How to Rationalize Radical Expressions? (+FREE Worksheet!)

As you may know, radical expressions cannot be in the denominator, so in this article, we will teach you how to get rid of them by rationalizing radical expressions.
Related Topics
- How to Add and Subtract Radical Expressions
- How to Multiply Radical Expressions
- How to Solve Radical Equations
- How to Simplify Radical Expressions
- How to Find Domain and Range of Radical Functions
A step-by-step guide to Rationalizing Radical Expressions
- Radical expressions cannot be in the denominator. (number in the bottom)
- To get rid of the radicals in the denominator, multiply both the numerator and the denominator by the radical in the denominator.
- If there is a radical and another integer in the denominator, multiply both numerator and denominator by the conjugate of the denominator.
- The conjugate of \((a+b)\) is \((a-b)\) and vice versa.
Examples
Rationalizing Radical Expressions – Example 1:
Simplify. \(\frac{5}{\sqrt{6}- 4}\)
Solution:
Multiply by the conjugate: \(\frac{\sqrt{6}+ 4}{\sqrt{6}+ 4 } →\frac{5}{\sqrt{6}- 4 }×\frac{\sqrt{6}+ 4}{\sqrt{6}+ 4 }\)
\( (\sqrt{6}- 4)(\sqrt{6}+ 4)=-10\) then: \(\frac{5}{\sqrt{6}- 4 }×\frac{\sqrt{6}+ 4}{\sqrt{6}+ 4 }=\frac{5(\sqrt{6}+ 4)}{-10 }\)
Use the fraction rule: \(\frac{a}{-b}=-\frac{a}{b}→\frac{5(\sqrt{6}+ 4)}{-10 }=-\frac{5(\sqrt{6}+ 4)}{10 }=-\frac{1}{2 }(\sqrt{6}+ 4)\)
Rationalizing Radical Expressions – Example 2:
Simplify. \(\frac{2}{\sqrt{3}- 1 }\)
Solution:
Multiply by the conjugate: \(\frac{\sqrt{3}+1}{\sqrt{3}+1}\) → \(\frac{2}{\sqrt{3}-1 }×\frac{\sqrt{3}+1}{\sqrt{3}+1}\)
\(({\sqrt{3}- 1 )}\) \(({\sqrt{3}+1 )}\) \(=2\) then: \(\frac{2}{\sqrt{3}-1 }×\frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2 × (\sqrt{3}+1)}{2} =\sqrt{3}+1\)
Rationalizing Radical Expressions – Example 3:
Simplify. \(\frac{1}{\sqrt{5}- 2 }\)
Solution:
Multiply by the conjugate: \(\frac{\sqrt{5}+ 2}{\sqrt{5}+ 2 } →\frac{1}{\sqrt{5}- 2 }×\frac{\sqrt{5}+ 2}{\sqrt{5}+ 2 } \)
\((\sqrt{5}- 2)(\sqrt{5}+ 2)=1\) then: \(\frac{1}{\sqrt{5}– 2 }×\frac{\sqrt{5}+ 2}{\sqrt{5}+ 2}=\frac{1 × (\sqrt{5}+ 2)}{1 }= \sqrt{5}+ 2\)
Rationalizing Radical Expressions – Example 4:
Simplify. \(\frac{4}{\sqrt{13}- 3 }\)
Solution:
Multiply by the conjugate: \(\frac{\sqrt{13}+ 3}{\sqrt{13}+ 3 } →\frac{4}{\sqrt{13}- 3 }×\frac{\sqrt{13}+ 3}{\sqrt{13}+ 3 } \)
\((\sqrt{13}- 3)(\sqrt{13}+ 3)=4\) then: \(\frac{4}{\sqrt{13}– 3 }×\frac{\sqrt{13}+ 3}{\sqrt{13}+ 3}=\frac{4×(\sqrt{13}+ 3)}{4 }= \sqrt{13}+ 3\)
Exercises for Rationalizing Radical Expressions
- \(\color{blue}{\frac{15}{\sqrt{5}-2}}\)
- \(\color{blue}{\frac{\sqrt{3}+\sqrt{6}}{6-\sqrt{5}}}\)
- \(\color{blue}{\frac{4+\sqrt{2}}{\sqrt{2}-\sqrt{7}}}\)
- \(\color{blue}{\frac{2+\sqrt{8}}{\sqrt{3}-\sqrt{2}}}\)
- \(\color{blue}{\frac{\sqrt{9c}}{\sqrt{c^5}}}\)
- \(\color{blue}{\frac{10}{7-\sqrt{6}}}\)

- \(\color{blue}{15(\sqrt{5}+2)}\)
- \(\color{blue}{\frac{(\sqrt{3}+\sqrt{6})(6+\sqrt{5})}{31}}\)
- \(\color{blue}{-\frac{4\sqrt{2}+4\sqrt{7}+2+\sqrt{14}}{5}}\)
- \(\color{blue}{2\sqrt{3}+2\sqrt{2}+2\sqrt{6}+4}\)
- \(\color{blue}{\frac{3}{c^2}}\)
- \(\color{blue}{\frac{10(7+\sqrt{6})}{43}}\)
The Absolute Best Book for the Algebra Test
Related to This Article
More math articles
- Calculus: Navigating the Pathways of Particles
- The College Mathematics Exam Overview
- Students Effective Use of Math Skills in Other Academic Disciplines
- Using Number Lines to Compare and Order Rational Numbers
- SHSAT Math – Test Day Tips
- Algebra Puzzle – Challenge 40
- Top 10 Free Websites for ALEKS Math Preparation
- FREE CLEP College Algebra Practice Test
- Full-Length CLEP College Algebra Practice Test
- 7th Grade OST Math Practice Test Questions
What people say about "How to Rationalize Radical Expressions? (+FREE Worksheet!) - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.