Page 1.5: Rationalizing Expressions

Introduction

An algebraic expression may sometimes have a radical in the denominator (or numerator). In calculus, a helpful technique would be to get rid of the radical in the denominator (or numerator). The technique used to rid the expression of the radical is called rationalizing. We will first discuss the idea of rationalizing the denominator, followed by rationalizing the numerator.

What is Rationalizing a Denominator?

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 because 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 let’s examine how to get from irrational to rational denominators.

Rationalizing Denominators with One Term

Let’s start with the fraction \(\ \frac{1}{\sqrt{2}}\). Its denominator is \(\ \sqrt{2}\), an irrational number.

Notice that multiplying the fraction by 1 does not change the value and so equivalently, multiplying the fraction by \(\ \frac{\sqrt{2}}{\sqrt{2}}\) also does not change it's value. Allow us to now see 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 \(\ \sqrt{2} \cdot \sqrt{2}=\sqrt{4}=2\)?

Here are some more examples. Notice how the value of the fraction is not changed at all; it is simply being multiplied by another name for 1.

Example

Rationalize the denominator. \(\ \frac{2+\sqrt{3}}{\sqrt{3}}\)

Solution

\(\ \frac{2+\sqrt{3}}{\sqrt{3}}\)	The denominator of this fraction is \(\ \sqrt{3}\). To make it into a rational number, multiply it by \(\ \sqrt{3}\), since \(\ \sqrt{3} \cdot \sqrt{3}=3\).
\(\ \frac{2+\sqrt{3}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\)	Multiply the entire fraction by another name for 1, \(\ \frac{\sqrt{3}}{\sqrt{3}}\).
\(\ \frac{\sqrt{3}(2+\sqrt{3})}{\sqrt{3} \cdot \sqrt{3}}\)
\(\ \frac{2 \sqrt{3}+\sqrt{3} \cdot \sqrt{3}}{\sqrt{9}}\)	Use the Distributive Property to multiply \(\ \sqrt{3}(2+\sqrt{3})\).
\(\ \frac{2 \sqrt{3}+\sqrt{9}}{\sqrt{9}}\)	Simplify the radicals, where possible. \(\ \sqrt{9}=3\).

\(\ \frac{2+\sqrt{3}}{\sqrt{3}}=\frac{2 \sqrt{3}+3}{3}\)

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 another name for 1, you can eliminate a radical in the denominator without changing the value of the expression itself.

Example

Rationalize the denominator. \(\ \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}}\), where \(\ x \neq 0\)

Solution

\(\ \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}}\)	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.
\(\ \frac{\sqrt{x}(\sqrt{x}+\sqrt{y})}{\sqrt{x} \cdot \sqrt{x}}\)
\(\ \frac{\sqrt{x} \cdot \sqrt{x}+\sqrt{x} \cdot \sqrt{y}}{\sqrt{x} \cdot \sqrt{x}}\)	Use the Distributive Property. Simplify the radicals, where possible. Remember that \(\ \sqrt{x} \cdot \sqrt{x}=x\).

\(\ \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}}=\frac{x+\sqrt{x y}}{x}\)

Example

Rationalize the denominator and simplify. \(\ \sqrt{\frac{100 x}{11 y}}\), where \(\ y \neq 0\)

Solution

\(\ \frac{\sqrt{100 x}}{\sqrt{11 y}}\)	Rewrite \(\ \sqrt{\frac{a}{b}}\) as \(\ \frac{\sqrt{a}}{\sqrt{b}}\).
\(\ \frac{\sqrt{100 x}}{\sqrt{11 y}} \cdot \frac{\sqrt{11 y}}{\sqrt{11 y}}\)	The denominator is \(\ \sqrt{11 y}\), so multiplying the entire expression by \(\ \frac{\sqrt{11 y}}{\sqrt{11 y}}\) will rationalize the denominator.
\(\ \frac{\sqrt{100 \cdot 11 x y}}{\sqrt{11 y} \cdot \sqrt{11 y}}\)	Multiply and simplify the radicals, where possible.
\(\ \frac{\sqrt{100} \cdot \sqrt{11 x y}}{\sqrt{11 y} \cdot \sqrt{11 y}}\)	100 is a perfect square. Remember that \(\ \sqrt{100}=10\) and \(\ \sqrt{x} \cdot \sqrt{x}=x\).

\(\ \sqrt{\frac{100 x}{11 y}}=\frac{10 \sqrt{11 x y}}{11 y}\)

Exercise

Rationalize the denominator and simplify. \(\ \frac{3}{\sqrt{7}}\)

1. \(\ \frac{3 \sqrt{7}}{\sqrt{7}}\)
2. \(\ 3\)
3. \(\ \frac{7 \sqrt{3}}{3}\)
4. \(\ \frac{3 \sqrt{7}}{7}\)

Answer

1. Incorrect. After you rationalize a denominator, the denominator should not be in radical form. Try multiplying the fraction by \(\ \frac{\sqrt{7}}{\sqrt{7}}\). The correct answer is \(\ \frac{3 \sqrt{7}}{7}\).
2. Incorrect. The process of rationalizing does not remove the denominator, it just changes the fraction so that the denominator is no longer a radical. Try multiplying the numerator and denominator by \(\ \frac{\sqrt{7}}{\sqrt{7}}\). The correct answer is \(\ \frac{3 \sqrt{7}}{7}\).
3. Incorrect. You may have rationalized the fraction \(\ \frac{7}{\sqrt{3}}\) instead of \(\ \frac{3}{\sqrt{7}}\). Try multiplying the numerator and denominator by \(\ \frac{\sqrt{7}}{\sqrt{7}}\). The correct answer is \(\ \frac{3 \sqrt{7}}{7}\).
4. Correct. To rationalize the denominator, multiply \(\ \frac{3}{\sqrt{7}}\) by \(\ \frac{\sqrt{7}}{\sqrt{7}}\). This will create an integer in the denominator, so the fraction will have been rationalized.

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, but 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 that when binomials of the form \(\ (a+b)(a-b)\) are multiplied, the product is \(\ a^{2}-b^{2}\). So, for example, \(\ (x+3)(x-3)=x^{2}-3 x+3 x-9=x^{2}-9\); notice that the terms \(\ -3 x\) and \(\ +3 x\) combine to 0. Now for the connection to rationalizing denominators: what if you replaced \(\ x\) with \(\ \sqrt{2}\)?

Look at the examples below. Just as \(\ -3 x+3 x\) combines to 0 in the top example, \(\ -3 \sqrt{2}+3 \sqrt{2}\) combines to 0 in the bottom example.

\(\ \begin{array}{l}
(x+3)(x-3) \\
=x^{2}-3 x+3 x-9 \\
=x^{2}-9
\end{array}\)

\(\ \begin{array}{l}
(\sqrt{2}+3)(\sqrt{2}-3) \\
=(\sqrt{2})^{2}-3 \sqrt{2}+3 \sqrt{2}-9 \\
=(\sqrt{2})^{2}-9 \\
=2-9 \\
=-7
\end{array}\)

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 a conjugate, 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
\(\ \sqrt{2}+3\)	\(\ \sqrt{2}-3\)	\(\ (\sqrt{2}+3)(\sqrt{2}-3)=(\sqrt{2})^{2}-(3)^{2}=2-9=-7\)
\(\ \sqrt{x}-5\)	\(\ \sqrt{x}+5\)	\(\ (\sqrt{x}-5)(\sqrt{x}+5)=(\sqrt{x})^{2}-(5)^{2}=x-25\)
\(\ 8-2 \sqrt{x}\)	\(\ 8+2 \sqrt{x}\)	\(\ (8-2 \sqrt{x})(8+2 \sqrt{x})=(8)^{2}-(2 \sqrt{x})^{2}=64-4 x\)
\(\ 1+\sqrt{x y}\)	\(\ 1-\sqrt{x y}\)	\(\ (1+\sqrt{x y})(1-\sqrt{x y})=(1)^{2}-(\sqrt{x y})^{2}=1-x y\)

More generally, if a fraction has a denominator of the form \(a + b \sqrt{c} \), then the conjugate is \(a - b \sqrt{c} \) and vice-versa.

Example

Rationalize the denominator and simplify. \(\ \frac{5-\sqrt{7}}{3+\sqrt{5}}\)

Solution

\(\ \frac{5-\sqrt{7}}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}\)	Find the conjugate of \(\ 3+\sqrt{5}\). Then multiply the entire expression by \(\ \frac{3-\sqrt{5}}{3-\sqrt{5}}\).
\(\ \frac{(5-\sqrt{7})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}\)
\(\ \frac{5 \cdot 3-5 \sqrt{5}-3 \sqrt{7}+\sqrt{7} \cdot \sqrt{5}}{3 \cdot 3-3 \sqrt{5}+3 \sqrt{5}-\sqrt{5} \cdot \sqrt{5}}\)	Use the Distributive Property to multiply the binomials in the numerator and denominator.
\(\ \frac{15-5 \sqrt{5}-3 \sqrt{7}+\sqrt{35}}{9-3 \sqrt{5}+3 \sqrt{5}-\sqrt{25}}\)	Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to 0.
\(\ \begin{array}{l} \frac{15-5 \sqrt{5}-3 \sqrt{7}+\sqrt{35}}{9-\sqrt{25}} \\ \frac{15-5 \sqrt{5}-3 \sqrt{7}+\sqrt{35}}{9-5} \end{array}\)	Simplify radicals where possible.

\(\ \frac{5-\sqrt{7}}{3+\sqrt{5}}=\frac{15-5 \sqrt{5}-3 \sqrt{7}+\sqrt{35}}{4}\)

Example

Rationalize the denominator and simplify. \(\ \frac{\sqrt{x}}{\sqrt{x}+2}\)

Solution

\(\ \frac{\sqrt{x}}{\sqrt{x}+2} \cdot \frac{\sqrt{x}-2}{\sqrt{x}-2}\)	Find the conjugate of \(\ \sqrt{x}+2\). Then multiply the numerator and denominator by \(\ \frac{\sqrt{x}-2}{\sqrt{x}-2}\).
\(\ \frac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}\)
\(\ \frac{\sqrt{x} \cdot \sqrt{x}-2 \sqrt{x}}{\sqrt{x} \cdot \sqrt{x}-2 \sqrt{x}+2 \sqrt{x}-2 \cdot 2}\)	Use the Distributive Property to multiply the binomials in the numerator and denominator.
\(\ \frac{\sqrt{x} \cdot \sqrt{x}-2 \sqrt{x}}{\sqrt{x} \cdot \sqrt{x}-2 \sqrt{x}+2 \sqrt{x}-4}\)	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.

\(\ \frac{\sqrt{x}}{\sqrt{x}+2}=\frac{x-2 \sqrt{x}}{x-4}\)

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.

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

\(\ \sqrt[3]{100}\) 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.

Exercise

Identify the conjugate of the denominator. \(\ \frac{x+2}{\sqrt{2 x}-7}\)

1. \(\ \sqrt{2 x}+7\)
2. \(\ \sqrt{2 x}-7\)
3. \(\ 2 x+\sqrt{7}\)
4. \(\ (2 x)^{2}+7\)

Answer

1. Correct. The conjugate will be the binomial that, when multiplied by the denominator, eliminates the radical. The conjugate is \(\ \sqrt{2 x}+7\).
2. Incorrect. The denominator is not the conjugate. Look for the binomial that, when multiplied by the denominator, eliminates the radical. The correct answer is \(\ \sqrt{2 x}+7\).
3. Incorrect. Multiplying the denominator by \(\ 2 x+\sqrt{7}\) will not remove the radical. Look for the binomial that follows the pattern \(\ (a+b)(a-b)=a^{2}-b^{2}\). The correct answer is \(\ \sqrt{2 x}+7\).
4. Incorrect. Multiplying the denominator by \(\ (2 x)^{2}+7\) will not remove the radical. Look for the binomial that follows the pattern \(\ (a+b)(a-b)=a^{2}-b^{2}\). The correct answer is \(\ \sqrt{2 x}+7\).

Rationalizing The Numerator

The process of rationalizing the numerator is extremely similar to that of rationalizing the denominator.

Exercise \(\PageIndex{1}\)

Rationalize the numerator: \[ \frac{\sqrt{4+h}-2}{h} \nonumber\ \] You may assume that \(h \neq 0 \).

Answer

\begin{align*} \frac{\sqrt{4+h}-2}{h} &= \frac{\sqrt{4+h}-2}{h} \times \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2} \\ &= \frac{1}{\sqrt{4+h}+2} \end{align*}

Exercise \(\PageIndex{1}\)

Rationalize the numerator: \[ \frac{\sqrt{t^2+9}-3}{t} \nonumber\ \] You may assume that \(t \neq 0 \).

Answer

\[ \frac{\sqrt{t^2+9}-3}{t} = \frac{\sqrt{t^2+9}-3}{t} \times \frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3} = \frac{t}{\sqrt{t^2+9}+3} \nonumber\ \]