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Section 9.3 Properties of Radicals

Subsection Product 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.

Example 9.23.

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.

Product Rule for Radicals

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

Quotient Rule for Radicals

\begin{equation*} \text{If}~~a \ge 0,~b \gt 0~~~\text{then}~~~~\blert{\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}} \end{equation*}
Caution 9.24.

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*}
Example 9.25.

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.

Subsection Simplifying 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.
Example 9.26.

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*}
Caution 9.27.

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.

Reading Questions Reading Questions

1.

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

Subsection Square 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.

Example 9.28.

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*}

Reading Questions Reading Questions

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.

Caution 9.29.

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.

Example 9.30.

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*}

Reading Questions Reading Questions

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.

Example 9.31.

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*}

Reading Questions Reading Questions

4.

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

Subsection Sums and Differences

What about sums and differences of radicals? We cannot add or subtract radicals by combining their radicands. That is,

\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*}

Square roots with identical radicands are called like radicals.

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*}

Thus, we may add like radicals by adding their coefficients. The same idea applies to subtraction.

Reading Questions Reading Questions

5.

What are like radicals?

Example 9.32.

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.

Reading Questions Reading Questions

6.

How do we add or subtract like radicals?

7.

How do we add or subtract unlike radicals?

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

Example 9.33.

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*}

Subsection Skills Warm-Up

Exercises Exercises

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.
\(\dfrac{a^2+4a^4}{8}=\dfrac{a^2}{4} \cdot ?\)

Solutions Answers to Skills Warm-Up

Exercises Exercises

Exercises Homework 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.

\(3\sqrt{6}\)