Lesson 44: Rationalizing Denominators

New Concept

If $a$, $b$, $c$, and $d$ are rational numbers, then $a\sqrt{b} + c\sqrt{d}$ and $a\sqrt{b} - c\sqrt{d}$ are conjugates of each other.

What’s next

Next, you’ll use conjugates to rationalize binomial denominators, a key technique for simplifying complex radical expressions you'll see in future math and physics problems.

To rationalize the denominator of an expression, write an equivalent expression so that there are no radicals in any denominator and no denominators in any radical. An expression is in simplest form when no radicand has a perfect square root factor and there are no irrational radical expressions in any denominator.

Example 1: $\frac{3}{2\sqrt{5}} = \frac{3}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{10}$
Example 2: $\sqrt{\frac{1}{12}} = \sqrt{\frac{1 \cdot 3}{12 \cdot 3}} = \frac{\sqrt{3}}{\sqrt{36}} = \frac{\sqrt{3}}{6}$
Example 3: $\frac{7}{\sqrt{x}} = \frac{7}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{7\sqrt{x}}{x}$

Think of an irrational denominator like wearing socks with sandals—it's just not done in the world of simplified math! Rationalizing is a style makeover for your fractions. We multiply the top and bottom by a special number to kick that pesky radical out of the denominator, making the expression clean, tidy, and officially in its simplest form.

If $a$, $b$, $c$, and $d$ are rational numbers, then $a\sqrt{b} + c\sqrt{d}$ and $a\sqrt{b} - c\sqrt{d}$ are conjugates of each other, and their product is a rational number.

Example 1: The conjugate of $-4 + \sqrt{3}$ is $-4 - \sqrt{3}$.
Example 2: The conjugate of $\sqrt{5} + \sqrt{2}$ is $\sqrt{5} - \sqrt{2}$.
Example 3: $\frac{5}{\sqrt{6} + \sqrt{3}} = \frac{5(\sqrt{6} - \sqrt{3})}{(\sqrt{6})^2 - (\sqrt{3})^2} = \frac{5(\sqrt{6} - \sqrt{3})}{3}$

Radical conjugates are like secret agent twins! They look almost identical but have one opposite sign. When they multiply together, they use their special power—the difference of squares formula, $(a+b)(a-b)=a^2-b^2$—to eliminate the radicals completely. This trick is your go-to move for cleaning up binomials in the denominator, leaving behind a nice, rational number.

For $a > 0$ and $b > 0$, $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$.

Example 1: $\sqrt{\frac{2}{16}} = \frac{\sqrt{2}}{\sqrt{16}} = \frac{\sqrt{2}}{4}$
Example 2: $\sqrt{\frac{5}{49}} = \frac{\sqrt{5}}{\sqrt{49}} = \frac{\sqrt{5}}{7}$
Example 3: $\sqrt{\frac{9}{10}} = \frac{\sqrt{9}}{\sqrt{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$

This property is like a license to split! If you have a fraction chilling inside a radical sign, you can break it up into two separate radicals: one for the numerator and one for the denominator. This move is super helpful for isolating the pesky radical in the denominator so you can then focus on rationalizing it away.