Lesson Objectives
  • Demonstrate an understanding of the simplified form of a radical
  • Learn how to rationalize the denominator

Simplifying Radicals - Rationalizing the Denominator


A few lessons ago, we learned how to simplify radicals. A simplified radical does not contain any radicals in the denominator. The process we use to clear a denominator of its radical is called "rationalizing the denominator". Let's suppose we wanted to simplify the following: $$\require{cancel}\frac{\sqrt{3}}{\sqrt{6}}$$ At this point, our expression is not considered simplified since we have a radical in the denominator. What steps can we take to clear the radical from the denominator? Recall with fractions we can change the appearance without changing the value. If we multiply both numerator and denominator by the square root of 6, the radical will be cleared from the denominator. $$\frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{9} \cdot \sqrt{2}}{6} = \frac{\cancel{3}\sqrt{2}}{2\cancel{6}} = \frac{\sqrt{2}}{2}$$ Let's look at a few examples.
Example 1: Simplify each $$\frac{\sqrt{15}}{8\sqrt{30}}$$ To rationalize the denominator, we can multiply the numerator and denominator by the square root of 30: $$\frac{\sqrt{15}}{8\sqrt{30}} \cdot \frac{\sqrt{30}}{\sqrt{30}} = \frac{\sqrt{225} \cdot \sqrt{2}}{8 \cdot 30} = \frac{\cancel{15}\sqrt{2}}{\cancel{15} \cdot 16} = \frac{\sqrt{2}}{16}$$ Example 2: Simplify each $$\frac{12x}{4\sqrt{20x^3}}$$ Let's start by simplifying: $$\frac{12x}{4\sqrt{20x^3}} = \frac{12x}{4 \cdot \sqrt{4x^2} \cdot \sqrt{5x}} = \frac{3\cancel{12x}}{2\cancel{8x}\sqrt{5x}} = \frac{3}{2\sqrt{5x}}$$ Now let's rationalize the denominator. $$\frac{3}{2\sqrt{5x}}$$ To rationalize the denominator, we can multiply the numerator and denominator by the square root of 5x: $$\frac{3}{2\sqrt{5x}} \cdot \frac{\sqrt{5x}}{\sqrt{5x}} = \frac{3\sqrt{5x}}{10x}$$ Example 3: Simplify each $$\frac{3x + 5\sqrt{5x^4}}{2\sqrt{3x^2}}$$ Let's start by simplifying: $$\frac{3x + 5\sqrt{5x^4}}{2\sqrt{3x^2}} = \frac{3x + 5x^2\sqrt{5}}{2x\sqrt{3}}$$ Now let's rationalize the denominator. $$\frac{3x + 5x^2\sqrt{5}}{2x\sqrt{3}} = \frac{\cancel{x}(3 + 5x\sqrt{5})}{2\cancel{x}\sqrt{3}} = \frac{3 + 5x\sqrt{5}}{2\sqrt{3}}$$ To rationalize the denominator, we can multiply the numerator and denominator by the square root of 3. $$\frac{3 + 5x\sqrt{5}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3} + 5x\sqrt{15}}{6}$$

Rationalizing the Denominator with Higher Roots

Let's look at an example that deals with rationalizing a denominator with a higher root.
Example 4: Simplify each $$\frac{\sqrt[3]{15x^2y^3}}{3\sqrt[3]{-48x^4y^3}}$$ Let's start by simplifying: $$\frac{\sqrt[3]{15x^2y^3}}{3\sqrt[3]{-48x^4y^3}} = \frac{\sqrt[3]{y^3} \cdot \sqrt[3]{3} \cdot \sqrt[3]{x} \cdot \sqrt[3]{5x}}{3 \cdot \sqrt[3]{y^3} \cdot \sqrt[3]{-8} \cdot \sqrt[3]{x^3} \cdot \sqrt[3]{x} \cdot \sqrt[3]{3} \cdot \sqrt[3]{2}} = $$ $$\frac{\cancel{\sqrt[3]{y^3}} \cdot \cancel{\sqrt[3]{3}} \cdot \cancel{\sqrt[3]{x}} \cdot \sqrt[3]{5x}}{-6x \cdot \cancel{\sqrt[3]{y^3}} \cdot \cancel{\sqrt[3]{x}} \cdot \cancel{\sqrt[3]{3}} \cdot \sqrt[3]{2}} = \frac{\sqrt[3]{5x}}{-6x\sqrt[3]{2}}$$ Now let's rationalize the denominator. Since we are dealing with a cube root, we want our radicand in the denominator to be a perfect cube. To achieve this goal, we can multiply both numerator and denominator by the cube root of 4. This will give us the cube root of 8 in the denominator, which is 2. $$\frac{\sqrt[3]{5x}}{-6x\sqrt[3]{2}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}} = -\frac{\sqrt[3]{20x}}{12x}$$