In <<Unresolved xref, reference "section-rationalizing-denominator"; check spelling or use "provisional" attribute>>, we learned how to rationalize the denominator in simple expressions like \(\frac{1}{\sqrt{2}}\text{.}\) In this section, we will extend the concept and handle more complicated expressions.
To remove radicals from the denominator of \(\frac{1}{\sqrt{2}}\text{,}\) we multiply the numerator and denominator by \(\sqrt{2}\text{,}\) and we have:
\begin{align*} \frac{1}{\sqrt{2}}\amp=\frac{1\multiplyright{\sqrt{2}}}{\sqrt{2}\multiplyright{\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*}In this section, we assume all variables are positive, so we don't need to use the absolute value symbols.
We can extend this concept and use the following properties to rationalize denominators with radicals of other degrees:
\begin{align*} \sqrt[3]{x^3}\amp=x\\ \sqrt[4]{x^4}\amp=x\\ \amp\ldots\\ \sqrt[n]{x^n}\amp=x \end{align*}Let's look at a few examples.
Rationalize the denominator in \(\frac{1}{\sqrt[3]{2}}\text{.}\)
To remove the radical \(\sqrt[3]{2}\text{,}\) we will use the property \(\sqrt[3]{2^3}=2\text{.}\) This implies we will multiply the numerator and denominator by \(\sqrt[3]{2^2}\text{:}\)
\begin{align*} \frac{1}{\sqrt[3]{2}}\amp=\frac{1\multiplyright{\sqrt[3]{2^2}}}{\sqrt[3]{2}\multiplyright{\sqrt[3]{2^2}}}\\ \amp=\frac{\sqrt[3]{2^2}}{\sqrt[3]{2^3}}\\ \amp=\frac{\sqrt[3]{4}}{2} \end{align*}Rationalize the denominator in \(\frac{\sqrt{3}}{3\sqrt{2x}}\text{.}\) Assume all variables are positive.
In the example above, to remove the radical \(\sqrt{2x}\text{,}\) we did \(\sqrt{2x}\cdot\sqrt{2x}=2x\text{.}\) In the next example, to remove the radical \(\sqrt[3]{2x^2}\text{,}\) we do \(\sqrt[3]{2x}\cdot\sqrt[3]{2^2x^2}=2x\text{.}\)
Rationalize the denominator in \(\frac{\sqrt[3]{y}}{3\sqrt[3]{2x}}\text{.}\) Assume all variables are positive.
When a radicand is a fraction, we should break it into two radicals, and then rationalize the denominator, like in the next two examples.
Rationalize the denominator in \(\sqrt[3]{\frac{2}{3}}\text{.}\)
Rationalize the denominator in \(\sqrt[3]{\frac{r}{s^2}}\text{.}\) Assume all variables are positive.
Before we rationalize the denominator, we should simplify the radical if possible, like in the next example.
Rationalize the denominator in \(\frac{1}{\sqrt[3]{54x^5}}\text{.}\) Assume all variables are positive.
Rationalizing the Denominator
How can be remove the radical in the denominator of \(\frac{1}{\sqrt{2}+1}\text{?}\) Let's try by multiplying the numerator and denominator by \(\sqrt{2}\text{:}\)
\begin{align*} \frac{1}{\sqrt{2}+1}\amp=\frac{1\multiplyright{\sqrt{2}}}{(\sqrt{2}+1)\multiplyright{\sqrt{2}}}\\ \amp=\frac{\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}+1\cdot\sqrt{2}}\\ \amp=\frac{\sqrt{2}}{2+\sqrt{2}} \end{align*}We removed one radical in the denominator, but created another. We need to find another method, using the Difference of Squares formula:
\begin{equation*} (a+b)(a-b)=a^2-b^2 \end{equation*}Those two squares in \(a^2-b^2\) will remove square roots. To remove the radical in \(\sqrt{2}+1\) with the Difference of Squares formula, we will multiply it with \(\sqrt{2}-1\text{,}\) and we have:
\begin{align*} (\sqrt{2}+1)(\sqrt{2}-1)\amp=(\sqrt{2})^2-(1)^2\\ \amp=2-1\\ \amp=1 \end{align*}To remove the radical in 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\multiplyright{(\sqrt{2}-1)}}{(\sqrt{2}+1)\multiplyright{(\sqrt{2}-1)}}\\ \amp=\frac{\sqrt{2}-1}{(\sqrt{2})^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.
Rationalize the denominator in \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\text{.}\)
To remove radicals in \(\sqrt{5}+\sqrt{3}\) with the Difference of Squares formula, we multiply it with \(\sqrt{5}-\sqrt{3}\text{.}\) The solution is:
\begin{align*} \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\amp=\frac{(\sqrt{5}-\sqrt{3})\multiplyright{(\sqrt{5}-\sqrt{3})}}{(\sqrt{5}+\sqrt{3})\multiplyright{(\sqrt{5}-\sqrt{3})}}\\ \amp=\frac{\sqrt{5}\cdot\sqrt{5}-\sqrt{5}\cdot\sqrt{3}-\sqrt{3}\cdot\sqrt{5}+(-\sqrt{3})(-\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}\\ \amp=\frac{5-\sqrt{15}-\sqrt{15}+3}{5-3}\\ \amp=\frac{8-2\sqrt{15}}{2}\\ \amp=\frac{2(4-\sqrt{15})}{2}\\ \amp=4-\sqrt{15} \end{align*}Rationalize the denominator in \(\frac{\sqrt{3}}{3-2\sqrt{3}}\text{.}\)
To remove the radical in \(3-2\sqrt{3}\) with the Difference of Squares formula, we multiply it with \(3+2\sqrt{3}\text{.}\) The solution is:
\begin{align*} \frac{\sqrt{3}}{3-2\sqrt{3}}\amp=\frac{\sqrt{3}\multiplyright{(3+2\sqrt{3})}}{(3-2\sqrt{3})\multiplyright{(3+2\sqrt{3})}}\\ \amp=\frac{3\sqrt{3}+2\sqrt{3}\cdot\sqrt{3}}{(3)^2-(2\sqrt{3})^2}\\ \amp=\frac{3\sqrt{3}+2\cdot3}{9-2^2(\sqrt{3})^2}\\ \amp=\frac{3\sqrt{3}+6}{9-4(3)}\\ \amp=\frac{3(\sqrt{3}+2)}{9-12}\\ \amp=\frac{3(\sqrt{3}+2)}{-3}\\ \amp=\frac{\sqrt{3}+2}{-1}\\ \amp=-\sqrt{3}-2 \end{align*}Rationalize the denominator in \(\frac{\sqrt{m}+\sqrt{n}}{\sqrt{m}-\sqrt{n}}\text{.}\) Assume all variables are positive.
To remove radicals in \(\sqrt{m}-\sqrt{n}\) with the Difference of Squares formula, we multiply it with \(\sqrt{m}+\sqrt{n}\text{.}\) The solution is:
\begin{align*} \frac{\sqrt{m}+\sqrt{n}}{\sqrt{m}-\sqrt{n}}\amp=\frac{(\sqrt{m}+\sqrt{n})\multiplyright{(\sqrt{m}+\sqrt{n})}}{(\sqrt{m}-\sqrt{n})\multiplyright{(\sqrt{m}+\sqrt{n})}}\\ \amp=\frac{\sqrt{m}\cdot\sqrt{m}+\sqrt{m}\cdot\sqrt{n}+\sqrt{n}\cdot\sqrt{m}+\sqrt{n}\cdot\sqrt{n}}{(\sqrt{m})^2-(\sqrt{n})^2}\\ \amp=\frac{m+\sqrt{mn}+\sqrt{mn}+n}{m-n}\\ \amp=\frac{m+2\sqrt{mn}+n}{m-n} \end{align*}Rationalizing the Denominator by Difference of Squares Formula
