How to Rationalize Radical Expressions

How to Rationalize Radical Expressions

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

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

  1. \(\color{blue}{\frac{15}{\sqrt{5}-2}}\)
  2. \(\color{blue}{\frac{\sqrt{3}+\sqrt{6}}{6-\sqrt{5}}}\)
  3. \(\color{blue}{\frac{4+\sqrt{2}}{\sqrt{2}-\sqrt{7}}}\)
  4. \(\color{blue}{\frac{2+\sqrt{8}}{\sqrt{3}-\sqrt{2}}}\)
  5. \(\color{blue}{\frac{\sqrt{9c}}{\sqrt{c^5}}}\)
  6. \(\color{blue}{\frac{10}{7-\sqrt{6}}}\)
This image has an empty alt attribute; its file name is answers.png
  1. \(\color{blue}{15(\sqrt{5}+2)}\)
  2. \(\color{blue}{\frac{(\sqrt{3}+\sqrt{6})(6+\sqrt{5})}{31}}\)
  3. \(\color{blue}{-\frac{4\sqrt{2}+4\sqrt{7}+2+\sqrt{14}}{5}}\)
  4. \(\color{blue}{2\sqrt{3}+2\sqrt{2}+2\sqrt{6}+4}\)
  5. \(\color{blue}{\frac{3}{c^2}}\)
  6. \(\color{blue}{\frac{10(7+\sqrt{6})}{43}}\)

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