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 [latex]\dfrac{1}{3}[/latex] and [latex]\dfrac{2}{5}[/latex] are since the denominators are integers, but what about the quantities [latex]\dfrac{1}{\sqrt{2}}[/latex] and [latex]\dfrac{1}{\sqrt{5}}[/latex]? 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 from the denominator, which makes the expression easier to understand.

The idea of rationalizing a denominator makes a bit more sense if you consider the definition of “rationalize.” Recall that the numbers [latex]5[/latex], [latex]\dfrac{1}{2}[/latex], and [latex]0.75[/latex] are all known as rational numbers—they can each be expressed as a ratio of two integers ([latex]\dfrac{5}{1},\dfrac{1}{2}[/latex], and [latex]\dfrac{3}{4}[/latex] 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. Note that the expressions in each row are equal, so they are different ways of representing the same value. You can verify with a calculator that both expressions have the same decimal approximation.

Now we learn how to turn an expression with an irrational denominator into one with a rational denominator.

Rationalizing Denominators with One Term

Let us start with the fraction [latex]\dfrac{1}{\sqrt{2}}[/latex]. Its denominator is [latex]\sqrt{2}[/latex], an irrational number. This makes it difficult to figure out what the value of [latex]\dfrac{1}{\sqrt{2}}[/latex] is.

You can rewrite this fraction without changing its value if you multiply it by a factor equal to [latex]1[/latex]. In this case, let that factor be [latex]\dfrac{\sqrt{2}}{\sqrt{2}}[/latex]. Watch what happens.

[latex]\dfrac{1}{\sqrt{2}}\cdot 1=\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}[/latex]

The denominator of the new fraction is no longer a radical (notice, however, that the numerator is).

So why choose to multiply [latex]\dfrac{1}{\sqrt{2}}[/latex] by [latex]\dfrac{\sqrt{2}}{\sqrt{2}}[/latex]? Our goal was to make the radicand in the denominator a perfect square so we could simplify it. We did this by introducing another copy of the same radical. In algebraic terms, this idea is represented by [latex]\sqrt{x}\cdot \sqrt{x}=x[/latex] (where [latex]x[/latex] is nonnegative.)

Here are some examples. Remember that the value of the fraction is not changed at all. It is similar to the process of getting common denominators.

You can use the same method of rationalizing denominators to simplify fractions with radicals that contain a variable. As long as you multiply the original expression by a quantity equivalent to [latex]1[/latex], you can eliminate a radical in the denominator without changing the value of the expression.

We can use a similar process to rationalize denominators with higher index radicals. The goal is still to make the radicand in the denominator a perfect power.  Let’s consider a quick example.

[latex]\dfrac{1}{\sqrt[3]{2}}[/latex]

To eliminate the radical from the denominator, this time we need the radicand to be a perfect cube. This suggests we need to introduce two more factors of [latex]2[/latex] in the radicand instead of just one:

[latex]\begin{align}\frac{1}{\sqrt[3]{2}}&=\frac{1}{\sqrt[3]{2}}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}\\ &=\frac{\sqrt[3]{4}}{\sqrt[3]{8}}\\ &=\frac{\sqrt[3]{4}}{2}\end{align}[/latex]

This is the final answer since the denominator has no radical. Consider the following similar example. Try thinking about how you might rationalize the denominator before you see the solution.

[latex]\dfrac{1}{\sqrt[3]{4}}[/latex]

The radicand in the denominator already has two factors of [latex]2.[/latex] Introducing just one more factor of [latex]2[/latex] will make it a perfect cube.

[latex]\begin{align}\frac{1}{\sqrt[3]{4}}&=\frac{1}{\sqrt[3]{4}}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}\\ &=\frac{\sqrt[3]{2}}{\sqrt[3]{8}}\\ &=\frac{\sqrt[3]{2}}{2}\end{align}[/latex]

The denominator is now rationalized. It is also possible to approach the problem by multiplying twice by [latex]\dfrac{\sqrt[3]{4}}{\sqrt[3]{4}}[/latex] like in the first example, but it requires extra simplification steps at the end:

[latex]\begin{align}\frac{1}{\sqrt[3]{4}}&=\frac{1}{\sqrt[3]{4}}\cdot\frac{\sqrt[3]{4}}{\sqrt[3]{4}}\cdot\frac{\sqrt[3]{4}}{\sqrt[3]{4}}\\ &=\frac{\sqrt[3]{16}}{\sqrt[3]{64}}\\ &=\frac{\sqrt[3]{8\cdot 2}}{4}\\ &=\frac{2\sqrt[3]{2}}{4}\\ &=\frac{\sqrt[3]{2}}{2}\end{align}[/latex]

Here are a few more examples to try.

In the following video, we show examples of rationalizing the denominator of a radical expression that contains integer radicands.

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 [latex]\dfrac{3}{\sqrt{2}+3}[/latex] where the denominator is composed of two terms, [latex]\sqrt{2}[/latex] and [latex]+3[/latex].

Unfortunately, you cannot rationalize these denominators the same way you rationalize single-term denominators. If you multiply [latex](\sqrt{2}+3)\cdot\sqrt{2}[/latex], you get [latex]2+3\sqrt{2}[/latex]. The original [latex]\sqrt{2}[/latex] is gone, but now the quantity [latex]3\sqrt{2}[/latex] has appeared. This is no better!

We saw an example in the previous section in which we multiplied two binomial radical expressions and got a whole number. That example had two expressions of the form [latex](a+b), (a-b)[/latex] which we called conjugates, and the two expressions form a conjugate pair. The product is [latex](a+b)(a-b) = a^2-b^2[/latex] (the Difference of Squares formula), and if  [latex]a[/latex] and/or [latex]b[/latex] are square roots, squaring will eliminate the roots!  Here is an example of multiplying [latex]\sqrt{2}+3[/latex] by its conjugate, resulting in a whole number as desired:

[latex]\begin{align}&\quad\left( \sqrt{2}+3 \right)\left( \sqrt{2}-3 \right)\\&={{\left( \sqrt{2} \right)}^{2}}-3^2&&\color{blue}{\textsf{use Difference of Squares}}\\&=2-9\\&=-7\end{align}[/latex]

There you have it! Multiplying [latex]\sqrt{2}+3[/latex] by [latex]\sqrt{2}-3[/latex] removed the radical without adding another. 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 (we have assumed all variables in the table represent nonnegative quantities).

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

[latex]\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}[/latex]

[latex]\sqrt[3]{100}[/latex] cannot be simplified since its prime factorization is [latex]2\cdot 2\cdot 5\cdot 5,[/latex] so it has no perfect cube factors. Multiplying [latex]\sqrt[3]{10}+5[/latex] 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.

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 [latex]\dfrac{1}{\sqrt{5}}[/latex], multiply numerator and denominator by a radical that will make the denominator a perfect power. When the denominator contains two terms, as in[latex]\dfrac{2}{\sqrt{5}+3}[/latex], identify the conjugate of the denominator, here[latex]\sqrt{5}-3[/latex], and multiply both numerator and denominator by the conjugate.