Rationalizing Denominators
Learning Outcome
 Remove radicals from a single term denominator
 Rationalize denominators with multiple terms
Irrational 
Rational 


[latex] \frac{1}{\sqrt{2}}[/latex] 
= 
[latex] \frac{\sqrt{2}}{2}[/latex] 
[latex] \frac{2+\sqrt{3}}{\sqrt{3}}[/latex] 
= 
[latex] \frac{2\sqrt{3}+3}{3}[/latex] 
Rationalizing Denominators with One Term
Let us start with the fraction [latex] \frac{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] \frac{1}{\sqrt{2}}[/latex] is. You can rename this fraction without changing its value if you multiply it by a quantity equal to [latex]1[/latex]. In this case, let that quantity be [latex] \frac{\sqrt{2}}{\sqrt{2}}[/latex]. Watch what happens.[latex] \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}[/latex]
The denominator of the new fraction is no longer a radical (notice, however, that the numerator is). So why choose to multiply [latex] \frac{1}{\sqrt{2}}[/latex] by [latex] \frac{\sqrt{2}}{\sqrt{2}}[/latex]? You knew that the square root of a number times itself will be a whole number. In algebraic terms, this idea is represented by [latex] \sqrt{x}\cdot \sqrt{x}=x[/latex]. Look back to the denominators in the multiplication of [latex] \frac{1}{\sqrt{2}}\cdot 1[/latex]. Do you see where [latex] \sqrt{2}\cdot \sqrt{2}=\sqrt{4}=2[/latex]? 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 [latex]1[/latex].Example
Rationalize the denominator.[latex] \frac{6\sqrt{6}}{\sqrt{3}}[/latex]
Answer: The denominator of this fraction is [latex] \sqrt{3}[/latex]. To make it into a rational number, multiply it by [latex] \sqrt{3}[/latex], since [latex] \sqrt{3}\cdot \sqrt{3}=3[/latex]. This means we need to multiply the entire fraction by [latex] \frac{\sqrt{3}}{\sqrt{3}}[/latex] because it is equal to [latex]1[/latex].
[latex]\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}[/latex]
Simplify the coefficients and the radicals, where possible.[latex]\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]
Answer
[latexdisplay] \frac{6\sqrt{6}}{\sqrt{3}}=6\sqrt{2}[/latexdisplay]Example
Rationalize the denominator.[latex] \frac{2+\sqrt{3}}{\sqrt{3}}[/latex]
Answer: The denominator of this fraction is [latex] \sqrt{3}[/latex]. To make it into a rational number, multiply it by [latex] \sqrt{3}[/latex], since [latex] \sqrt{3}\cdot \sqrt{3}=3[/latex].
[latex] \frac{2+\sqrt{3}}{\sqrt{3}}[/latex]
Multiply the entire fraction by a quantity which simplifies to [latex]1[/latex]: [latex] \frac{\sqrt{3}}{\sqrt{3}}[/latex].[latex]\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}[/latex]
Use the Distributive Property to multiply [latex] \sqrt{3}(2+\sqrt{3})[/latex].[latex] \frac{2\sqrt{3}+\sqrt{3}\cdot \sqrt{3}}{\sqrt{9}}[/latex]
[latex] \frac{2\sqrt{3}+\sqrt{9}}{\sqrt{9}}[/latex]
Simplify the radicals, where possible. [latex] \sqrt{9}=3[/latex]. The answer is [latex]\frac{2\sqrt{3}+3}{3}[/latex].Try It
[ohm_question]2765[/ohm_question]Example
Rationalize the denominator.[latex] \frac{\sqrt{2y}}{\sqrt{4x}},\text{ where }x\ne \text{0}[/latex]
Answer: The denominator is [latex] \sqrt{4x}[/latex], so the entire expression can be multiplied by [latex] \frac{\sqrt{4x}}{\sqrt{4x}}[/latex] to get rid of the radical in the denominator.
[latex] \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}}}{4x}\end{array}[/latex]
Simplify the numerator. [latexdisplay]\frac{2\sqrt{2\cdot{x}\cdot{y}}}{4x}\\\\=\frac{2\sqrt{2xy}}{4x}[/latexdisplay]Answer
[latexdisplay]\frac{\sqrt{2y}}{\sqrt{4x}}=\frac{2\sqrt{2xy}}{4x}[/latexdisplay]Example
Rationalize the denominator.[latex] \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}},\text{ where }x\ne \text{0}[/latex]
Answer: The denominator is [latex] \sqrt{x}[/latex], so the entire expression can be multiplied by [latex] \frac{\sqrt{x}}{\sqrt{x}}[/latex] to get rid of the radical in the denominator.
[latex] \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}[/latex]
Use the Distributive Property. Simplify the radicals where possible. Remember that [latex] \sqrt{x}\cdot \sqrt{x}=x[/latex].[latex] \frac{\sqrt{x}\cdot \sqrt{x}+\sqrt{x}\cdot \sqrt{y}}{\sqrt{x}\cdot \sqrt{x}}[/latex]
The answer is [latex]\frac{x+\sqrt{xy}}{x}[/latex].Example
Rationalize the denominator and simplify.[latex] \sqrt{\frac{100x}{11y}},\text{ where }y\ne \text{0}[/latex]
Answer: Use the property [latex] \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}[/latex] to rewrite the radical.
[latex] \frac{\sqrt{100x}}{\sqrt{11y}}[/latex]
The denominator is [latex] \sqrt{11y}[/latex], so multiplying the entire expression by [latex] \frac{\sqrt{11y}}{\sqrt{11y}}[/latex] will rationalize the denominator.[latex] \frac{\sqrt{100x}\cdot\sqrt{11y}}{\sqrt{11y}\cdot\sqrt{11y}}[/latex]
Multiply and simplify the radicals where possible.[latex] \frac{\sqrt{100\cdot 11xy}}{\sqrt{11y}\cdot \sqrt{11y}}[/latex]
100 is a perfect square. Remember that[latex] \sqrt{100}=10[/latex] and [latex] \sqrt{x}\cdot \sqrt{x}=x[/latex].[latex] \frac{\sqrt{100}\cdot \sqrt{11xy}}{\sqrt{11y}\cdot \sqrt{11y}}[/latex]
The answer is [latex]\frac{10\sqrt{11xy}}{11y}[/latex].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] \frac{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 singleterm denominators. If you multiply [latex] \sqrt{2}+3[/latex] by [latex] \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! 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 [latex] (a+b)(ab)[/latex] are multiplied. So, for example, [latex] (x+3)(x3)={{x}^{2}}3x+3x9={{x}^{2}}9[/latex]; notice that the terms [latex]−3x[/latex] and [latex]+3x[/latex] combine to 0. Now for the connection to rationalizing denominators: what if you replaced x with [latex] \sqrt{2}[/latex]? Look at the side by side examples below. Just as [latex] 3x+3x[/latex] combines to [latex]0[/latex] on the left, [latex] 3\sqrt{2}+3\sqrt{2}[/latex] combines to [latex]0[/latex] on the right.[latex] \begin{array}{l}(x+3)(x3)\\={{x}^{2}}3x+3x9\\={{x}^{2}}9\end{array}[/latex]  [latex] \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\\=29\\=7\end{array}[/latex] 
Term  Conjugate  Product 

[latex] \sqrt{2}+3[/latex]  [latex] \sqrt{2}3[/latex]  [latex] \left( \sqrt{2}+3 \right)\left( \sqrt{2}3 \right)={{\left( \sqrt{2} \right)}^{2}}{{\left( 3 \right)}^{2}}=29=7[/latex] 
[latex] \sqrt{x}5[/latex]  [latex] \sqrt{x}+5[/latex]  [latex] \left( \sqrt{x}5 \right)\left( \sqrt{x}+5 \right)={{\left( \sqrt{x} \right)}^{2}}{{\left( 5 \right)}^{2}}=x25[/latex] 
[latex] 82\sqrt{x}[/latex]  [latex] 8+2\sqrt{x}[/latex]  [latex] \left( 82\sqrt{x} \right)\left( 8+2\sqrt{x} \right)={{\left( 8 \right)}^{2}}{{\left( 2\sqrt{x} \right)}^{2}}=644x[/latex] 
[latex] 1+\sqrt{xy}[/latex]  [latex] 1\sqrt{xy}[/latex]  [latex] \left( 1+\sqrt{xy} \right)\left( 1\sqrt{xy} \right)={{\left( 1 \right)}^{2}}{{\left( \sqrt{xy} \right)}^{2}}=1xy[/latex] 
Example
Rationalize the denominator and simplify.[latex] \frac{5\sqrt{7}}{3+\sqrt{5}}[/latex]
Answer: Find the conjugate of [latex] 3+\sqrt{5}[/latex]. Then multiply the entire expression by [latex] \frac{3\sqrt{5}}{3\sqrt{5}}[/latex].
[latex] \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}[/latex]
Use the Distributive Property to multiply the binomials in the numerator and denominator.[latex] \frac{5\cdot 35\sqrt{5}3\sqrt{7}+\sqrt{7}\cdot \sqrt{5}}{3\cdot 33\sqrt{5}+3\sqrt{5}\sqrt{5}\cdot \sqrt{5}}[/latex]
Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to [latex]0[/latex].[latex] \frac{155\sqrt{5}3\sqrt{7}+\sqrt{35}}{93\sqrt{5}+3\sqrt{5}\sqrt{25}}[/latex]
Simplify radicals where possible.[latex] \begin{array}{c}\frac{155\sqrt{5}3\sqrt{7}+\sqrt{35}}{9\sqrt{25}}\\\\\frac{155\sqrt{5}3\sqrt{7}+\sqrt{35}}{95}\end{array}[/latex]
The answer is [latex]\frac{155\sqrt{5}3\sqrt{7}+\sqrt{35}}{4}[/latex].Example
Rationalize the denominator and simplify.[latex] \frac{\sqrt{x}}{\sqrt{x}+2}[/latex]
Answer: Find the conjugate of [latex] \sqrt{x}+2[/latex]. Then multiply the numerator and denominator by [latex] \frac{\sqrt{x}2}{\sqrt{x}2}[/latex].
[latex] \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}[/latex]
Use the Distributive Property to multiply the binomials in the numerator and denominator.[latex] \frac{\sqrt{x}\cdot \sqrt{x}2\sqrt{x}}{\sqrt{x}\cdot \sqrt{x}2\sqrt{x}+2\sqrt{x}2\cdot 2}[/latex]
Simplify. Remember that [latex] \sqrt{x}\cdot \sqrt{x}=x[/latex]. Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to [latex]0[/latex].[latex] \frac{\sqrt{x}\cdot \sqrt{x}2\sqrt{x}}{\sqrt{x}\cdot \sqrt{x}2\sqrt{x}+2\sqrt{x}4}[/latex]
The answer is [latex]\frac{x2\sqrt{x}}{x4}[/latex].Try It
[ohm_question]196062[/ohm_question][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 any further since its prime factors are [latex] 2\cdot 2\cdot 5\cdot 5[/latex]. There are no cubed numbers to pull out! Multiplying [latex] \sqrt[3]{10}+5[/latex] by its conjugate does not result in a radicalfree expression. In the following video, we show more examples of how to rationalize a denominator using the conjugate. https://youtu.be/vINRIRgeKqUSummary
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] \frac{1}{\sqrt{5}}[/latex], multiplying the fraction by [latex] \frac{\sqrt{5}}{\sqrt{5}}[/latex] will remove the radical from the denominator. When the denominator contains two terms, as in[latex] \frac{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.Contribute!
Licenses & Attributions
CC licensed content, Shared previously
 Ex 1: Rationalize the Denominator of a Radical Expression. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
 Ex: Rationalize the Denominator of a Radical Expression  Conjugate. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
 College Algebra. Provided by: OpenStax Authored by: Abramson, Jay. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at: http://cnx.org/contents/[email protected]:1/Prefac.
 Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology Located at: https://www.nroc.org/. License: CC BY: Attribution.