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# Rationalizing the Denominator

In mathematics, a "radical expression" is a mathematical expression that contains a radical symbol, such as the square root √ symbol. The term underneath the radical symbol is called the radicand. The radical symbol can be used to denote a square root, cube root, fourth root, or any other root by adding a superscript number. For example, $\sqrt[3]{8}$ represents the cube root of 8.

When dealing with radical expressions, we apply the technique of rationalization. It's generally considered good practice in algebra classes to give your final answer without any radical signs in the denominator. In most cases, this can be accomplished by multiplying by a well-chosen 1.

## Rationalize the denominator and simplify

To simplify the denominator, you multiply both the numerator and denominator by the well-chosen 1, which will be the conjugate. The conjugate is the expression with the same terms as the original but with the sign between the terms changed. For example, the conjugate of $\sqrt{a}+b$ is $\sqrt{a}-b$ . The identities below can be applied when dealing with a denominator involving radicals. Using these identities, we can simplify the terms in both the numerator and the denominator.

 Expression Conjugate Product $\sqrt{a}+b$ $\sqrt{a}-b$ $\left(\sqrt{a}+b\right)\left(\sqrt{a}–b\right)=a–{b}^{2}$ $\sqrt{a}-b$ $\sqrt{a}+b$ $\left(\sqrt{a}-b\right)\left(\sqrt{a}+b\right)=a–{b}^{2}$ $a+\sqrt{b}$ $a-\sqrt{b}$ $\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)={a}^{2}–b$ $a-\sqrt{b}$ $a+\sqrt{b}$ $\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)={a}^{2}–b$ $\sqrt{a}+\sqrt{b}$ $\sqrt{a}-\sqrt{b}$ $\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b$ $\sqrt{a}-\sqrt{b}$ $\sqrt{a}+\sqrt{b}$ $\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=a-b$

## How to rationalize the denominator with one term

Rationalizing the denominator is the process of turning an expression whose denominator has a term with a number under a radical sign into an expression whose denominator is a rational number.

Let's consider the expression $\frac{1}{\sqrt{a}}$ , where a is a positive number. To rationalize the denominator, we multiply both the numerator and the denominator by $\sqrt{a}$ .

$\frac{1}{\sqrt{a}}×\frac{\sqrt{a}}{\sqrt{a}}=\frac{\sqrt{a}}{a}$

Note that the denominator is now a rational number, which is what we aimed for. However, this expression is still an irrational number. But the important thing to note is that it can be simplified and written in a cleaner form.

Example 1

Rationalize the denominator of $\frac{1}{\sqrt{3}}$ .

The goal is to write $\frac{1}{\sqrt{3}}$ as an equivalent expression in which the denominator is a rational number.

Thus, by multiplying $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ , we can get the required equivalent expression.

$\frac{1}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

Note that there can be radical terms in the numerator before or after rationalizing the denominator.

Example 2

Rationalize the denominator of $\frac{\sqrt{5}}{\sqrt{6}}$ .

Multiply by 1 in the form of $\frac{\sqrt{6}}{\sqrt{6}}$ .

Simplify.

$\frac{\sqrt{5}}{\sqrt{6}}×\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{30}}{6}$

So when you rationalize the denominator of $\frac{\sqrt{5}}{\sqrt{6}}$ , you get $\frac{\sqrt{30}}{6}$ .

## How to rationalize the denominator with two terms

The following are the steps to perform rationalization on denominators containing two terms.

Step 1. Multiply both the numerator and the denominator by the denominator's conjugate, our well-chosen 1.

Step 2. Distribute or use the FOIL technique for both the numerator and denominator.

Step 3. Now simplify the radical terms and combine the obtained like terms.

Step 4. Finally, convert the fraction to a simplified term if possible.

Example 3

$\frac{2}{\sqrt{8}-\sqrt{7}}$

As we know, the conjugate of $\left(\sqrt{8}-\sqrt{7}\right)$ is $\left(\sqrt{8}+\sqrt{7}\right)$ (since the conjugate of $\sqrt{a}-\sqrt{b}$ is $\sqrt{a}+\sqrt{b}$ ).

Multiplying the numerator and denominator with $\left(\sqrt{8}+\sqrt{7}\right)$ , we get:

$=\frac{2}{\sqrt{8}-\sqrt{7}}×\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}$

By applying the FOIL technique and suitable identities,

$=2×\frac{\sqrt{8}+\sqrt{7}}{{\sqrt{8}}^{2}-{\sqrt{7}}^{2}}$

On simplification,

$=\frac{2\sqrt{8}+2\sqrt{7}}{8-7}$

$=\frac{2\sqrt{8}+2\sqrt{7}}{1}$

$=2\sqrt{8}+2\sqrt{7}$

The denominator (1) obtained here is a rational number, as desired.

In some situations, if the denominator has a rational part and an irrational part, you can still use the conjugate.

Example 4

Simplify.

$\frac{1}{4-3\sqrt{7}}$

First, multiply both the numerator and the denominator by the conjugate of the denominator.

$\frac{1}{4-3\sqrt{7}}=\frac{1}{4-3\sqrt{7}}×\frac{4+3\sqrt{7}}{4+3\sqrt{7}}$

The denominator is now a difference of squares.

$=\frac{4+3\sqrt{7}}{{4}^{2}-{\left(3\sqrt{7}\right)}^{2}}$

Use the power of a product property in the denominator.

$=\frac{4+3\sqrt{7}}{{4}^{2}-9×7}$

$=\frac{4+3\sqrt{7}}{16-63}$

$=\frac{-4-3\sqrt{7}}{47}$

## Flashcards covering the Rationalizing the Denominator

Algebra II Flashcards

## Get help learning about rationalizing the denominator

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