Lesson Objectives
  • Demonstrate an understanding of how to simplify a radical
  • Learn how to rationalize a binomial denominator

How to Rationalize a Binomial Denominator


In the last lesson, we learned how to rationalize the denominator. In this lesson, we will build on this process and learn how to rationalize a binomial denominator. Before we jump into the lesson, 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 1: Simplify each $$\frac{4x^2\sqrt{2}}{4 - 4x\sqrt{3x}}$$ 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 2: Simplify each $$\frac{3 + 2\sqrt{x}}{2x - 2\sqrt{x}}$$ 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. $$\frac{10x + 6\sqrt{x} + 4x\sqrt{x}}{4x^2 - 4x} = \frac{\cancel{2}(5x + 3\sqrt{x} + 2x\sqrt{x})}{\cancel{2}(2x^2 - 2x)} = \frac{5x + 3\sqrt{x} + 2x\sqrt{x}}{2x^2 - 2x}$$