Rationalizing Denominators Lesson

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. Assume all variables represent positive real numbers.$$\frac{12x}{4\sqrt{20x^3}}$$ 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. Assume all variables represent positive real numbers.$$\frac{3x + 5\sqrt{5x^4}}{2\sqrt{3x^2}}$$ Let's start by simplifying: $$\frac{3x + 5\sqrt{5x^4}}{2\sqrt{3x^2}}=\frac{3x + 5x^2\sqrt{5}}{2x\sqrt{3}}$$ $$\frac{3x + 5x^2\sqrt{5}}{2x\sqrt{3}}=\frac{\cancel{x}(3 + 5x\sqrt{5})}{2\cancel{x}\sqrt{3}}=\frac{3 + 5x\sqrt{5}}{2\sqrt{3}}$$ Now let's rationalize the denominator.$$\frac{3 + 5x\sqrt{5}}{2\sqrt{3}}$$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. Assume all variables represent positive real numbers.$$\frac{\sqrt[3]{15x^2y^3}}{3\sqrt[3]{-48x^4y^3}}$$ 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. $$\frac{\sqrt[3]{5x}}{-6x\sqrt[3]{2}}\cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}}=-\frac{\sqrt[3]{20x}}{12x}$$

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How to Rationalize a Binomial Denominator

Before we jump into this topic, we need to understand some key vocabulary. Conjugates are two binomials where the first terms are the same and the last terms are the same. The only difference is the sign between the terms. In one case, we will have a sum and in the other case, we will have a difference.
(a + b)(a - b)
(x + y)(x - y)
(z + q)(z - q)
When we multiply conjugates together using FOIL, the O and I step will cancel. We are left with the square of the first term minus the square of the last term.
(a + b)(a - b) = a² - b²
(x + y)(x - y) = x² - y²
(z + q)(z - q) = z² - q²
When a radical expression has a sum or difference with square root radicals in the denominator, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator. Let's look at a few examples.
Example 5: Simplify each. Assume all variables represent positive real numbers.$$\frac{4x^2\sqrt{2}}{4 - 4x\sqrt{3x}}$$ To rationalize the denominator, we will multiply the numerator and the denominator by the conjugate of the denominator. $$\frac{4x^2\sqrt{2}}{4 - 4x\sqrt{3x}}\cdot \frac{4 + 4x\sqrt{3x}}{4 + 4x\sqrt{3x}}=\frac{16x^2\sqrt{2}+ 16x^3\sqrt{6x}}{16 - 48x^3}$$ Although we have rationalized the denominator, we can still simplify further. We can factor out a 16 from the numerator and the denominator and cancel. $$\require{cancel}\frac{16x^2\sqrt{2}+ 16x^3\sqrt{6x}}{16 - 48x^3}=\frac{\cancel{16}(x^2\sqrt{2}+ x^3\sqrt{6x})}{\cancel{16}(1 - 3x^3)}=\frac{x^2\sqrt{2}+ x^3\sqrt{6x}}{1 - 3x^3}$$ Example 6: Simplify each. Assume all variables represent positive real numbers.$$\frac{3 + 2\sqrt{x}}{2x - 2\sqrt{x}}$$ To rationalize the denominator, we will multiply the numerator and the denominator by the conjugate of the denominator. $$\frac{3 + 2\sqrt{x}}{2x - 2\sqrt{x}}\cdot \frac{2x + 2\sqrt{x}}{2x + 2\sqrt{x}}=\frac{10x + 6\sqrt{x}+ 4x\sqrt{x}}{4x^2 - 4x}$$ Although we have rationalized the denominator, we can still simplify further. We can factor out a 2 from the numerator and the denominator and cancel. $$\frac{10x + 6\sqrt{x}+ 4x\sqrt{x}}{4x^2 - 4x}=\frac{\cancel{2}(5x + 3\sqrt{x}+ 2x\sqrt{x})}{\cancel{2}(2x^2 - 2x)}=\frac{5x + 3\sqrt{x}+ 2x\sqrt{x}}{2x^2 - 2x}$$

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