### SubsectionProduct and Quotient Rules

Radicals with the same value may be written in different forms. Consider the following calculation:

\begin{align*} \left(2\sqrt{2}\right)^2 \amp = 2^2\left(\sqrt{2}\right)^2~~~~~~~~\blert{\text{By the fourth law of exponents.}}\\ \amp = 4(2)=8 \end{align*}

This calculation shows that $2\sqrt{2}$ is equal to $\sqrt{8}\text{,}$ because the square of $2\sqrt{2}$ is 8. You can verify on your calculator that $2\sqrt{2}$ and $\sqrt{8}$ have the same decimal approximation, 2.828.

###### Example9.23.

Show that $~~\sqrt{45} = 3\sqrt{5}\text{.}$

Solution

$\sqrt{45}$ is a number whose square is 45. We can square $3\sqrt{5}$ as follows:

\begin{align*} (3\sqrt{5})^2 \amp = 3^2(\sqrt{5})^2\\ \amp = 9(5)=45 \end{align*}

Because the square of $3\sqrt{5}$ is equal to 45, it is the case that $3\sqrt{5}=\sqrt{45}\text{.}$

It is usually helpful to write a radical expression as simply as possible. The expression $2\sqrt{2}$ is considered simpler than $\sqrt{8}\text{,}$ because the radicand is a smaller number. Similarly, $3\sqrt{5}$ is simpler than $\sqrt{45}\text{.}$ In this section we discover properties of radicals that help us simplify radical expressions.

In the Activities we will verify the following properties of radicals.

\begin{equation*} \text{If}~~a,~b \ge 0,~~~\text{then}~~~~\blert{\sqrt{ab}=\sqrt{a}\sqrt{b}} \end{equation*}

\begin{equation*} \text{If}~~a \ge 0,~b \gt 0~~~\text{then}~~~~\blert{\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}} \end{equation*}
###### Caution9.24.

It is just as important to remember that we do not have a sum or difference rule for radicals. That is, in general,

\begin{gather*} \sqrt{a+b} \not= \sqrt{a}+\sqrt{b}\\ \sqrt{a-b} \not= \sqrt{a}-\sqrt{b} \end{gather*}
###### Example9.25.

Decide which statement is true and which is false.

1. $\sqrt{4} + \sqrt{9} = \sqrt{13}?$
2. $\sqrt{4}\sqrt{9} = \sqrt{36}?$
Solution
1. $\sqrt{4} + \sqrt{9} = 2+3 = 5\text{,}$ but $\sqrt{13} \not= 5\text{,}$ so the first statement is false.

2. $\sqrt{4}\sqrt{9} = 2(3) = 6\text{,}$ and $\sqrt{36} = 6\text{,}$ so the second statement is true.

### SubsectionSimplifying Square Roots

We can use the product rule for radicals to write $\sqrt{12} = \sqrt{4}\sqrt{3}\text{.}$ Now, $\sqrt{4}=2\text{,}$ so we can simplify $\sqrt{12}$ as

\begin{equation*} \sqrt{12} = \sqrt{4}\sqrt{3} = 2\sqrt{3} \end{equation*}

The expression $2\sqrt{3}$ is a simplified form for $\sqrt{12}\text{.}$ The factor of 4, which is a perfect square, has been removed from the radical. This example illustrates a strategy for simplifying radicals.

###### To Simplify a Square Root.
1. Factor any perfect squares from the radicand.
2. Use the product rule to write the radical as a product of two square roots.
3. Simplify the square root of the perfect square.
###### Example9.26.

Simplify $~\sqrt{45}\text{.}$

Solution

We look for a perfect square that divides evenly into 45. The largest perfect square that divides 45 is 9, so we factor 45 as $9 \cdot 5$ Then we use the product rule to write

\begin{equation*} \sqrt{45}=\sqrt{9 \cdot 5} = \sqrt{9}\sqrt{5} \end{equation*}

Finally, we simplify $\sqrt{9}$ to get

\begin{equation*} \sqrt{45}= \sqrt{9}\sqrt{5} =3\sqrt{5} \end{equation*}
###### Caution9.27.

Finding a decimal approximation for a radical is not the same as simplifying the radical. In Example 9.26, we can use a calculator to find

\begin{equation*} \sqrt{45} \approx 6.708 \end{equation*}

but 6.708 is not the exact value for $\sqrt{45}\text{.}$ For long calculations, too much accuracy may be lost by approximating each radical. However, $3\sqrt{5}$ is equivalent to $\sqrt{45}\text{,}$ which means that their values are exactly the same. We can replace one expression by the other without losing accuracy.

###### 1.

What is the difference between simplifying a square root and approximating a square root?

### SubsectionSquare Root of a Variable Expression

To square a power of a variable we double the exponent. For example, the square of $x^5$ is

\begin{equation*} \left(x^5\right)^2 = x^{5 \cdot 2} = x^{10} \end{equation*}

Because taking the square root of a number is the opposite of squaring a number, to take the square root of an even power we divide the exponent by 2.

###### Example9.28.

If $x$ is a nonnegative number,

\begin{equation*} \sqrt{x^{10}} = x^5~~~~~\text{because}~~~~~~(x^5)^2 = x^{10} \end{equation*}

Similarly, if $a$ is not negative,

\begin{equation*} \sqrt{a^6} = a^3~~~~~\text{because}~~~~~~(a^3)^2 = a^6 \end{equation*}

###### 2.

How do we simplify the square root of a variable raised to an even exponent?

For the rest of this section we'll assume that all variables are nonnegative.

###### Caution9.29.

Note that $\sqrt{a^{16}}$ is not equal to $a^4\text{.}$ Compare the two radicals:

\begin{equation*} \sqrt{16} = 4 ~~~~~~\text{but}~~~~~~ \sqrt{a^{16}} = a^8 \end{equation*}
###### Look Closer.

How can we simplify the square root of an odd power? We write the power as a product of two factors, one having an even exponent and one having exponent 1.

###### Example9.30.

Simplify $~~\sqrt{x^7}$

Solution

We factor $x^7$ as $x^6 \cdot x\text{.}$ Then we use the product rule to write

\begin{equation*} \sqrt{x^7} = \sqrt{x^6 \cdot x} = \sqrt{x^6} \cdot \sqrt{x} \end{equation*}

Finally, we simplify the square root of $x^6$ to get

\begin{equation*} \sqrt{x^7} = \sqrt{x^6} \cdot \sqrt{x} = x^3\sqrt{x} \end{equation*}

###### 3.

How do we simplify the square root of a variable raised to an odd exponent?

If the radicand contains more than one variable or a coefficient, we consider the constants and each variable separately. We try to remove the largest factors possible from the radicand.

###### Example9.31.

Simplify $~~\sqrt{20x^2y^3}$

Solution

We look for the largest perfect square that divides 20; it is 4. We write the radicand as the product of two factors, one of which contains the perfect square and even powers of the variables. That is,

\begin{equation*} 20x^2y^3 = 4x^2y^2 \cdot 5y~~~~~\blert{\text{Factor into perfect squares and "leftovers."}} \end{equation*}

Now we write the radical as a product.

\begin{equation*} \sqrt{20x^2y^3}=\sqrt{4x^2y^2 \cdot 5y} = \sqrt{4x^2y^2} \cdot \sqrt{5y} \end{equation*}

Finally, we simplify the first of the two factors to find

\begin{align*} \sqrt{20x^2y^3} \amp =\sqrt{4x^2y^2} \cdot \sqrt{5y}~~~~~\blert{\text{Take square root of the first factor.}}\\ \amp = 2xy\sqrt{5y} \end{align*}

###### 4.

In Example 9.31, what will we get if we square $2xy\sqrt{5y}\text{?}$

### SubsectionSums and Differences

\begin{equation*} \alert{\sqrt{a} + \sqrt{b} \not= \sqrt{a+b}} \end{equation*}

You can verify some examples on your calculator:

• $\sqrt{3} + \sqrt{5} \not= \sqrt{8}$

• $\sqrt{4} + \sqrt{16} \not= \sqrt{20}$

• $\sqrt{7} + \sqrt{7} \not= \sqrt{14}$

So, we cannot simplify a sum or difference if the expressions under the radical are different. However, we can combine radicals with the same radicand. For example, we can write

\begin{equation*} \sqrt{7}+\sqrt{7} = 2\sqrt{7} \end{equation*}

just as we write

\begin{equation*} x+x=2x \end{equation*}

###### Look Closer.

We can add or subtract like radicals in the same way that we add or subtract like terms, namely by adding or subtracting their coefficients. For example,

\begin{equation*} 2r+3r=5r \end{equation*}

where $r$ is a variable that can stand for any real number. In particular, if $r=\sqrt{2}\text{,}$ we have

\begin{equation*} 2\sqrt{2}+3\sqrt{2} = 5\sqrt{2} \end{equation*}

###### Example9.32.

Simplify if possible.

1. $7\sqrt{3}-2\sqrt{3}$
2. $3\sqrt{2}+4\sqrt{3}$
Solution
1. Because $7\sqrt{3}$ and $2\sqrt{3}$ are like radicals, we can combine them as

\begin{equation*} 7\sqrt{3}-2\sqrt{3} = 5\sqrt{3} \end{equation*}
2. However, $3\sqrt{2}$ and $4\sqrt{3}$ are not like radicals. We cannot simplify sums or differences of unlike radicals. Thus, $3\sqrt{2}+4\sqrt{3}$ cannot be combined into a single term.

###### 7.

Sometimes we must simplify the square roots in a sum or difference before we can recognize like radicals.

###### Example9.33.

Simplify $~~\sqrt{20} - 3\sqrt{50} + 2\sqrt{45}$

Solution

We simplify each square root by removing perfect squares from the radicals.

\begin{align*} \sqrt{20} - 3\sqrt{50} + 2\sqrt{45} \amp = \sqrt{\blert{4} \cdot 5} - 3\sqrt{\blert{25} \cdot 2} + 2\sqrt{\blert{9} \cdot 5}\\ \amp = \blert{2}\sqrt{5} - 3 \cdot \blert{5}\sqrt{2} + 2 \cdot \blert{3}\sqrt{5}\\ \amp = 2\sqrt{5} - 15\sqrt{2} + 6\sqrt{5} \end{align*}

Then we combine the like radicals $2\sqrt{5}$ and $6\sqrt{5}$ to get

\begin{equation*} \sqrt{20} - 3\sqrt{50} + 2\sqrt{45} = 8\sqrt{5} - 15\sqrt{2} \end{equation*}

### SubsectionSkills Warm-Up

#### ExercisesExercises

Find the missing factor.

###### 1.
$60x^9 = 3x^3 \cdot ?$
###### 2.
$16z^{16} = 4z^4 \cdot ?$
###### 3.
$108a^5b^2 = 36a^4b^2 \cdot ?$
###### 4.
$\dfrac{20}{7}m^7=4m^6 \cdot ?$
###### 5.
$\dfrac{5k^5}{9n}=\dfrac{k^4}{9} \cdot ?$
###### 6.
$\dfrac{a^2+4a^4}{8}=\dfrac{a^2}{4} \cdot ?$

### ExercisesHomework 9.3

For Problems 1–3, decide whether the statement is true or false. Then use a calculator to verify your answer.

###### 1.

$\sqrt{6}=\sqrt{2}\sqrt{3}$

###### 2.

$\sqrt{16}=\sqrt{18}-\sqrt{2}$

###### 3.

$\sqrt{5}+\sqrt{5}=\sqrt{10}$

For Problems 4–9, find the square root.

###### 4.

$\sqrt{y^8}$

###### 5.

$\sqrt{n^{36}}$

###### 6.

$\pm \sqrt{16x^4}$

###### 7.

$-\sqrt{121a^2b^6}$

###### 8.

$\sqrt{9(x+y)^2}$

###### 9.

$-\sqrt{\dfrac{64}{b^6}}$

For Problems 10–15, simplify the square root.

###### 10.

$\sqrt{8}$

###### 11.

$-\sqrt{20}$

###### 12.

$\sqrt{125}$

###### 13.

$\sqrt{x^3}$

###### 14.

$-\sqrt{b^{11}}$

###### 15.

$\sqrt{p^{25}}$

For Problems 16–24, simplify the square root.

###### 16.

$\sqrt{8a^3}$

###### 17.

$\pm \sqrt{72m^9}$

###### 18.

$\sqrt{\dfrac{x^8}{27}}$

###### 19.

$\sqrt{48c^6d}$

###### 20.

$-\sqrt{\dfrac{45}{4}b^2d^3}$

###### 21.

$\sqrt{\dfrac{9w^3}{28z}}$

###### 22.

$3\sqrt{4x^3}$

###### 23.

$-2a\sqrt{50a^3b^2}$

###### 24.

$-\dfrac{2}{3k}\sqrt{9b^3k^5}$

For Problems 25–31, simplify if possible.

###### 25.

$\sqrt{3}+2\sqrt{3}$

###### 26.

$3\sqrt{5}-5\sqrt{7}$

###### 27.

$2\sqrt{6}-9\sqrt{6}$

###### 28.

$\sqrt{20}+\sqrt{45}-2\sqrt{80}$

###### 29.

$\sqrt{3}-2\sqrt{12}-\sqrt{18}$

###### 30.

$\sqrt{8a}+\sqrt{18a}-7\sqrt{2a}$

###### 31.

$2\sqrt{5x^3}-x\sqrt{125x}-3\sqrt{20x^2}$

For Problems 32–34, find and correct the error.

###### 32.

$\sqrt{36+64} \rightarrow 6+8$

###### 33.

$\sqrt{3}+\sqrt{3} \rightarrow \sqrt{6}$

###### 34.

$\sqrt{9+x^2} \rightarrow 3+x$

Mental Exercise: For Problems 35–37, choose the best approximation for the square root. Do not use pencil, paper, or calculator.

###### 35.

$\sqrt{13}$

1. $6$
2. $6.5$
3. $3.5$
4. $4$
###### 36.

$\sqrt{72}$

1. $64$
2. $9$
3. $36$
4. $81$
###### 37.

$\sqrt{125.6}$

1. $11$
2. $12$
3. $15$
4. $25$

For Problems 38–40, write the expression as the square root of an integer. Hint: square the expression.

###### 38.

$2\sqrt{5}$

###### 39.

$2\sqrt{3}$

###### 40.

$3\sqrt{6}$