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# Rationalizing Denominators

### Learning Outcome

• Remove radicals from a single term denominator
• Rationalize denominators with multiple terms
Although radicals follow the same rules that integers do, it is often difficult to figure out the value of an expression containing radicals. For example, you probably have a good sense of how much $\frac{4}{8},\ 0.75$ and $\frac{6}{9}$ are, but what about the quantities $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{5}}$? These are much harder to visualize. That said, sometimes you have to work with expressions that contain many radicals. Often the value of these expressions is not immediately clear. In cases where you have a fraction with a radical in the denominator, you can use a technique called rationalizing a denominator to eliminate the radical. The point of rationalizing a denominator is to make it easier to understand what the quantity really is by removing radicals from the denominators. The idea of rationalizing a denominator makes a bit more sense if you consider the definition of “rationalize.” Recall that the numbers $5$, $\frac{1}{2}$, and $0.75$ are all known as rational numbers—they can each be expressed as a ratio of two integers ($\frac{5}{1},\frac{1}{2}$, and $\frac{3}{4}$ respectively). Some radicals are irrational numbers because they cannot be represented as a ratio of two integers. As a result, the point of rationalizing a denominator is to change the expression so that the denominator becomes a rational number. Here are some examples of irrational and rational denominators.

Irrational

Rational

$\frac{1}{\sqrt{2}}$

=

$\frac{\sqrt{2}}{2}$

$\frac{2+\sqrt{3}}{\sqrt{3}}$

=

$\frac{2\sqrt{3}+3}{3}$

Now examine how to get from irrational to rational denominators.

## Rationalizing Denominators with One Term

Let us start with the fraction $\frac{1}{\sqrt{2}}$. Its denominator is $\sqrt{2}$, an irrational number. This makes it difficult to figure out what the value of $\frac{1}{\sqrt{2}}$ is. You can rename this fraction without changing its value if you multiply it by a quantity equal to $1$. In this case, let that quantity be $\frac{\sqrt{2}}{\sqrt{2}}$. Watch what happens.

$\frac{1}{\sqrt{2}}\cdot 1=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2\cdot 2}}=\frac{\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{2}}{2}$

The denominator of the new fraction is no longer a radical (notice, however, that the numerator is). So why choose to multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$? You knew that the square root of a number times itself will be a whole number. In algebraic terms, this idea is represented by $\sqrt{x}\cdot \sqrt{x}=x$. Look back to the denominators in the multiplication of $\frac{1}{\sqrt{2}}\cdot 1$. Do you see where $\sqrt{2}\cdot \sqrt{2}=\sqrt{4}=2$? In the following video, we show examples of rationalizing the denominator of a radical expression that contains integer radicands. https://youtu.be/K7NdhPLVl7g Here are some more examples. Notice how the value of the fraction is not changed at all—it is simply being multiplied by $1$.

### Example

Rationalize the denominator.

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

Answer: The denominator of this fraction is $\sqrt{3}$. To make it into a rational number, multiply it by $\sqrt{3}$, since $\sqrt{3}\cdot \sqrt{3}=3$. This means we need to multiply the entire fraction by $\frac{\sqrt{3}}{\sqrt{3}}$ because it is equal to $1$.

$\begin{array}{r}\frac{-6\sqrt{6}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\\\\=\frac{-6\cdot\sqrt{6\cdot3}}{\sqrt{3\cdot3}}\\\\=\frac{-6\cdot\sqrt{6\cdot3}}{3}\end{array}$

Simplify the coefficients and the radicals, where possible.

$\begin{array}{c}\frac{-6\cdot\sqrt{6\cdot3}}{3}\\\\=-2\sqrt{3^2\cdot2}\\\\=-2\cdot3\sqrt{2}\\\\=-6\sqrt{2}\end{array}$

[latex-display] \frac{-6\sqrt{6}}{\sqrt{3}}=-6\sqrt{2}[/latex-display]

### Example

Rationalize the denominator.

$\frac{2+\sqrt{3}}{\sqrt{3}}$

Answer: The denominator of this fraction is $\sqrt{3}$. To make it into a rational number, multiply it by $\sqrt{3}$, since $\sqrt{3}\cdot \sqrt{3}=3$.

$\frac{2+\sqrt{3}}{\sqrt{3}}$

Multiply the entire fraction by a quantity which simplifies to $1$: $\frac{\sqrt{3}}{\sqrt{3}}$.

$\begin{array}{r}\frac{2+\sqrt{3}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\\\\\frac{\sqrt{3}(2+\sqrt{3})}{\sqrt{3}\cdot \sqrt{3}}\end{array}$

Use the Distributive Property to multiply $\sqrt{3}(2+\sqrt{3})$.

$\frac{2\sqrt{3}+\sqrt{3}\cdot \sqrt{3}}{\sqrt{9}}$

$\frac{2\sqrt{3}+\sqrt{9}}{\sqrt{9}}$

Simplify the radicals, where possible. $\sqrt{9}=3$. The answer is $\frac{2\sqrt{3}+3}{3}$.

### Try It

[ohm_question]2765[/ohm_question]
In the video example that follows, we show more examples of how to rationalize a denominator with an integer radicand. https://youtu.be/K7NdhPLVl7g You can use the same method to rationalize denominators to simplify fractions with radicals that contain a variable. As long as you multiply the original expression by a quantity that simplifies to $1$, you can eliminate a radical in the denominator without changing the value of the expression itself.

### Example

Rationalize the denominator.

$\frac{\sqrt{2y}}{\sqrt{4x}},\text{ where }x\ne \text{0}$

Answer: The denominator is $\sqrt{4x}$, so the entire expression can be multiplied by $\frac{\sqrt{4x}}{\sqrt{4x}}$ to get rid of the radical in the denominator.

$\begin{array}{c}\frac{\sqrt{2y}}{\sqrt{4x}}\cdot \frac{\sqrt{4x}}{\sqrt{4x}}\\\\=\frac{\sqrt{2\cdot{y}\cdot{4}\cdot{x}}}{\sqrt{4^2\cdot{x^2}}}\\\\=\frac{\sqrt{2\cdot{2^2}\cdot{y}\cdot{x}}}{4|x|}\end{array}$

Simplify the numerator. [latex-display]\frac{2\sqrt{2\cdot{x}\cdot{y}}}{4|x|}\\\\=\frac{2\sqrt{2xy}}{4|x|}[/latex-display]

[latex-display]\frac{\sqrt{2y}}{\sqrt{4x}}=\frac{2\sqrt{2xy}}{4|x|}[/latex-display]

### Example

Rationalize the denominator.

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}},\text{ where }x\ne \text{0}$

Answer: The denominator is $\sqrt{x}$, so the entire expression can be multiplied by $\frac{\sqrt{x}}{\sqrt{x}}$ to get rid of the radical in the denominator.

$\begin{array}{c}\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}}\cdot \frac{\sqrt{x}}{\sqrt{x}}\\\\\frac{\sqrt{x}(\sqrt{x}+\sqrt{y})}{\sqrt{x}\cdot \sqrt{x}}\end{array}$

Use the Distributive Property. Simplify the radicals where possible. Remember that $\sqrt{x}\cdot \sqrt{x}=x$.

$\frac{\sqrt{x}\cdot \sqrt{x}+\sqrt{x}\cdot \sqrt{y}}{\sqrt{x}\cdot \sqrt{x}}$

The answer is $\frac{x+\sqrt{xy}}{x}$.

### Example

Rationalize the denominator and simplify.

$\sqrt{\frac{100x}{11y}},\text{ where }y\ne \text{0}$

Answer: Use the property $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ to rewrite the radical.

$\frac{\sqrt{100x}}{\sqrt{11y}}$

The denominator is $\sqrt{11y}$, so multiplying the entire expression by $\frac{\sqrt{11y}}{\sqrt{11y}}$ will rationalize the denominator.

$\frac{\sqrt{100x}\cdot\sqrt{11y}}{\sqrt{11y}\cdot\sqrt{11y}}$

Multiply and simplify the radicals where possible.

$\frac{\sqrt{100\cdot 11xy}}{\sqrt{11y}\cdot \sqrt{11y}}$

100 is a perfect square. Remember that$\sqrt{100}=10$ and $\sqrt{x}\cdot \sqrt{x}=x$.

$\frac{\sqrt{100}\cdot \sqrt{11xy}}{\sqrt{11y}\cdot \sqrt{11y}}$

The answer is $\frac{10\sqrt{11xy}}{11y}$.

The video that follows shows more examples of how to rationalize a denominator with a monomial radicand. https://youtu.be/EBUzRctmgyk

## Rationalizing Denominators with Two Terms

Denominators do not always contain just one term as shown in the previous examples. Sometimes, you will see expressions like $\frac{3}{\sqrt{2}+3}$ where the denominator is composed of two terms, $\sqrt{2}$ and $+3$. Unfortunately, you cannot rationalize these denominators the same way you rationalize single-term denominators. If you multiply $\sqrt{2}+3$ by $\sqrt{2}$, you get $2+3\sqrt{2}$. The original $\sqrt{2}$ is gone, but now the quantity $3\sqrt{2}$ has appeared...this is no better! In order to rationalize this denominator, you want to square the radical term and somehow prevent the integer term from being multiplied by a radical. Is this possible? It is possible—and you have already seen how to do it! Recall what the product is when binomials of the form $(a+b)(a-b)$ are multiplied. So, for example, $(x+3)(x-3)={{x}^{2}}-3x+3x-9={{x}^{2}}-9$; notice that the terms $−3x$ and $+3x$ combine to 0. Now for the connection to rationalizing denominators: what if you replaced x with $\sqrt{2}$? Look at the side by side examples below. Just as $-3x+3x$ combines to $0$ on the left, $-3\sqrt{2}+3\sqrt{2}$ combines to $0$ on the right.
 $\begin{array}{l}(x+3)(x-3)\\={{x}^{2}}-3x+3x-9\\={{x}^{2}}-9\end{array}$ $\begin{array}{l}\left( \sqrt{2}+3 \right)\left( \sqrt{2}-3 \right)\\={{\left( \sqrt{2} \right)}^{2}}-3\sqrt{2}+3\sqrt{2}-9\\={{\left( \sqrt{2} \right)}^{2}}-9\\=2-9\\=-7\end{array}$
There you have it! Multiplying $\sqrt{2}+3$ by $\sqrt{2}-3$ removed one radical without adding another. In this example, $\sqrt{2}-3$ is known as the conjugate of $\sqrt{2}+3$, and $\sqrt{2}+3$ and $\sqrt{2}-3$ are known as a conjugate pair. To find the conjugate of a binomial that includes radicals, change the sign of the second term to its opposite as shown in the table below.
Term Conjugate Product
$\sqrt{2}+3$ $\sqrt{2}-3$ $\left( \sqrt{2}+3 \right)\left( \sqrt{2}-3 \right)={{\left( \sqrt{2} \right)}^{2}}-{{\left( 3 \right)}^{2}}=2-9=-7$
$\sqrt{x}-5$ $\sqrt{x}+5$ $\left( \sqrt{x}-5 \right)\left( \sqrt{x}+5 \right)={{\left( \sqrt{x} \right)}^{2}}-{{\left( 5 \right)}^{2}}=x-25$
$8-2\sqrt{x}$ $8+2\sqrt{x}$ $\left( 8-2\sqrt{x} \right)\left( 8+2\sqrt{x} \right)={{\left( 8 \right)}^{2}}-{{\left( 2\sqrt{x} \right)}^{2}}=64-4x$
$1+\sqrt{xy}$ $1-\sqrt{xy}$ $\left( 1+\sqrt{xy} \right)\left( 1-\sqrt{xy} \right)={{\left( 1 \right)}^{2}}-{{\left( \sqrt{xy} \right)}^{2}}=1-xy$

### Example

Rationalize the denominator and simplify.

$\frac{5-\sqrt{7}}{3+\sqrt{5}}$

Answer: Find the conjugate of $3+\sqrt{5}$. Then multiply the entire expression by $\frac{3-\sqrt{5}}{3-\sqrt{5}}$.

$\begin{array}{c}\frac{5-\sqrt{7}}{3+\sqrt{5}}\cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}\\\\\frac{\left( 5-\sqrt{7} \right)\left( 3-\sqrt{5} \right)}{\left( 3+\sqrt{5} \right)\left( 3-\sqrt{5} \right)}\end{array}$

Use the Distributive Property to multiply the binomials in the numerator and denominator.

$\frac{5\cdot 3-5\sqrt{5}-3\sqrt{7}+\sqrt{7}\cdot \sqrt{5}}{3\cdot 3-3\sqrt{5}+3\sqrt{5}-\sqrt{5}\cdot \sqrt{5}}$

Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to $0$.

$\frac{15-5\sqrt{5}-3\sqrt{7}+\sqrt{35}}{9-3\sqrt{5}+3\sqrt{5}-\sqrt{25}}$

$\begin{array}{c}\frac{15-5\sqrt{5}-3\sqrt{7}+\sqrt{35}}{9-\sqrt{25}}\\\\\frac{15-5\sqrt{5}-3\sqrt{7}+\sqrt{35}}{9-5}\end{array}$

The answer is $\frac{15-5\sqrt{5}-3\sqrt{7}+\sqrt{35}}{4}$.

### Example

Rationalize the denominator and simplify.

$\frac{\sqrt{x}}{\sqrt{x}+2}$

Answer: Find the conjugate of $\sqrt{x}+2$. Then multiply the numerator and denominator by $\frac{\sqrt{x}-2}{\sqrt{x}-2}$.

$\begin{array}{c}\frac{\sqrt{x}}{\sqrt{x}+2}\cdot \frac{\sqrt{x}-2}{\sqrt{x}-2}\\\\\frac{\sqrt{x}\left( \sqrt{x}-2 \right)}{\left( \sqrt{x}+2 \right)\left( \sqrt{x}-2 \right)}\end{array}$

Use the Distributive Property to multiply the binomials in the numerator and denominator.

$\frac{\sqrt{x}\cdot \sqrt{x}-2\sqrt{x}}{\sqrt{x}\cdot \sqrt{x}-2\sqrt{x}+2\sqrt{x}-2\cdot 2}$

Simplify. Remember that $\sqrt{x}\cdot \sqrt{x}=x$. Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to $0$.

$\frac{\sqrt{x}\cdot \sqrt{x}-2\sqrt{x}}{\sqrt{x}\cdot \sqrt{x}-2\sqrt{x}+2\sqrt{x}-4}$

The answer is $\frac{x-2\sqrt{x}}{x-4}$.

### Try It

[ohm_question]196062[/ohm_question]
One word of caution: this method will work for binomials that include a square root, but not for binomials with roots greater than $2$. This is because squaring a root that has an index greater than 2 does not remove the root, as shown below.

$\begin{array}{l}\left( \sqrt[3]{10}+5 \right)\left( \sqrt[3]{10}-5 \right)\\={{\left( \sqrt[3]{10} \right)}^{2}}-5\sqrt[3]{10}+5\sqrt[3]{10}-25\\={{\left( \sqrt[3]{10} \right)}^{2}}-25\\=\sqrt[3]{100}-25\end{array}$

$\sqrt[3]{100}$ cannot be simplified any further since its prime factors are $2\cdot 2\cdot 5\cdot 5$. There are no cubed numbers to pull out! Multiplying $\sqrt[3]{10}+5$ by its conjugate does not result in a radical-free expression. In the following video, we show more examples of how to rationalize a denominator using the conjugate. https://youtu.be/vINRIRgeKqU

## Summary

When you encounter a fraction that contains a radical in the denominator, you can eliminate the radical by using a process called rationalizing the denominator. To rationalize a denominator, you need to find a quantity that, when multiplied by the denominator, will create a rational number (no radical terms) in the denominator. When the denominator contains a single term, as in $\frac{1}{\sqrt{5}}$, multiplying the fraction by $\frac{\sqrt{5}}{\sqrt{5}}$ will remove the radical from the denominator. When the denominator contains two terms, as in$\frac{2}{\sqrt{5}+3}$, identify the conjugate of the denominator, here$\sqrt{5}-3$, and multiply both numerator and denominator by the conjugate.

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