## Section14.3More on Rationalizing the Denominator

In Section 8.2, 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.

### Subsection14.3.1A 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*}
###### Example14.3.2

Rationalize the denominator of the expressions.

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

2. $\frac{\sqrt{5}}{\sqrt{72}}$

Explanation
1. 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}}{6}\\ \amp=\frac{\sqrt{6}}{2} \end{align*}
2. 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 2}\\ \amp=\frac{\sqrt{10}}{12} \end{align*}

### Subsection14.3.2Rationalize Denominator with Difference of Squares Formula

How can be 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.

###### Example14.3.3

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*}
###### Example14.3.4

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*}

### Subsection14.3.3Exercises

###### 1

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{\sqrt{6}} = }$

###### 2

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{\sqrt{7}} = }$

###### 3

Rationalize the denominator and simplify the expression.

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

###### 4

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{20}{\sqrt{10}} = }$

###### 5

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{{\sqrt{28}}} = }$

###### 6

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{{\sqrt{45}}} = }$

###### 7

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{8}{{\sqrt{180}}} = }$

###### 8

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{9}{{\sqrt{72}}} = }$

###### 9

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{3}{\sqrt{m}} = }$

###### 10

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{\sqrt{n}} = }$

###### 11

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{13}{14}} = }$

###### 12

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{14}{15}} = }$

###### 13

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{11}{72}} = }$

###### 14

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{5}{24}} = }$

###### 15

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{3}{\sqrt{11}+8}=}$

###### 16

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{4}{\sqrt{2}+3}=}$

###### 17

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{5}{\sqrt{17}+6}=}$

###### 18

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{3}{\sqrt{6}+5}=}$

###### 19

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{2}-12}{\sqrt{11}+10}=}$

###### 20

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{5}-13}{\sqrt{13}+8}=}$

###### 21

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{2}-14}{\sqrt{7}+5}=}$

###### 22

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{5}-15}{\sqrt{13}+3}=}$