Section13.4Rationalizing Monomial Square Root Denominators

Traditionally, expressions containing square roots in the denominator are not considered simplified. There are two common types of expressions that arise where square roots naturally occur in the denominator. In this section we deal with expressions where the denominator is a monomial (one term) and in the final section we deal we expressions where the denominator is a binomial (two terms).

Let's consider $\frac{1}{\sqrt{2}}\text{.}$ We need to determine a legitimate maneuver that will result in the square root factor taking up residence outside of the denominator of the fraction. If we multiply the denominator by $\sqrt{2}\text{,}$ the resultant product in the denominator will be $2\text{.}$ Of course, multiplying anything by $\sqrt{2}$ does not result in an equivalent expression unless some sort of balancing act is performed simultaneously. In this case the balancing act is to also multiply the numerator by $\sqrt{2}\text{.}$ To wit:

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

Now it may not seem to you that the expression $\frac{\sqrt{2}}{2}$ is any simpler than the expression $\frac{1}{\sqrt{2}}\text{,}$ and in a sense you are correct in that assertion. As math folks, though, we find use in rewriting the expression with the square root in the numerator. On pay-off is that it can help us identifying square root terms that can be combined (see next section). Historically, rewriting the expression with the radical in the numerator dramatically simplified any attendant arithmetic. Consider that $\sqrt{2}$ is approximately $1.414\text{.}$ Which would you rather compute by hand: $\frac{1.414}{2}$ or $\frac{1}{1.414}\text{?}$ If it's not apparent to you, try both.

Finally, it is frequently the case that clearing the square root away from the denominator results in a simpler expression in all senses of the term. Let's make one small tweak to the last example.

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

When the original radicand contains a prefect square factor, we need to be sure that we remember to factor that from the radical. We also need to make sure to remember to carry along any other factors and completely simplify the resultant expression. Several examples follow.

Example13.4.1.

Rationalize the denominator and simplify the result for the expression $\frac{4}{\sqrt{12}}\text{.}$

Solution

We begin by multiplying both the numerator and denominator of the expression by $\sqrt{12}\text{.}$ We then simplify $\sqrt{12}$ and then reduce the resultant fraction.

\begin{align*} \frac{4}{\sqrt{12}}\amp=\frac{4}{\sqrt{12}} \cdot \highlight{\frac{\sqrt{12}}{\sqrt{12}}}\\ \amp=\frac{4 \cdot \highlight{\sqrt{12}}}{12}\\ \amp=\frac{4 \cdot \highlight{2\sqrt{3}}}{12}\\ \amp=\frac{2\sqrt{3}}{3} \end{align*}
Example13.4.2.

Rationalize the denominator and simplify the result for the expression $\frac{1}{\sqrt{500}}\text{.}$

Solution

We begin by multiplying both the numerator and denominator of the expression by $\sqrt{500}\text{.}$ We then simplify $\sqrt{500}$ and then reduce the resultant fraction.

\begin{align*} \frac{1}{\sqrt{500}}\amp=\frac{1}{\sqrt{500}} \cdot \highlight{\frac{\sqrt{500}}{\sqrt{500}}}\\ \amp=\frac{\highlight{\sqrt{500}}}{500}\\ \amp=\frac{\highlight{10\sqrt{5}}}{500}\\ \amp=\frac{\sqrt{5}}{50} \end{align*}
Example13.4.3.

Rationalize the denominator and simplify the result for the expression $-\frac{4}{\sqrt{44}}\text{.}$

Solution

We begin by multiplying both the numerator and denominator of the expression by $\sqrt{44}\text{.}$ We then simplify $\sqrt{44}$ and then reduce the resultant fraction.

\begin{align*} -\frac{4}{\sqrt{44}}\amp=-\frac{4}{\sqrt{44}} \cdot \highlight{\frac{\sqrt{44}}{\sqrt{44}}}\\ \amp=-\frac{4 \cdot \highlight{\sqrt{44}}}{44}\\ \amp=-\frac{4 \cdot \highlight{2\sqrt{11}}}{44}\\ \amp=-\frac{2\sqrt{11}}{11} \end{align*}
Example13.4.4.

Rationalize the denominator and simplify the result for the expression $\frac{1}{2\sqrt{128}}\text{.}$

Solution

We begin by multiplying both the numerator and denominator of the expression by $\sqrt{128}\text{.}$ We then simplify $\sqrt{128}$ and then reduce the resultant fraction.

\begin{align*} \frac{1}{2\sqrt{128}}\amp=\frac{1}{2 \cdot \sqrt{128}} \cdot \highlight{\frac{\sqrt{128}}{\sqrt{128}}}\\ \amp=\frac{\highlight{\sqrt{128}}}{2 \cdot 128}\\ \amp=\frac{\highlight{8\sqrt{2}}}{256}\\ \amp=\frac{\sqrt{2}}{32} \end{align*}

You can use Figure 13.4.5 to generate several more examples/practice problems with step by step solutions.

ExercisesExercises

Rationalize each denominator. Completely simplify each result.

1.

$\frac{7}{\sqrt{7}}$

Solution

We begin by multiplying both the numerator and the denominator of the expression by $\sqrt{7}\text{.}$ We then reduce the resultant fraction.

\begin{align*} \frac{7}{\sqrt{7}}\amp=\frac{7}{\sqrt{7}} \cdot \highlight{\frac{\sqrt{7}}{\sqrt{7}}}\\ \amp=\frac{7\sqrt{7}}{7}\\ \amp=\sqrt{7} \end{align*}
2.

$\frac{30}{\sqrt{10}}$

Solution

We begin by multiplying both the numerator and the denominator of the expression by $\sqrt{10}\text{.}$ We then reduce the resultant fraction.

\begin{align*} \frac{30}{\sqrt{10}}\amp=\frac{30}{\sqrt{10}} \cdot \highlight{\frac{\sqrt{10}}{\sqrt{10}}}\\ \amp=\frac{30\sqrt{10}}{10}\\ \amp=3\sqrt{10} \end{align*}
3.

$\frac{1}{\sqrt{20}}$

Solution

We begin by multiplying both the numerator and the denominator of the expression by $\sqrt{20}\text{.}$ We then simplify $\sqrt{20}$ and reduce the resultant fraction.

\begin{align*} \frac{1}{\sqrt{20}}\amp=\frac{1}{\sqrt{20}} \cdot \highlight{\frac{\sqrt{20}}{\sqrt{20}}}\\ \amp=\frac{\highlight{\sqrt{20}}}{20}\\ \amp=\frac{\highlight{2\sqrt{5}}}{20}\\ \amp=\frac{\sqrt{5}}{10} \end{align*}
4.

$\frac{25}{\sqrt{50}}$

Solution

We begin by multiplying both the numerator and the denominator of the expression by $\sqrt{50}\text{.}$ We then reduce the resultant fraction and simplify $\sqrt{50}\text{.}$

\begin{align*} \frac{25}{\sqrt{50}}\amp=\frac{25}{\sqrt{50}} \cdot \highlight{\frac{\sqrt{50}}{\sqrt{50}}}\\ \amp=\frac{25\sqrt{50}}{50}\\ \amp=\frac{\highlight{\sqrt{50}}}{2}\\ \amp=\frac{\highlight{5\sqrt{2}}}{2} \end{align*}
5.

$\frac{1}{6\sqrt{3}}$

Solution

We begin by multiplying both the numerator and the denominator of the expression by $\sqrt{3}\text{.}$ We then reduce the resultant fraction.

\begin{align*} \frac{1}{6\sqrt{3}}\amp=\frac{1}{6\sqrt{3}} \cdot \highlight{\frac{\sqrt{3}}{\sqrt{3}}}\\ \amp=\frac{\sqrt{3}}{6 \cdot 3}\\ \amp=\frac{\sqrt{3}}{18} \end{align*}
6.

$\frac{121}{11\sqrt{11}}$

Solution

We begin by multiplying both the numerator and the denominator of the expression by $\sqrt{11}\text{.}$ We then reduce the resultant fraction.

\begin{align*} \frac{121}{11\sqrt{11}}\amp=\frac{121}{11\sqrt{11}} \cdot \highlight{\frac{\sqrt{11}}{\sqrt{11}}}\\ \amp=\frac{121\sqrt{11}}{11 \cdot 11}\\ \amp=\sqrt{11} \end{align*}