Rationalize the Denominator Lesson

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

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

So far, we have learned how to do everything except clear a radical from the denominator. What happens if we see a problem such as: $$\frac{5}{\sqrt{2}}$$ We can see that the square root of 2 is in the denominator, this violates our rules for a simplified radical. To clean up the problem, we use a process known as rationalizing the denominator. To do this, we just need to multiply both numerator and denominator by the square root of 2: $$\frac{5}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{2}$$ Since we are multiplying by a complex form of 1, the value does not change. We have simply changed the form of the number to comply with our rules for a simplified radical. Notice how the denominator went from an irrational number (square root of 2) to a rational number (2). This is where the term "rationalizing the denominator" comes from. Let's look at some examples.
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 = 2⁵
If we had one additional factor of 2, we would have 6 factors of 2. This would also be 4 cubed:
64 = 4³
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 x², we need two additional factors of x:
x² • x² = x⁴
We can rationalize our denominator by multiplying the numerator and denominator by the fourth root of 25x²: $$\frac{\sqrt[4]{3}}{\sqrt[4]{25x^2}}\cdot \frac{\sqrt[4]{25x^2}}{\sqrt[4]{25x^2}}=$$ $$\frac{\sqrt[4]{3 \cdot 25x^2}}{\sqrt[4]{625x^4}}=\frac{\sqrt[4]{75x^2}}{5x}$$

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