10.4: Rationalize Denominators
- Page ID
- 45137
Rationalizing the denominator is the process for obtaining denominators without radicals.
When given a quotient with radicals, it is common practice to leave an expression without a radical in the denominator. After simplifying an expression, if there is a radical in the denominator, we will rationalize it so that the denominator is left without any radicals. We start by rationalizing denominators with square roots, and then extend this idea to higher roots.
Rationalizing Denominators with Square Roots
To rationalize the denominator with a square root, multiply the numerator and denominator by the exact radical in the denominator, e.g, \[\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}}\nonumber\]
Simplify: \(\dfrac{\sqrt{6}}{\sqrt{5}}\)
Solution
We see the expression is irreducible and that the denominator contains \(\sqrt{5}\). We rationalize the denominator so that the denominator is left without radicals.
\[\begin{array}{rl}\dfrac{\sqrt{6}}{\sqrt{5}}&\text{Rationalize the denominator} \\ \dfrac{\sqrt{6}}{\sqrt{5}}\cdot\color{blue}{\dfrac{\sqrt{5}}{\sqrt{5}}}&\color{black}{\text{Multiply fractions}} \\ \dfrac{\sqrt{6}\cdot\color{blue}{\sqrt{5}}}{\sqrt{5}\cdot\color{blue}{\sqrt{5}}}&\color{black}{\text{Apply product rule}} \\ \dfrac{\sqrt{30}}{\sqrt{25}}&\text{Simplify radicals} \\ \dfrac{\sqrt{30}}{5}&\text{Simplified expression}\end{array}\nonumber\]
Notice, the expression is simplified completely and there are no longer any radicals in the denominator. This is the goal for these problems.
Simplify: \(\dfrac{6\sqrt{14}}{12\sqrt{22}}\)
Solution
We see the expression isn’t reduced. We will reduce the fraction by applying the quotient rule, then rationalize the denominator, if needed.
\[\begin{array}{rl}\dfrac{6\sqrt{14}}{12\sqrt{22}}&\text{Apply quotient rule} \\ \dfrac{6}{12}\cdot\sqrt{\dfrac{14}{22}}&\text{Reduce fractions} \\ \dfrac{1}{2}\cdot\sqrt{\dfrac{7}{11}}&\text{Rewrite as one fraction} \\ \dfrac{\sqrt{7}}{2\sqrt{11}}&\text{Rationalize the denominator} \\ \dfrac{\sqrt{7}}{2\sqrt{11}}\cdot \color{blue}{\dfrac{\sqrt{11}}{\sqrt{11}}}&\text{Multiple fractions} \\ \dfrac{\sqrt{7}\cdot\color{blue}{\sqrt{11}}}{2\cdot\sqrt{11}\cdot\color{blue}{\sqrt{11}}}&\text{Apply product rule} \\ \dfrac{\sqrt{77}}{2\cdot\sqrt{121}}&\text{Simplify radicals} \\ \dfrac{\sqrt{77}}{2\cdot 11}&\text{Simplify radicals} \\ \dfrac{\sqrt{77}}{22}&\text{Simplified expression}\end{array}\nonumber\]
Simplify: \(\dfrac{\sqrt{3}-9}{2\sqrt{6}}\)
Solution
We see the expression is irreducible and that the denominator contains \(\sqrt{6}\). We rationalize the denominator so that the denominator is left without radicals.
\[\begin{array}{rl}\dfrac{\sqrt{3}-9}{2\sqrt{6}}&\text{Rationalize the denominator} \\ \dfrac{(\sqrt{3}-9)}{2\sqrt{6}}\cdot\color{blue}{\dfrac{\sqrt{6}}{\sqrt{6}}}&\color{black}{\text{Multiply fractions}} \\ \dfrac{\color{blue}{\sqrt{6}}\color{black}{(}\sqrt{3}-9)}{2\sqrt{6}\cdot\color{blue}{\sqrt{6}}}&\color{black}{\text{Distribute and apply product rule}} \\ \dfrac{\sqrt{18}-9\sqrt{6}}{2\cdot\sqrt{36}}&\text{Rewrite the radicand }18 \\ \dfrac{\sqrt{9\cdot 2}-9\sqrt{6}}{12}&\text{Simplify radicals} \\ \dfrac{3\sqrt{2}-9\sqrt{6}}{12}&\text{Factor a GCF from the numerator} \\ \dfrac{\cancel{3}(\sqrt{2}-3\sqrt{6})}{\cancelto{4}{12}}&\text{Reduce by a factor of }3 \\ \dfrac{(\sqrt{2}-3\sqrt{6})}{4}&\text{Simplified expression}\end{array}\nonumber\]
Rationalizing Denominators with Higher Roots
Radicals with higher roots in the denominators are a bit more challenging. Notice, rationalizing the denominator with square roots works out nicely because we are only trying to obtain a radicand that is a perfect square in the denominator. Here, we are trying to obtain radicands that are perfect cubes or higher in the denominator. Let’s try an example.
Simplify: \(\dfrac{4\sqrt[3]{2}}{7\sqrt[3]{25}}\)
Solution
We see the expression is irreducible and that the denominator contains \(\sqrt[3]{25}\). We rationalize the denominator so that the denominator is left without radicals. Notice we need a radicand that is a perfect cube in the denominator.
\[\begin{array}{rl}\dfrac{4\sqrt[3]{2}}{7\sqrt[3]{25}}&\text{Rationalize the denominator} \\ \dfrac{4\sqrt[3]{2}}{7\sqrt[3]{25}}\cdot\color{blue}{\dfrac{\sqrt[3]{5}}{\sqrt[3]{5}}}&\color{black}{\text{Multiply fractions}} \\ \dfrac{4\cdot\sqrt[3]{2}\cdot\color{blue}{\sqrt[3]{5}}}{7\cdot\sqrt[3]{25}\cdot\color{blue}{\sqrt[3]{5}}}&\color{black}{\text{Apply product rule}} \\ \dfrac{4\sqrt[3]{10}}{7\sqrt[3]{125}}&\text{Simplify radicals} \\ \dfrac{4\sqrt[3]{10}}{7\cdot 5}&\text{Simplify radicals} \\ \dfrac{4\sqrt[3]{10}}{35}&\text{Simplified expression}\end{array}\nonumber\]
We choose to multiply by \(\sqrt[3]{5}\) because we noticed \(\sqrt[3]{25}=\sqrt[3]{5^2}\), and all we needed was an additional factor of \(5\) to make a perfect cube in the denominator. Since \(7\) is a coefficient and not a part of the radicand, we do not include it when rationalizin
Simplify: \(\dfrac{3\sqrt[4]{11}}{\sqrt[4]{2}}\)
Solution
We see the expression is irreducible and that the denominator contains \(\sqrt[4]{2}\). We rationalize the denominator so that the denominator is left without radicals. Notice we need a radicand that is a perfect fourth power in the denominator.
\[\begin{array}{rl}\dfrac{3\sqrt[4]{11}}{\sqrt[4]{2}}&\text{Rationalize the denominator} \\ \dfrac{3\sqrt[4]{11}}{\sqrt[4]{2}}\cdot\color{blue}{\dfrac{\sqrt[4]{8}}{\sqrt[4]{8}}}&\color{black}{\text{Multiply fractions}} \\ \dfrac{3\cdot\sqrt[4]{11}\cdot\color{blue}{\sqrt[4]{8}}}{\sqrt[4]{2}\cdot\color{blue}{\sqrt[4]{8}}}&\color{black}{\text{Apply product rule}} \\ \dfrac{3\sqrt[4]{88}}{\sqrt[4]{16}}&\text{Simplify radicals} \\ \dfrac{3\sqrt[4]{88}}{2}&\text{Simplified expression}\end{array}\nonumber\]
We choose to multiply by \(\sqrt[4]{8}\) because we noticed \(\sqrt[4]{2}\), and all we needed was three additional factors of \(2\) to make a perfect fourth power in the denominator.
Rationalize Denominators Using the Conjugate
There are times where the given denominator is not just one term. Often, in the denominator, we have a difference or sum of two terms in which one or both terms are square roots. In order to rationalize these denominators, we use the idea from a difference of two squares:
\[(a+b)(a-b)=a^2-b^2\nonumber\]
Notice, with the difference of two squares, we are left without any outer or inner product terms- just the squares of the first and last terms. Since these denominators take the form of a binomial, we have a special name for the factor we use when rationalizing the denominator. The factor is called the conjugate.
We rationalize denominators of the type \(a\pm\sqrt{b}\) by multiplying the numerator and denominator by their conjugates, e.g.,
\[\dfrac{1}{a+\sqrt{b}}\cdot\dfrac{a-\sqrt{b}}{a-\sqrt{b}}\nonumber\]
The conjugate for
- \(a+\sqrt{b}\) is \(a-\sqrt{b}\)
- \(a-\sqrt{b}\) is \(a+\sqrt{b}\)
The case is similar for when there is something like \(\sqrt{a}\pm\sqrt{b}\) in the denominator.
Putting all these ideas together, let's try an example.
Simplify: \(\dfrac{2}{\sqrt{3}-5}\)
Solution
We notice the difference in the denominator and so we know we will use the conjugate to rationalize the denominator.
\[\begin{array}{rl}\dfrac{2}{\sqrt{3}-5}&\text{Rationalize the denominator} \\ \dfrac{2}{(\sqrt{3}-5)}\cdot\color{blue}{\dfrac{(\sqrt{3}+5)}{(\sqrt{3}+5)}}&\color{black}{\text{Multiply fractions}} \\ \dfrac{2\color{blue}{(\sqrt{3}+5)}}{(\sqrt{3}-5)\color{blue}{(\sqrt{3}+5)}}&\color{black}{\text{Distribute and FOIL}} \\ \dfrac{2\sqrt{3}+10}{\sqrt{9}+\cancel{5\sqrt{3}}-\cancel{5\sqrt{3}}-25}&\text{Simplify} \\ \dfrac{2\sqrt{3}+10}{3-25}&\text{Subtract} \\ \dfrac{2\sqrt{3}+10}{-22}&\text{Factor a GCF from the numerator} \\ \dfrac{2(\sqrt{3}+5)}{-22}&\text{Reduce by a factor of }2 \\ \dfrac{\sqrt{3}+5}{-11}&\text{Rewrite} \\ -\dfrac{\sqrt{3}+5}{11}&\text{Simplified expression}\end{array}\nonumber\]
Simplify: \(\dfrac{3-\sqrt{5}}{2-\sqrt{3}}\)
Solution
We notice the difference in the denominator and so we know we will use the conjugate to rationalize the denominator.
\[\begin{array}{rl}\dfrac{3-\sqrt{5}}{2-\sqrt{3}}&\text{Rationalize the denominator} \\ \dfrac{(3-\sqrt{5})}{(2-\sqrt{3})}\cdot\color{blue}{\dfrac{(2+\sqrt{3})}{(2+\sqrt{3})}}&\color{black}{\text{Multiply fractions}} \\ \dfrac{(3-\sqrt{5})\color{blue}{(2+\sqrt{3})}}{(2-\sqrt{3})\color{blue}{(2+\sqrt{3})}}&\color{black}{\text{FOIL}} \\ \dfrac{6+3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{4+\cancel{2\sqrt{3}}-\cancel{2\sqrt{3}}-\sqrt{9}}&\text{Simplify} \\ \dfrac{6+3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{4-3}&\text{Subtract} \\ \dfrac{6+3\sqrt{3}-2\sqrt{5}-\sqrt{15}}{1}&\text{Rewrite} \\ 6+3\sqrt{3}-2\sqrt{5}-\sqrt{15}&\text{Simplified expression}\end{array}\nonumber\]
During the \(5^{\text{th}}\) century BC in India, Aryabhata published a treatise on astronomy. His work included a method for finding the square root of numbers that have many digits.
Simplify: \(\dfrac{2\sqrt{5}-3\sqrt{7}}{5\sqrt{6}+4\sqrt{2}}\)
Solution
We notice the sum in the denominator and so we know we will use the conjugate to rationalize the denominator.
\[\begin{array}{rl}\dfrac{2\sqrt{5}-3\sqrt{7}}{5\sqrt{6}+4\sqrt{2}}&\text{Rationalize the denominator} \\ \dfrac{(2\sqrt{5}-3\sqrt{7})}{(5\sqrt{6}+4\sqrt{2})}\cdot\color{blue}{\dfrac{(5\sqrt{6}-4\sqrt{2})}{(5\sqrt{6}-4\sqrt{2})}}&\color{black}{\text{Multiply fractions}} \\ \dfrac{(2\sqrt{5}-3\sqrt{7})\color{blue}{(5\sqrt{6}-4\sqrt{2})}}{(5\sqrt{6}+4\sqrt{2})\color{blue}{(5\sqrt{6}-4\sqrt{2})}}&\color{black}{\text{FOIL}} \\ \dfrac{10\sqrt{30}-8\sqrt{10}-15\sqrt{42}-12\sqrt{14}}{25\sqrt{36}-\cancel{20\sqrt{12}}+\cancel{20\sqrt{12}}-16\sqrt{4}}&\text{Simplify} \\ \dfrac{10\sqrt{30}-8\sqrt{10}-15\sqrt{42}-12\sqrt{14}}{25\cdot 6-16\cdot 2}&\text{Subtract} \\ \dfrac{10\sqrt{30}-8\sqrt{10}-15\sqrt{42}-12\sqrt{14}}{118}&\text{Simplified expression}\end{array}\nonumber\]
Rationalize Denominators Homework
Simplify.
\(\dfrac{2\sqrt{4}}{3\sqrt{3}}\)
\(\dfrac{\sqrt[3]{5}}{4\sqrt[3]{4}}\)
\(\dfrac{\sqrt{12}}{\sqrt{3}}\)
\(\dfrac{\sqrt{2}}{3\sqrt{5}}\)
\(\dfrac{4\sqrt{3}}{\sqrt{15}}\)
\(\dfrac{4+2\sqrt{3}}{\sqrt{9}}\)
\(\dfrac{4+2\sqrt{3}}{5\sqrt{4}}\)
\(\dfrac{2-5\sqrt{5}}{4\sqrt{13}}\)
\(\dfrac{\sqrt{2}-3\sqrt{3}}{\sqrt{3}}\)
\(\dfrac{5}{3\sqrt{5}+\sqrt{2}}\)
\(\dfrac{2}{5+\sqrt{2}}\)
\(\dfrac{3}{4-3\sqrt{3}}\)
\(\dfrac{4}{3+\sqrt{5}}\)
\(-\dfrac{4}{4-4\sqrt{2}}\)
\(\dfrac{1}{1+\sqrt{2}}\)
\(\dfrac{\sqrt{14}-2}{\sqrt{7}-\sqrt{2}}\)
\(\dfrac{\sqrt{ab}-a}{\sqrt{b}-\sqrt{a}}\)
\(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
\(\dfrac{2+\sqrt{6}}{2+\sqrt{3}}\)
\(\dfrac{a-\sqrt{b}}{a+\sqrt{b}}\)
\(\dfrac{6}{3\sqrt{2}-2\sqrt{3}}\)
\(\dfrac{a-b}{a\sqrt{b}-b\sqrt{a}}\)
\(\dfrac{2-\sqrt{5}}{-3+\sqrt{5}}\)
\(\dfrac{-4+\sqrt{3}}{4\sqrt{9}}\)
\(\dfrac{2\sqrt{3}-2}{2\sqrt{16}}\)
\(\dfrac{\sqrt{5}+4}{4\sqrt{17}}\)
\(\dfrac{\sqrt{5}-\sqrt{2}}{3\sqrt{6}}\)
\(\dfrac{5}{\sqrt{3}+4\sqrt{5}}\)
\(\dfrac{5}{2\sqrt{3}-\sqrt{2}}\)
\(\dfrac{4}{\sqrt{2}-2}\)
\(\dfrac{2}{2\sqrt{5}+2\sqrt{3}}\)
\(\dfrac{4}{4\sqrt{3}-\sqrt{5}}\)
\(\dfrac{3+\sqrt{3}}{\sqrt{3}-1}\)
\(\dfrac{2+\sqrt{10}}{\sqrt{2}+\sqrt{5}}\)
\(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{14}+\sqrt{7}}\)
\(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
\(\dfrac{2\sqrt{5}+\sqrt{3}}{1-\sqrt{3}}\)
\(\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(\dfrac{ab}{a\sqrt{b}-b\sqrt{a}}\)
\(\dfrac{4\sqrt{2}+3}{3\sqrt{2}+\sqrt{3}}\)
\(\dfrac{-1+\sqrt{5}}{2\sqrt{5}+5\sqrt{2}}\)
\(\dfrac{5\sqrt{2}+\sqrt{3}}{5+5\sqrt{2}}\)
\(\dfrac{\sqrt{3}+\sqrt{2}}{2\sqrt{3}-\sqrt{2}}\)