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Section 9.4 Operations on Radicals

Subsection Products

We have used the product rule for radicals,

\begin{equation*} \sqrt{ab} = \sqrt{a} \sqrt{b} \end{equation*}

to simplify square roots. We can also use the product rule to multiply radicals together.

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

For example,

\begin{equation*} \sqrt{2} \cdot \sqrt{3}=\sqrt{6}~~~~~~\text{and}~~~~~ \sqrt{5} \cdot \sqrt{x}=\sqrt{5x} \end{equation*}
Example 9.34.

Find the product and simplify \(\sqrt{2x}\sqrt{10xy}\)

Solution

First, we apply the product rule.

\begin{align*} \sqrt{2x}\sqrt{10xy} \amp =\sqrt{20x^2y}~~~~~~~~~~\blert{\text{Factor out perfect squares.}}\\ \amp = \sqrt{4x^2}\sqrt{5y} = 2x\sqrt{5y} \end{align*}

Reading Questions Reading Questions

1.

Which of these statements is correct:

  1. \(\sqrt{x^2+4} = \sqrt{x^2}+\sqrt{4}\)
  2. \(\sqrt{4x^2} = \sqrt{4}\sqrt{x^2}\)
  3. \(\sqrt{x^2-4} = \sqrt{x^2}-\sqrt{4}\)
  4. \(\sqrt{\dfrac{x^2}{4}} = \dfrac{\sqrt{x^2}}{\sqrt{4}}\)

We use the distributive law to remove parentheses from products involving radicals. Compare the product involving radicals on the left with the more familiar use of the distributive law on the right.

\begin{equation*} 6(\sqrt{3} + \sqrt{2}) = 6\sqrt{3} + 6\sqrt{2}~~~~~~~~6(x+y)=6x+6y \end{equation*}
Example 9.35.

Multiply \(~~5(3x+4\sqrt{2})\)

Solution

Apply the distributive law to find

\begin{equation*} \blert{5}(3x+4\sqrt{2}) = \blert{5} \cdot 3x + \blert{5} \cdot 4\sqrt{2} = 15x + 20\sqrt{2} \end{equation*}

Reading Questions Reading Questions

2.

Explain why this statement is incorrect: \(5 \cdot 4\sqrt{2} = 20 \cdot 5 \sqrt{2}\)

In the next example, we use the distributive law along with the product rule to simplify the product.

Example 9.36.

Multiply \(~~2\sqrt{5}(3+\sqrt{3})\)

Solution

Apply the distributive law to obtain

\begin{align*} \blert{2\sqrt{5}}(3+\sqrt{3}) \amp =\blert{2\sqrt{5}} \cdot 3 + \blert{2\sqrt{5}} \cdot \sqrt{3}\\ \amp ~~~~~~~~~~\blert{\text{Apply the product rule to the second term.}}\\ \amp = 6\sqrt{5} + 2\sqrt{15} \end{align*}
Caution 9.37.

Note the difference between the products

\begin{equation*} \sqrt{5} \cdot 3 = 3\sqrt{5} ~~~~~\text{and}~~~~~ \sqrt{5} \cdot \sqrt{3} = \sqrt{15} \end{equation*}

The rule \(\sqrt{a}\sqrt{b} = \sqrt{ab}\) applies to the second product, but not the first.

Reading Questions Reading Questions

3.

Explain how to multiply \((3\sqrt{2})(4\sqrt{2})\)

To multiply binomials, we use the FOIL method.

Example 9.38.

Multiply \(~~\left(2+\sqrt{3}\right)\left(1-2\sqrt{3}\right)\text{,}\) then simplify.

Solution
\begin{align*} \left(2+\sqrt{3}\right)\left(1-2\sqrt{3}\right) \amp = 2 \cdot 1-2 \cdot 2\sqrt{3}+\sqrt{3} \cdot 1-\sqrt{3}\cdot 2\sqrt{3}\\ \amp ~~~~~~~~~~\blert{\text{F}~~~~~~~~~~~~~~\text{O}~~~~~~~~~~~~~~~~~~\text{I}~~~~~~~~~~~~~~~~~~\text{L}}\\ \amp = 2-4\sqrt{3}+\sqrt{3}-2\cdot 3 ~~~~~~~~\blert{\text{Note that}~~\sqrt{3} \cdot \sqrt{3} = 3}\\ \amp = 2-4\sqrt{3}+\sqrt{3}-6 ~~~~~~~~~~~~~~\blert{\text{Combine like terms.}}\\ \amp = -4-3\sqrt{3} \end{align*}

Reading Questions Reading Questions

4.

Explain why \(\sqrt{b}\sqrt{b}=b\)

Subsection Quotients

We use the quotient rule to simplify square roots of fractions by writing \(\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}\text{.}\) We can also use the quotient rule to simplify quotients of square roots.

Quotient Rule.
\begin{equation*} \text{If}~~a \ge 0~~\text{and}~~b \gt 0,~~\text{then}~~~~\blert{ \dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}} } \end{equation*}
Example 9.39.

Simplify \(~~\dfrac{2\sqrt{30a}}{\sqrt{6a}}\)

Solution

We use the quotient rule to write the expression as a single radical. Then we simplify the fraction inside the radical.

\begin{align*} \dfrac{2\sqrt{30a}}{\sqrt{6a}} \amp = 2\sqrt{\dfrac{30a}{6a}}~~~~~~~~\blert{\text{Reduce the fraction.}}\\ \amp = 2\sqrt{5} \end{align*}

Reading Questions Reading Questions

5.

Is it true that we can only simplify a product or quotient of radicals if they are like radicals?

Subsection Fractions

When we solve quadratic equations, the solutions are often fractions that involve square roots. Many such fractions can be simplified using properties of radicals.

Example 9.40.

One of the solutions of the equation \(4x^2-8x=1\) is \(\dfrac{8+\sqrt{80}}{8}\text{.}\) Simplify this radical expression.

Solution

First, simplify the square root: \(\sqrt{80}=\sqrt{16 \cdot 5} = 4\sqrt{5}\text{.}\) Thus,

\begin{align*} \dfrac{8+\sqrt{80}}{8} \amp = \dfrac{8+4\sqrt{5}}{8} \amp \amp \blert{\text{Factor numerator and denominator.}}\\ \amp = \dfrac{\cancel{\blert{4}}(2+\sqrt{5})}{\cancel{\blert{4}} \cdot 2} \amp \amp \blert{\text{Divide out common factors.}}\\ \amp = \dfrac{2+\sqrt{5}}{2} \end{align*}
Caution 9.41.

In Example 9.40, note that

\begin{equation*} 8+4\sqrt{5} \not= 12\sqrt{5} \end{equation*}

because \(8\) and \(4\sqrt{5}\) are not like terms. Also note that

\begin{equation*} \dfrac{8+\sqrt{80}}{8} \not= \sqrt{80} \end{equation*}

Because 8 is a term of the numerator, not a factor, we cannot cancel the 8's.

Reading Questions Reading Questions

6.

Which of these cannot be reduced: \(~~\dfrac{4+\sqrt{6}}{2}~~\) or \(~~\dfrac{4\sqrt{6}}{2}~~\) Why?

We can add or subtract fractions involving radicals.

Example 9.42.

Subtract \(~~\dfrac{1}{2}-\dfrac{\sqrt{3}}{3}\)

Solution

The LCD for the two fractions is 6. We build each fraction to an equivalent one with denominator 6, then combine the numerators.

\begin{align*} \dfrac{1}{2}-\dfrac{\sqrt{3}}{3} \amp = \dfrac{1 \cdot \blert{3}}{2\cdot \blert{3}}-\dfrac{\sqrt{3}\cdot \blert{2}}{3\cdot \blert{2}}\\ \amp = \dfrac{3}{6}-\dfrac{2\sqrt{3}}{6} = \dfrac{3-2\sqrt{3}}{6} \end{align*}

Subsection Rationalizing the Denominator

For some applications, it is easier to work with expressions that do not have radicals in the denominators of fractions. For example, the fraction \(~\dfrac{\sqrt{2}}{\sqrt{3}}~\) is equivalent to

\begin{equation*} \dfrac{\sqrt{2}\cdot \blert{\sqrt{3}}}{\sqrt{3}\cdot \blert{\sqrt{3}}} = \dfrac{\sqrt{6}}{3} \end{equation*}

We multiplied the numerator and denominator by \(\sqrt{3}\text{,}\) the same root that appeared in the denominator originally, and thus eliminated the radical from the denominator. This process is called rationalizing the denominator.

Example 9.43.

Rationalize the denominator of \(~~\dfrac{10}{\sqrt{50}}\)

Solution

We simplify the radical before attempting to rationalize the denominator.

\begin{equation*} \dfrac{10}{\sqrt{50}} = \dfrac{10}{5\sqrt{2}}=\dfrac{2}{\sqrt{2}} \end{equation*}

Now we multiply top and bottom of the fraction by \(\sqrt{2}\text{.}\)

\begin{equation*} \dfrac{2 \cdot \blert{\sqrt{2}}}{\sqrt{2}\cdot \blert{\sqrt{2}}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2} \end{equation*}

Reading Questions Reading Questions

7.

What does it mean to rationalize the denominator?

8.

What should you do before trying to rationalize a denominator?

Subsection Skills Warm-Up

Exercises Exercises

Each simplification is incorrect. Give the correct simplification, or write c.b.s. if the expression cannot be simplified.

1.

\(2+4\sqrt{5} \rightarrow 6\sqrt{5}\)

2.

\(\dfrac{2+3\sqrt{5}}{2} \rightarrow 3\sqrt{5}\)

3.

\(3\sqrt{5}+3\sqrt{2} \rightarrow 3\sqrt{7}\)

4.

\(\dfrac{4+\sqrt{6}}{2} \rightarrow 2+\sqrt{3}\)

5.

\(2(3\sqrt{5}) \rightarrow 6\sqrt{10}\)

6.

\(2(3+\sqrt{5}) \rightarrow 6+\sqrt{10}\)

7.

\((3\sqrt{5})(4\sqrt{5}) \rightarrow 12\sqrt{5}\)

8.

\(3\sqrt{5}+4\sqrt{5} \rightarrow 7\sqrt{10}\)

Solutions Answers to Skills Warm-Up

Exercises Exercises

Exercises Homework 9.4

For Problems 1–3, simplify the product.

1.

\(\sqrt{8}\sqrt{2}\)

2.

\(\sqrt{2x}\sqrt{3x}\)

3.

\(\left(3\sqrt{8a}\right)\left(a\sqrt{18a}\right)\)

For Problems 4–6, multiply.

4.

\(\sqrt{2}(3+\sqrt{3})\)

5.

\(\sqrt{5}(4+2\sqrt{15})\)

6.

\(\left(2\sqrt{p}\right)\left(\sqrt{2p}-p\sqrt{2}\right)\)

For Problems 7–10, multiply.

7.

\((3+\sqrt{2})(1-\sqrt{2})\)

8.

\((4-\sqrt{a})(4+\sqrt{a})\)

9.

\((2+\sqrt{3})^2\)

10.

\((2\sqrt{w}+\sqrt{5})(\sqrt{w}-2\sqrt{5})\)

For Problems 11–13, simplify the quotient.

11.

\(\dfrac{\sqrt{18}}{\sqrt{2}}\)

12.

\(\dfrac{\sqrt{75x^3}}{\sqrt{3x}}\)

13.

\(\dfrac{\sqrt{48b}}{\sqrt{27b}}\)

For Problems 14–16, reduce if possible.

14.

\(\dfrac{9-3\sqrt{5}}{3}\)

15.

\(\dfrac{-8+\sqrt{8}}{4}\)

16.

\(\dfrac{6a-\sqrt{18}}{6a}\)

For Problems 17–20, write the expression as a single fraction in simplest form.

17.

\(\dfrac{5}{4}+\dfrac{3\sqrt{2}}{2}\)

18.

\(\dfrac{3}{2a}+\dfrac{\sqrt{3}}{6a}\)

19.

\(\dfrac{3\sqrt{3}}{2}+3\)

20.

\(\dfrac{3}{4}-2\sqrt{y}\)

For Problems 21–24, find and correct the error in the calculation.

21.

\(8\sqrt{7}-2\sqrt{5} \rightarrow 6\sqrt{2}\)

22.

\(6\sqrt{8} \rightarrow \sqrt{48}\)

23.

\(\dfrac{5-10\sqrt{3}}{5} \rightarrow -10\sqrt{3}\)

24.

\(\sqrt{x^2+16} \rightarrow x+4\)

For Problems 25–28, rationalize the denominator.

25.

\(\dfrac{5}{\sqrt{2}}\)

26.

\(\dfrac{a\sqrt{2}}{\sqrt{a}}\)

27.

\(\dfrac{b\sqrt{21}}{\sqrt{3b}}\)

28.

\(\sqrt{\dfrac{7x}{12}}\)

29.
  1. Write an expression for the height of an equilateral triangle of side \(w\text{.}\) (See the figure at right.)
  2. Write an expression for the area of the triangle in part (a).
triangle
30.
  1. Write an expression in terms of \(k\) for the height of the pyramid shown at right.
  2. Write an expressionin terms of \(k\) for the area of the pyramid.
pyramid

For Problems 31–33, verify by substitution that the given value is a solution of the equation.

31.

\(t^2-2\sqrt{3}t+3=0,~~~~~~~~t=\sqrt{3}\)

32.

\(s^2+1=4s,~~~~~~~~s=\sqrt{3}+2\)

33.

\(x^2+3x+1=0,~~~~~~~~x=\dfrac{-3+\sqrt{5}}{2}\)