Rationalize Denominators
Learning Outcome
- Rationalize denominators with one or 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 another quantity equal to [latex]1[/latex].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].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 single-term 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)(a-b)[/latex] are multiplied. So, for example, [latex] (x+3)(x-3)={{x}^{2}}-3x+3x-9={{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)(x-3)\\={{x}^{2}}-3x+3x-9\\={{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\\=2-9\\=-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}}=2-9=-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}}=x-25[/latex] |
[latex] 8-2\sqrt{x}[/latex] | [latex] 8+2\sqrt{x}[/latex] | [latex] \left( 8-2\sqrt{x} \right)\left( 8+2\sqrt{x} \right)={{\left( 8 \right)}^{2}}-{{\left( 2\sqrt{x} \right)}^{2}}=64-4x[/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}}=1-xy[/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 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}}[/latex]
Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to [latex]0[/latex].[latex] \frac{15-5\sqrt{5}-3\sqrt{7}+\sqrt{35}}{9-3\sqrt{5}+3\sqrt{5}-\sqrt{25}}[/latex]
Simplify radicals where possible.[latex] \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}[/latex]
The answer is [latex]\frac{15-5\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{x-2\sqrt{x}}{x-4}[/latex].[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 radical-free 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.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.