Section 6.2 Rationalizing the Denominator
A radical expression typically has several equivalent forms. For example, \(\frac{\sqrt{2}}{3}\) and \(\frac{2}{\sqrt{6}}\) are the same number. Mathematics has a preference for one of these forms over the other, and this section is about how to convert a given radical expression to that form.
Subsection 6.2.1 Rationalizing the Denominator
To simplify radical expressions, we have seen that it helps to make the radicand as small as possible. Another helpful principle is to not leave any irrational numbers, such as \(\sqrt{3}\) or \(2\sqrt{5}\text{,}\) in the denominator of a fraction. In other words, we want the denominator to be rational. The process of dealing with such numbers in the denominator is called rationalizing the denominator.
Let’s see how we can replace \(\frac{1}{\sqrt{5}}\) with an equivalent expression that has no radical expressions in its denominator. If we multiply a radical by itself, the result is the radicand, by Definition 6.1.2. As an example:
\begin{equation*}
\sqrt{5}\cdot\sqrt{5}=5
\end{equation*}
With \(\frac{1}{\sqrt{5}}\text{,}\) we may multiply both the numerator and denominator by the same nonzero number and have an equivalent expression. If we multiply the numerator and denominator by \(\sqrt{5}\text{,}\) we have:
\begin{align*}
\frac{1}{\sqrt{5}}\amp=\frac{1}{\sqrt{5}} \multiplyright{\frac{\sqrt{5}}{\sqrt{5}}}\\
\amp=\frac{1\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}\\
\amp=\frac{\sqrt{5}}{5}
\end{align*}
And voilà, we have an expression with no radical in its denominator. We can use a calculator to verify that \(\frac{1}{\sqrt{5}}\approx0.4472\text{,}\) and also \(\frac{\sqrt{5}}{5}\approx0.4472\text{.}\) They are equal.
Example 6.2.2.
Rationalize the denominator of the expressions.
- \(\displaystyle \frac{3}{\sqrt{6}}\)
- \(\displaystyle \frac{\sqrt{5}}{\sqrt{72}}\)
Explanation.
- To rationalize the denominator of \(\frac{3}{\sqrt{6}}\text{,}\) we multiply both the numerator and denominator by \(\frac{\sqrt{6}}{\sqrt{6}}\text{.}\)\begin{align*} \frac{3}{\sqrt{6}}\amp=\frac{3}{\sqrt{6}}\multiplyright{\frac{\sqrt{6}}{\sqrt{6}}}\\ \amp=\frac{3\sqrt{6}}{6}\\ \amp=\frac{\sqrt{6}}{2} \end{align*}Note that we reduced a fraction \(\frac{3}{6}\) whose numerator and denominator were no longer inside the radical.
- To rationalize the denominator of \(\frac{\sqrt{5}}{\sqrt{72}}\text{,}\) we could multiply both the numerator and denominator by \(\sqrt{72}\text{,}\) and it would be effective; however, we should note that the \(\sqrt{72}\) in the denominator can be reduced first. Doing this will simplify the arithmetic because there will be smaller numbers to work with.\begin{align*} \frac{\sqrt{5}}{\sqrt{72}}\amp=\frac{\sqrt{5}}{\sqrt{36\cdot 2}}\\ \amp=\frac{\sqrt{5}}{\sqrt{36}\cdot\sqrt{2}}\\ \amp=\frac{\sqrt{5}}{6\cdot\sqrt{2}}\\ \end{align*}Now all that remains is to multiply the numerator and denominator by \(\sqrt{2}\text{.}\)\begin{align*} \amp=\frac{\sqrt{5}}{6\cdot\sqrt{2}}\multiplyright{\frac{\sqrt{2}}{\sqrt{2}}}\\ \amp=\frac{\sqrt{10}}{6\cdot 2}\\ \amp=\frac{\sqrt{10}}{12} \end{align*}
Checkpoint 6.2.3.
Rationalize the denominator in \(\frac{2}{\sqrt{10}}\text{.}\)
Explanation.
We will rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{10}\text{:}\)
\begin{equation*}
\begin{aligned}
\frac{2}{\sqrt{10}}\amp=\frac{2}{\sqrt{10}}\multiplyright{\frac{\sqrt{10}}{\sqrt{10}}}\\
\amp=\frac{2\cdot\sqrt{10}}{\sqrt{10}\cdot\sqrt{10}}\\
\amp=\frac{2\sqrt{10}}{10}\\
\amp=\frac{\sqrt{10}}{5}
\end{aligned}
\end{equation*}
Again note that the fraction was simplified in the last step.
Example 6.2.4.
Rationalize the denominator in \(\sqrt{\frac{2}{7}}\text{.}\)
Explanation.
This example is slightly different. The entire fraction, including its denominator, is within a radical. Having a denominator within a radical is just as undesirable as having a radical in a denominator. So we want to do something to change the expression.
\begin{align*}
\sqrt{\frac{2}{7}}\amp=\frac{\sqrt{2}}{\sqrt{7}}\\
\amp=\frac{\sqrt{2}}{\sqrt{7}}\multiplyright{\frac{\sqrt{7}}{\sqrt{7}}}\\
\amp=\frac{\sqrt{2}\cdot\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}\\
\amp=\frac{\sqrt{14}}{7}
\end{align*}
Subsection 6.2.2 Rationalize the Denominator Using the Difference of Squares Formula
Conside the number \(\frac{1}{\sqrt{2}+1}\text{.}\) Its denominator is irrational, approximately \(2.414\ldots\text{.}\) Can we rewrite this as an equivalent expression where the denominator is rational? Let’s try multiplying the numerator and denominator by \(\sqrt{2}\text{:}\)
\begin{align*}
\frac{1}{\sqrt{2}+1}\amp=\frac{1}{\left(\sqrt{2}+1\right)}\multiplyright{\frac{\sqrt{2}}{\sqrt{2}}}\\
\amp=\frac{\sqrt{2}}{\sqrt{2}\cdot\highlight{\sqrt{2}}+1\cdot\highlight{\sqrt{2}}}\\
\amp=\frac{\sqrt{2}}{2+\sqrt{2}}
\end{align*}
We removed one radical from the denominator, but created another. We need to find another method. The difference of squares formula will help:
\begin{equation*}
(a+b)(a-b)=a^2-b^2
\end{equation*}
Those two squares in \(a^2-b^2\) can be used as a tool to annihilate radicals. Take \(\frac{1}{\sqrt{2}+1}\text{,}\) and multiply both the numerator and denominator by \(\sqrt{2}-1\text{:}\)
\begin{align*}
\frac{1}{\sqrt{2}+1}\amp=\frac{1}{\left(\sqrt{2}+1\right)}\multiplyright{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}}\\
\amp=\frac{\sqrt{2}-1}{\left(\sqrt{2}\right)^2-(1)^2}\\
\amp=\frac{\sqrt{2}-1}{2-1}\\
\amp=\frac{\sqrt{2}-1}{1}\\
\amp=\sqrt{2}-1
\end{align*}
Example 6.2.5.
Rationalize the denominator in \(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{5}+\sqrt{3}}\text{.}\)
Explanation.
To address the radicals in the denominator, we multiply both numerator and denominator by \(\sqrt{5}-\sqrt{3}\text{.}\)
\begin{align*}
\frac{\sqrt{7}-\sqrt{2}}{\sqrt{5}+\sqrt{3}}\amp=\frac{\sqrt{7}-\sqrt{2}}{\sqrt{5}+\sqrt{3}}\multiplyright{\frac{\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)}}\\
\amp=\frac{\sqrt{7}\multiplyright{\sqrt{5}}-\sqrt{7}\multiplyright{\sqrt{3}}-\sqrt{2}\multiplyright{\sqrt{5}}-\sqrt{2}\multiplyright{-\sqrt{3}}}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}\\
\amp=\frac{\sqrt{35}-\sqrt{21}-\sqrt{10}+\sqrt{6}}{5-3}\\
\amp=\frac{\sqrt{35}-\sqrt{21}-\sqrt{10}+\sqrt{6}}{2}
\end{align*}
Checkpoint 6.2.6.
Rationalize the denominator in \(\frac{\sqrt{3}}{3-2\sqrt{3}}\text{.}\)
Explanation.
To remove the radical in \(3-2\sqrt{3}\) with the difference of squares formula, we multiply it with \(3+2\sqrt{3}\text{.}\)
\begin{equation*}
\begin{aligned}
\frac{\sqrt{3}}{3-2\sqrt{3}}\amp=\frac{\sqrt{3}}{(3-2\sqrt{3})}\multiplyright{\frac{(3+2\sqrt{3})}{(3+2\sqrt{3})}}\\
\amp=\frac{\multiplyleft{3}\sqrt{3}+\multiplyleft{2\sqrt{3}}\sqrt{3}}{(3)^2-\left(2\sqrt{3}\right)^2}\\
\amp=\frac{3\sqrt{3}+2\cdot 3}{9-2^2\left(\sqrt{3}\right)^2}\\
\amp=\frac{3\sqrt{3}+6}{9-4(3)}\\
\amp=\frac{3\left(\sqrt{3}+2\right)}{9-12}\\
\amp=\frac{3\left(\sqrt{3}+2\right)}{-3}\\
\amp=\frac{\sqrt{3}+2}{-1}\\
\amp=-\sqrt{3}-2
\end{aligned}
\end{equation*}
Reading Questions 6.2.3 Reading Questions
1.
To rationalize a denominator in an expression like \(\frac{3}{\sqrt{5}}\text{,}\) explain the first step you will take.
2.
What is the special pattern from Section 5.5 that helps to rationalize the denominator in an expression like \(\frac{3}{2+\sqrt{5}}\text{?}\)
Exercises 6.2.4 Exercises
Exercise Group.
Simplify the expression. If the result has a denominator, it must be rationalized.
1.
\(\displaystyle{ \frac{1}{\sqrt{10}} }\)
2.
\(\displaystyle{ \frac{1}{\sqrt{2}} }\)
3.
\(\displaystyle{ \frac{15}{\sqrt{3}} }\)
4.
\(\displaystyle{ \frac{9}{\sqrt{3}} }\)
5.
\(\displaystyle{ \frac{1}{{\sqrt{18}}} }\)
6.
\(\displaystyle{ \frac{1}{{\sqrt{96}}} }\)
7.
\(\displaystyle{ \frac{10}{{\sqrt{72}}} }\)
8.
\(\displaystyle{ \frac{3}{{\sqrt{216}}} }\)
9.
\(\displaystyle{ \frac{9}{\sqrt{{64}}} }\)
10.
\(\displaystyle{ \frac{9}{\sqrt{{25}}} }\)
11.
\(\displaystyle{ \frac{9}{\sqrt{2}} }\)
12.
\(\displaystyle{ \frac{7}{\sqrt{2}} }\)
13.
\(\displaystyle{ \frac{5}{3\sqrt{3}} }\)
14.
\(\displaystyle{ \frac{7}{2\sqrt{5}} }\)
15.
\(\displaystyle{ \frac{20}{\sqrt{15}} }\)
16.
\(\displaystyle{ \frac{7}{\sqrt{21}} }\)
17.
\(\displaystyle{ \sqrt{\frac{15}{64}} }\)
18.
\(\displaystyle{ \sqrt{\frac{11}{81}} }\)
19.
\(\displaystyle{ \sqrt{\frac{9}{2}} }\)
20.
\(\displaystyle{ \sqrt{\frac{81}{2}} }\)
21.
\(\displaystyle{ \sqrt{\frac{3}{10}} }\)
22.
\(\displaystyle{ \sqrt{\frac{5}{13}} }\)
23.
\(\displaystyle{ \sqrt{\frac{63}{5}} }\)
24.
\(\displaystyle{ \sqrt{\frac{80}{13}} }\)
25.
\(\displaystyle{ \frac{1}{\sqrt{n}} }\)
26.
\(\displaystyle{ \frac{5}{\sqrt{a}} }\)
27.
\(\displaystyle{ \sqrt{\frac{11}{180}} }\)
28.
\(\displaystyle{ \sqrt{\frac{5}{72}} }\)
29.
\(\displaystyle{ \dfrac{2}{\sqrt{19}+8} }\)
30.
\(\displaystyle{ \dfrac{3}{\sqrt{11}+4} }\)
31.
\(\displaystyle{ \dfrac{6}{\sqrt{19}+6} }\)
32.
\(\displaystyle{ \dfrac{7}{\sqrt{3}+7} }\)
33.
\(\displaystyle{ \dfrac{\sqrt{3}-10}{\sqrt{11}+8} }\)
34.
\(\displaystyle{ \dfrac{\sqrt{2}-11}{\sqrt{13}+6} }\)
35.
\(\displaystyle{ \dfrac{\sqrt{5}-12}{\sqrt{11}+3} }\)
36.
\(\displaystyle{ \dfrac{\sqrt{3}-14}{\sqrt{13}+10} }\)
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