# 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.

## 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. $$\color{blue}{\frac{10}{7-\sqrt{6}}}$$
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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