want to find the largest factor that is a perfect square.

The factor pair with the largest perfect square is \(3\cdot4\text{.}\) We will use the property of multiplying radicals to separate the perfect square from the other factor. We write the perfect square first because it will end up in front of the radical.

\begin{align*} \sqrt{12}\amp=\sqrt{4}\cdot\sqrt{3}\\ \amp=2\cdot\sqrt{3}\\ \amp=2\sqrt{3} \end{align*}

Let's look at a few more examples.

Example8.2.6

Simplify \(\sqrt{72}\text{.}\)

Solution

Here is a table of factor pairs for the number \(72\text{.}\)

\begin{align*} 1\amp\cdot72\amp4\amp\cdot18\\ 2\amp\cdot36\amp6\amp\cdot12\\ 3\amp\cdot24\amp8\amp\cdot9 \end{align*}

The largest perfect square is \(36\) so we will rewrite \(72\) as \(36\cdot2\text{.}\)

\begin{align*} \sqrt{72}\amp=\sqrt{36\cdot2}\\ \amp=\sqrt{36}\cdot\sqrt{2}\\ \amp=6\sqrt{2} \end{align*}

Notice that if we had chosen \(4\cdot18\) we could simplify the radical partially but we would need to continue and find the perfect square of \(9\) in \(18\text{.}\)

Example8.2.7

Simplify \(\sqrt{125}\text{.}\)

Solution

Here is a table of factor pairs for the number \(125\text{.}\)

\begin{align*} 1\amp\cdot125\\ 5\amp\cdot25 \end{align*}

The largest perfect square is \(25\) so we will rewrite \(125\) as \(25\cdot5\text{.}\)

\begin{align*} \sqrt{125}\amp=\sqrt{25\cdot5}\\ \amp=\sqrt{25}\cdot\sqrt{5}\\ \amp=5\sqrt{5} \end{align*}
Example8.2.8

Simplify \(\sqrt{30}\text{.}\)

Solution

Here is a table of factor pairs for the number \(30\text{.}\)

\begin{align*} 1\amp\cdot30\amp3\amp\cdot10\\ 2\amp\cdot15\amp5\amp\cdot6 \end{align*}

The number \(30\) does not have any factors that are perfect squares so it cannot be simplifed further.

Subsection8.2.4Multiplying Square Root Expressions

In Theorem 8.2.3, we learned this property

\begin{equation*} \sqrt{x}\cdot\sqrt{y}=\sqrt{xy} \end{equation*}

which we can use to multiply square root expressions. We want to simplify each radical first to keep the radicands as small as possible. Let's look at a few examples.

Example8.2.9

Multiply \(\sqrt{8}\cdot\sqrt{54}\text{.}\)

Solution

We will simplify each radical first, and then multiply them together. We do not want to multiply \(8\cdot54\) because we will end up with a larger number that is harder to factor.

\begin{align*} \sqrt{8}\cdot\sqrt{54}\amp=\sqrt{4\cdot2}\cdot\sqrt{9\cdot6}\\ \amp=2\sqrt{2}\cdot3\sqrt{6}\\ \amp=2\cdot3\sqrt{2\cdot6}\\ \amp=2\cdot3\sqrt{2\cdot2\cdot3}\\ \amp=6\cdot2\sqrt{3}\\ \amp=12\sqrt{3} \end{align*}

We could have multiplied \(2\cdot6\) inside the radical to get \(12\) and then factored 12 into \(4\cdot3\text{.}\) Whenever you find a pair of identical factors that is a perfect square.

Example8.2.10

Multiply \(\sqrt{12}\cdot\sqrt{75}\text{.}\)

Solution

\begin{align*} \sqrt{12}\cdot\sqrt{75}\amp=\sqrt{4\cdot3}\cdot\sqrt{25\cdot3}\\ \amp=2\sqrt{3}\cdot5\sqrt{3}\\ \amp=10\sqrt{3\cdot3}\\ \amp=10\cdot3\\ \amp=30 \end{align*}
Example8.2.11

Multiply \(2\sqrt{7}\cdot3\sqrt{21}\text{.}\)

Solution

\begin{align*} 2\sqrt{7}\cdot3\sqrt{21}\amp=2\cdot3\cdot\sqrt{7}\cdot\sqrt{21}\\ \amp=6\cdot\sqrt{7}\cdot\sqrt{7\cdot3}\\ \amp=6\sqrt{7\cdot7\cdot3}\\ \amp=6\cdot7\cdot\sqrt{3}\\ \amp=42\sqrt{3} \end{align*}

Note that numbers outside the radical stay outside, while numbers inside the radical stay inside unless there is a perfect square.

Example8.2.12

Multiply \(\sqrt{\frac{6}{5}}\cdot\sqrt{\frac{3}{5}}\text{.}\)

Solution

\begin{align*} \sqrt{\frac{6}{5}}\cdot\sqrt{\frac{3}{5}}\amp=\sqrt{\frac{6}{5}\cdot\frac{3}{5}}\\ \amp=\sqrt{\frac{18}{25}}\\ \amp=\frac{\sqrt{18}}{\sqrt{25}}\\ \amp=\frac{\sqrt{9\cdot2}}{5}\\ \amp=\frac{3\sqrt{2}}{5} \end{align*}

Subsection8.2.5Simplifying Square Roots with Negative Radicands

Let's look at how to simplify a square root that has a negative radicand. Remember that \(\sqrt{16}=4\) because \(4^2=16\text{.}\) So what could \(\sqrt{-16}\) be equal to? There is no real number that we can square to get \(-16\text{,}\) because

\begin{equation*} 4^2=16\text{ and }(-4)^2=16 \end{equation*}

To handle this situation, mathematicans separate the factor of \(\sqrt{-1}\) and represent it with the letter \(i\text{,}\) which stands for imaginary number. Imaginary numbers are widely used in electrical engineering, computer science and other fields. Now we can simplify square roots with negative numbers inside. Let's look at a few examples.

Example8.2.13

Simplify \(\sqrt{-2}\text{.}\)

Solution

\begin{align*} \sqrt{-2}\amp=\sqrt{-1\cdot2}\\ \amp=\sqrt{-1}\cdot\sqrt{2}\\ \amp=i\sqrt{2} \end{align*}

We write the \(i\) first because it's difficult to tell the difference between \(\sqrt{2}i\) and \(\sqrt{2i}\text{.}\)

Example8.2.14

Simplify \(\sqrt{-72}\text{.}\)

Solution

\begin{align*} \sqrt{-72}\amp=\sqrt{-1\cdot36\cdot2}\\ \amp=\sqrt{-1}\cdot\sqrt{36}\cdot\sqrt{2}\\ \amp=6i\sqrt{2} \end{align*}

Subsection8.2.6Adding and Subtracting Square Root Expressions

In Theorem 8.2.3, we learned this property for multiplication

\begin{equation*} \sqrt{x}\cdot\sqrt{y}=\sqrt{xy} \end{equation*}

but we cannot do this with addition or subtraction. Here's an example to demonstrate this:

\begin{gather*} \sqrt{9+16}=\sqrt{25}=5\\ \end{gather*}

but

\begin{gather*} \sqrt{9+16}\stackrel{?}{=}\sqrt{9}+\sqrt{16}=3+4=7\text{ incorrect!} \end{gather*}

We do not get the same result if we separate the radicals, so we must complete any additions and subtractions inside the radical first. Here is another example:

\begin{equation*} \sqrt{100-64}=\sqrt{36}=6 \end{equation*}

To add and subtract radical expressions, remember that we can only add and subtract like terms. In this case, we will call them like radicals. Compare the following:

\begin{align*} x+x\amp=2x\\ \sqrt{5}+\sqrt{5}\amp=2\sqrt{5} \end{align*}

We know the first equation is true. When we substitute \(x=\sqrt{5}\) into \(x+x=2x\text{,}\) we have the second equation. Let's look at a few more examples.

Example8.2.15

Simplify \(\sqrt{2}+\sqrt{8}\text{.}\)

Solution

\begin{align*} \sqrt{2}+\sqrt{8}\amp=\sqrt{2}+\sqrt{4\cdot2}\\ \amp=\sqrt{2}+2\sqrt{2}\\ \amp=3\sqrt{2} \end{align*}

To help you understand \(\sqrt{2}+2\sqrt{2}=3\sqrt{2}\text{,}\) think of \(x+2x=3x\text{.}\)

Example8.2.16

Simplify \(2\sqrt{3}-3\sqrt{48}\text{.}\)

Solution

\begin{align*} 2\sqrt{3}-3\sqrt{48}\amp=2\sqrt{3}-3\sqrt{16\cdot3}\\ \amp=2\sqrt{3}-3\cdot4\sqrt{3}\\ \amp=2\sqrt{3}-12\sqrt{3}\\ \amp=-8\sqrt{3} \end{align*}
Example8.2.17

Simplify \(\sqrt{2}+\sqrt{27}\text{.}\)

Solution

\begin{align*} \sqrt{2}+\sqrt{27}\amp=\sqrt{2}+\sqrt{9\cdot3}\\ \amp=\sqrt{2}+3\sqrt{3} \end{align*}

We cannot simplify the expression further because \(\sqrt{2}\) and \(\sqrt{3}\) are not like radicals.

Example8.2.18

Simplify \(\sqrt{6}-\sqrt{18}\cdot\sqrt{12}\text{.}\)

Solution

\begin{align*} \sqrt{6}-\sqrt{18}\cdot\sqrt{12}\amp=\sqrt{6}-\sqrt{9\cdot2}\cdot\sqrt{4\cdot3}\\ \amp=\sqrt{6}-3\sqrt{2}\cdot2\sqrt{3}\\ \amp=\sqrt{6}-3\cdot2\cdot\sqrt{2}\cdot\sqrt{3}\\ \amp=\sqrt{6}-6\sqrt{6}\\ \amp=-5\sqrt{6} \end{align*}

Subsection8.2.7More Complicated Square Root Operations

In Section 6.3, we learned how to multiply polynomials like \(2(x+3)\) and \((x+2)(x+3)\text{.}\) All the methods we learned apply when we multiply square root expressions. We will look at a few examples done with different methods.

When simplifying radicals it is useful to keep in mind that

\begin{equation*} \sqrt{x}\cdot\sqrt{x}=(\sqrt{x})^2=x\text{.} \end{equation*}
Example8.2.19

Multiply \(\sqrt{2}(\sqrt{3}-\sqrt{2})\text{.}\)

Solution

We will use the distributive property to do this problem:

\begin{align*} \sqrt{2}(\sqrt{3}-\sqrt{2})\amp=\sqrt{2}\sqrt{3}-\sqrt{2}\sqrt{2}\\ \amp=\sqrt{6}-\sqrt{4}\\ \amp=\sqrt{6}-2 \end{align*}
Example8.2.20

Multiply \((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})\text{.}\)

Solution

First, we will use the FOIL method to do this problem:

\begin{align*} (\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})\amp=\sqrt{3}\sqrt{3}-\sqrt{3}\sqrt{2}+\sqrt{3}\sqrt{2}-\sqrt{2}\sqrt{2}\\ \amp=3-\sqrt{6}+\sqrt{6}-2\\ \amp=1 \end{align*}

Another way is to use the Difference of Squares Formula:

\begin{align*} (a+b)(a-b)\amp=a^2-b^2\\ (\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})\amp=(\sqrt{3})^2-(\sqrt{2})^2\\ \amp=3-2\\ \amp=1 \end{align*}
Example8.2.21

Multiply \((\sqrt{3}-\sqrt{2})^2\text{.}\)

Solution

We will use the Binomial Squared Formula to do this problem:

\begin{align*} (a-b)^2\amp=a^2-2ab+b^2\\ (\sqrt{3}-\sqrt{2})^2\amp=(\sqrt{3})^2-2(\sqrt{3})(\sqrt{2})+(\sqrt{2})^2\\ \amp=3-2\sqrt{6}+2\\ \amp=5-2\sqrt{6} \end{align*}
Example8.2.22

Multiply \((\sqrt{6}+\sqrt{12})(\sqrt{3}-\sqrt{2})\text{.}\)

Solution

We will use the FOIL Method to do this problem:

\begin{align*} (\sqrt{6}+\sqrt{12})(\sqrt{3}-\sqrt{2})\amp=\sqrt{6}\sqrt{3}-\sqrt{6}\sqrt{2}+\sqrt{12}\sqrt{3}-\sqrt{12}\sqrt{2}\\ \amp=\sqrt{18}-\sqrt{12}+\sqrt{36}-\sqrt{24}\\ \amp=3\sqrt{2}-2\sqrt{3}+6-2\sqrt{6} \end{align*}
Example8.2.23

Multiply \((2\sqrt{3}-3\sqrt{2})(3\sqrt{3}-2\sqrt{2})\text{.}\)

Solution

We will use generic rectangles to do this problem. First, we set up the rectangles:

<<SVG image is unavailable, or your browser cannot render it>>

Figure8.2.24Setting up Generic Rectangles to Multiply \((2\sqrt{3}-3\sqrt{2})(3\sqrt{3}-2\sqrt{2})\)

We will multiply the sides to find the area of each rectangle:

\begin{align*} \text{Top Left Rectangle's Area: }\amp=2\sqrt{3}\cdot3\sqrt{3}\\ \amp=2\cdot3\cdot\sqrt{3}\cdot\sqrt{3}\\ \amp=6\cdot3\\ \amp=18\\ \text{Top Right Rectangle's Area: }\amp=3\sqrt{3}\cdot(-3\sqrt{2})\\ \amp=-3\cdot3\cdot\sqrt{3}\cdot\sqrt{2}\\ \amp=-9\sqrt{6}\\ \text{Bottom Left Rectangle's Area: }\amp=2\sqrt{3}\cdot(-2\sqrt{2})\\ \amp=-2\cdot2\cdot\sqrt{3}\cdot\sqrt{2}\\ \amp=-4\sqrt{6}\\ \text{Bottom Right Rectangle's Area: }\amp=(-3\sqrt{2})\cdot(-2\sqrt{2})\\ \amp=3\cdot2\cdot\sqrt{2}\cdot\sqrt{2}\\ \amp=6\cdot2\\ \amp=12 \end{align*}

Now we can fill in the area if each rectangle:

<<SVG image is unavailable, or your browser cannot render it>>

Figure8.2.25Using Generic Rectangles to Multiply \((2\sqrt{3}-3\sqrt{2})(3\sqrt{3}-2\sqrt{2})\)

Finally, we add the area of all the rectangles to find the solution:

\begin{align*} (2\sqrt{3}-3\sqrt{2})(3\sqrt{3}-2\sqrt{2})\amp=18-9\sqrt{6}-4\sqrt{6}+12\\ \amp=30-13\sqrt{6} \end{align*}

Subsection8.2.8Rationalizing the Denominator

When simplifying square root expressions, we have seen that we need to make the radicand as small as possible. Another rule is that we do not leave any radicals, which are irrational numbers, in the denominator of a fraction. In other words, we want the denominator to be rational. This process is called rationalizing the denominator.

Let's see how we can remove the square root symbol from the denominator in \(\frac{1}{\sqrt{5}}\text{.}\) If we multiply the radical by itself, we will have a perfect square that we can simplify:

\begin{equation*} \sqrt{5}\cdot\sqrt{5}=\sqrt{25}=5 \end{equation*}

To make an equivalent fraction, though, we must multiply both the numerator and denominator by the same number. If we multiply the numerator and denominator by \(\sqrt{5}\text{,}\) we have:

\begin{align*} \frac{1}{\sqrt{5}}\amp=\frac{1\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}\\ \amp=\frac{\sqrt{5}}{5} \end{align*}

We can use a calculator to verify that \(\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}=0.4472\ldots\text{.}\) Let's look at a few more examples.

Example8.2.26

Rationalize the denominator in \(\frac{6}{\sqrt{3}}\text{.}\)

Solution

\begin{align*} \frac{6}{\sqrt{3}}\amp=\frac{6\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\\ \amp=\frac{6\sqrt{3}}{3}\\ \amp=2\sqrt{3} \end{align*}

We reduce any fractions that are outside the radical.

Example8.2.27

Rationalize the denominator in \(\frac{2}{\sqrt{12}}\text{.}\)

Solution

The first method is to multiply the numerator and denominator by \(\sqrt{12}\text{,}\) and then simplify the square root expression:

\begin{align*} \frac{2}{\sqrt{12}}\amp=\frac{2\cdot\sqrt{12}}{\sqrt{12}\cdot\sqrt{12}}\\ \amp=\frac{2\sqrt{4\cdot3}}{12}\\ \amp=\frac{2\cdot2\sqrt{3}}{12}\\ \amp=\frac{4\sqrt{3}}{12}\\ \amp=\frac{\sqrt{3}}{3} \end{align*}

The second method is to simplify \(\sqrt{12}\) first, and then rationalize the denominator:

\begin{align*} \frac{2}{\sqrt{12}}\amp=\frac{2}{\sqrt{4\cdot3}}\\ \amp=\frac{2}{2\sqrt{3}}\\ \amp=\frac{1}{\sqrt{3}}\\ \amp=\frac{1\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\\ \amp=\frac{\sqrt{3}}{3} \end{align*}
Example8.2.28

Rationalize the denominator in \(\sqrt{\frac{2}{7}}\text{.}\)

Solution

\begin{align*} \sqrt{\frac{2}{7}}\amp=\frac{\sqrt{2}}{\sqrt{7}}\\ \amp=\frac{\sqrt{2}\cdot\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}\\ \amp=\frac{\sqrt{14}}{7} \end{align*}

Subsection8.2.9Exercises

Introduction to Square Roots

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Multiplying Square Root Expressions

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Simplifying Square Roots

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Simplifying Square Root Expressions

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Pythagorean Theorem

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Simplifying Square Root with Negative Radicand

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Square Root Operations

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Rationalizing the Denominator

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Trigonometry Ratios

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