ORCCA More on Rationalizing the Denominator

Section 13.4 More on Rationalizing the Denominator

In Section 9.1, we learned how to rationalize the denominator in simple expressions like \(\frac{1}{\sqrt{2}}\text{.}\) We will briefly review this topic and then extend the concept to the next level.

Subsection 13.4.1 A Review of Rationalizing the Denominator

To remove radicals from the denominator of \(\frac{1}{\sqrt{2}}\text{,}\) we multiply the numerator and denominator by \(\sqrt{2}\text{:}\)

\begin{align*} \frac{1}{\sqrt{2}}\amp=\frac{1}{\sqrt{2}}\multiplyright{\frac{\sqrt{2}}{\sqrt{2}}}\\ \amp=\frac{\sqrt{2}}{2} \end{align*}

We used the property:

\begin{equation*} \sqrt{x}\cdot\sqrt{x}=x,\text{ where }x\text{ is positive} \end{equation*}

Example 13.4.1.

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 take the expression and multiply by a special version of \(\highlight{1}\) to make the radical in the denominator cancel.

\begin{align*} \frac{3}{\sqrt{6}}\amp=\frac{3}{\sqrt{6}}\multiplyright{\frac{\sqrt{6}}{\sqrt{6}}}\\ \amp=\frac{3\sqrt{6}}{\sqrt{36}}\\ \amp=\frac{3\sqrt{6}}{6}\\ \amp=\frac{\sqrt{6}}{2} \end{align*}
Rationalizing the denominator of \(\frac{\sqrt{5}}{\sqrt{72}}\) is slightly trickier. We could go the brute force method and 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. This will simplify future algebra.

\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\sqrt{4}}\\ \amp=\frac{\sqrt{10}}{6\cdot 2}\\ \amp=\frac{\sqrt{10}}{12} \end{align*}

Subsection 13.4.2 Rationalize Denominator with Difference of Squares Formula

How can we remove the radical from the denominator of \(\frac{1}{\sqrt{2}+1}\text{?}\) 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 remove square roots. To remove the radical from the denominator of \(\frac{1}{\sqrt{2}+1}\text{,}\) we multiply 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*}

Let's look at a few more examples.

Example 13.4.2.

Rationalize the denominator in \(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{5}+\sqrt{3}}\text{.}\)

Explanation

To remove radicals in \(\sqrt{5}+\sqrt{3}\) with the difference of squares formula, we multiply it with \(\sqrt{5}-\sqrt{3}\text{.}\)

\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*}

Example 13.4.3.

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{align*} \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{align*}