Lesson Objectives

- Demonstrate an understanding of how to simplify a radical
- Learn how to rationalize denominators with square roots
- Learn how to rationalize denominators with higher-level roots

## How to Rationalize the Denominator

In this lesson, we will learn how to completely simplify a radical and show how to rationalize the denominator.

Example 1: Simplify each $$\frac{4\sqrt{2}}{5\sqrt{5}}$$ We have a radical (square root of 5) in the denominator. We can multiply the numerator and denominator by the square root of 5. This will rationalize the denominator: $$\frac{4\sqrt{2}}{5\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{4\sqrt{10}}{25}$$ Example 2: Simplify each $$\frac{15\sqrt{2}}{8\sqrt{14}}$$ We have a radical (square root of 14) in the denominator. We can multiply the numerator and denominator by the square root of 14. This will rationalize the denominator: $$\frac{15\sqrt{2}}{8\sqrt{14}}\cdot \frac{\sqrt{14}}{\sqrt{14}}=$$ $$\frac{15 \cdot \sqrt{4}\cdot \sqrt{7}}{8 \cdot 14}=$$ $$\require{cancel}\frac{30\sqrt{7}}{112}=\frac{15 \cancel{30}\sqrt{7}}{56\cancel{112}}=\frac{15\sqrt{7}}{56}$$

4 = 2 • 2

This means we only need an additional factor of 2 or 8 to have a perfect cube. We can rationalize our denominator by multiplying both numerator and denominator by the cube root of 2: $$\frac{1}{\sqrt[3]{4}}\cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{1 \cdot \sqrt[3]{2}}{\sqrt[3]{8}}=\frac{\sqrt[3]{2}}{2}$$ Let's look at some additional examples.

Example 3: Simplify each $$\frac{\sqrt[3]{10}}{\sqrt[3]{32}}$$ Let's start by thinking about 32:

32 = 2

If we had one additional factor of 2, we would have 6 factors of 2. This would also be 4 cubed:

64 = 4

We can rationalize our denominator by multiplying the numerator and denominator by the cube root of 2: $$\frac{\sqrt[3]{10}}{\sqrt[3]{32}}\cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}}=$$ $$\frac{\sqrt[3]{10 \cdot 2}}{\sqrt[3]{64}}=\frac{\sqrt[3]{20}}{4}$$ Example 4: Simplify each $$\frac{\sqrt[4]{3}}{\sqrt[4]{25x^2}}$$ Let's start by thinking about 25:

25 = 5 • 5

We would need two additional factors of 5, to obtain a perfect fourth:

5 • 5 • 5 • 5 = 625

In terms of x

x

We can rationalize our denominator by multiplying the numerator and denominator by the fourth root of 25x

### Simplified Form of a Radical

- The radicand contains no factor (except 1) that is a:
- Perfect Square » Square Root
- Perfect Cube » Cube Root
- Perfect Fourth » Fourth Root
- So on and so forth...

- The radicand cannot contain fractions
- There is no radical present in any denominator

Example 1: Simplify each $$\frac{4\sqrt{2}}{5\sqrt{5}}$$ We have a radical (square root of 5) in the denominator. We can multiply the numerator and denominator by the square root of 5. This will rationalize the denominator: $$\frac{4\sqrt{2}}{5\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{4\sqrt{10}}{25}$$ Example 2: Simplify each $$\frac{15\sqrt{2}}{8\sqrt{14}}$$ We have a radical (square root of 14) in the denominator. We can multiply the numerator and denominator by the square root of 14. This will rationalize the denominator: $$\frac{15\sqrt{2}}{8\sqrt{14}}\cdot \frac{\sqrt{14}}{\sqrt{14}}=$$ $$\frac{15 \cdot \sqrt{4}\cdot \sqrt{7}}{8 \cdot 14}=$$ $$\require{cancel}\frac{30\sqrt{7}}{112}=\frac{15 \cancel{30}\sqrt{7}}{56\cancel{112}}=\frac{15\sqrt{7}}{56}$$

### Rationalizing Higher-Level Roots

Let's suppose we see a problem such as: $$\frac{1}{\sqrt[3]{4}}$$ When we encounter a cube root, we need to create a perfect cube. We currently have the cube root of 4. If we think about 4:4 = 2 • 2

This means we only need an additional factor of 2 or 8 to have a perfect cube. We can rationalize our denominator by multiplying both numerator and denominator by the cube root of 2: $$\frac{1}{\sqrt[3]{4}}\cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{1 \cdot \sqrt[3]{2}}{\sqrt[3]{8}}=\frac{\sqrt[3]{2}}{2}$$ Let's look at some additional examples.

Example 3: Simplify each $$\frac{\sqrt[3]{10}}{\sqrt[3]{32}}$$ Let's start by thinking about 32:

32 = 2

^{5}If we had one additional factor of 2, we would have 6 factors of 2. This would also be 4 cubed:

64 = 4

^{3}We can rationalize our denominator by multiplying the numerator and denominator by the cube root of 2: $$\frac{\sqrt[3]{10}}{\sqrt[3]{32}}\cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}}=$$ $$\frac{\sqrt[3]{10 \cdot 2}}{\sqrt[3]{64}}=\frac{\sqrt[3]{20}}{4}$$ Example 4: Simplify each $$\frac{\sqrt[4]{3}}{\sqrt[4]{25x^2}}$$ Let's start by thinking about 25:

25 = 5 • 5

We would need two additional factors of 5, to obtain a perfect fourth:

5 • 5 • 5 • 5 = 625

In terms of x

^{2}, we need two additional factors of x:x

^{2}• x^{2}= x^{4}We can rationalize our denominator by multiplying the numerator and denominator by the fourth root of 25x

^{2}: $$\frac{\sqrt[4]{3}}{\sqrt[4]{25x^2}}\cdot \frac{\sqrt[4]{25x^2}}{\sqrt[4]{25x^2}}=$$ $$\frac{\sqrt[4]{3 \cdot 25x^2}}{\sqrt[4]{625x^4}}=\frac{\sqrt[4]{75x^2}}{5x}$$ Ready for more?

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