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 numerator and 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: \(\frac{\sqrt{3}+1}{\sqrt{3}+1}\)
\(\frac{2}{\sqrt{3}-1 }×\frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2(\sqrt{3}+1)}{2}→ =(\sqrt{3}+1)\)
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: \(\frac{1}{\sqrt{5}– 2 }×\frac{\sqrt{5}+ 2}{\sqrt{5}+ 2}=\frac{(\sqrt{5}+ 2)}{1 }= \sqrt{5}+ 2\)
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
- \(\color{blue}{\frac{15}{\sqrt{5}-2}}\)
- \(\color{blue}{\frac{\sqrt{3}+\sqrt{6}}{6-\sqrt{5}}}\)
- \(\color{blue}{\frac{4+\sqrt{2}}{\sqrt{2}-\sqrt{7}}}\)
- \(\color{blue}{\frac{2+\sqrt{8}}{\sqrt{3}-\sqrt{2}}}\)
- \(\color{blue}{\frac{\sqrt{9c}}{\sqrt{c^5}}}\)
- \(\color{blue}{\frac{10}{7-\sqrt{6}}}\)
- \(\color{blue}{15(\sqrt{5}+2)}\)
- \(\color{blue}{\frac{(\sqrt{3}+\sqrt{6})(6+\sqrt{5})}{31}}\)
- \(\color{blue}{-\frac{4\sqrt{2}+4\sqrt{7}+2+\sqrt{14}}{5}}\)
- \(\color{blue}{2\sqrt{3}+2\sqrt{2}+2\sqrt{6}+4}\)
- \(\color{blue}{\frac{3}{c^2}}\)
- \(\color{blue}{\frac{10(7+\sqrt{6})}{43}}\)
