Rationalization is the process of removing radicals from the denominator of a fraction. To rationalize denominators we have to multiply the expression by a convenient value so that, when simplifying, we eliminate the radicals from the denominator. There are two main methods used to rationalize radicals depending on whether the denominator is a monomial or a binomial.

Here, we will look at a summary of these two methods along with several rationalization examples to fully master this process.

## Summary of rationalization of denominators

Rationalizing the denominator means eliminating the radical expressions in the denominator so that we do not have square roots, cubic roots, or any other roots. The main idea in rationalizing denominators is to multiply the original fraction by an appropriate value so that after simplifying, the denominator no longer contains radicals.

When the denominator is a monomial, we can apply the fact that:

$latex \sqrt{x} \cdot \sqrt{x}= \sqrt{{{x}^2}}=x$

Therefore, we can multiply both the numerator and denominator by the radical expression. After simplifying, we will obtain an expression without radicals in the denominator.

On the other hand, if the denominator is a binomial, we have to use the conjugate of the binomial. The conjugate of a binomial is equal to the same binomial, but with the sign of the middle changed.

For example, suppose we have the binomial $latex a+ \sqrt{b}$ in the denominator. The conjugate of this binomial is $latex a-\sqrt{b}$. The product of the binomial and its conjugate is:

$latex (a + \sqrt{b})(a-\sqrt{b})={{a}^2}-b$

## Rationalizing the denominator – Examples with answers

The rationalization process indicated above is used to rationalize both monomials and binomials in the following examples. Try to solve the exercises yourself before looking at the solution.

**EXAMPLE 1**

Rationalize the denominator of the expression $latex \frac{3}{\sqrt{2}}$.

##### Solution

The denominator contains a radical expression, the square root of 2. We can remove the radical from the denominator by multiplying it by itself since $latex \sqrt{2} \times \sqrt{2}=2$.

However, if we only multiply that expression to the denominator, we are changing the whole expression. Therefore, we also multiply the numerator to balance.

By multiplying the expression by $latex \frac{\sqrt{2}}{\sqrt{2}}$ we are actually multiplying by 1. Thus, we have:

$latex \frac{3}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{4}}$

$latex =\frac{3\sqrt{2}}{2}$

The expression we obtained no longer has a radical in the denominator, so we have already rationalized it.

**EXAMPLE 2**

Rationalize $latex \frac{8}{\sqrt{2}}$ and simplify if possible.

##### Solution

Similar to the previous exercise, we have a square root of 2 in the denominator. Therefore, we multiply the expression by $latex \frac{\sqrt{2}}{\sqrt{2}}$:

$latex \frac{8}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{8\sqrt{2}}{\sqrt{4}}$

$latex =\frac{8\sqrt{2}}{2}$

We have already rationalized the expression, but in this case, we can also simplify by canceling common factors:

$latex \frac{8\sqrt{2}}{2}=4\sqrt{2}$

**EXAMPLE 3**

Rationalize the expression $latex \sqrt{\frac{5}{3}}$.

##### Solution

In this case, we have a square root of a complete fraction. We can start by applying the rule of the quotient of square roots. This allows us to write the numerator and denominator with square roots separately. Therefore, we have:

$latex \sqrt{\frac{5}{3}}=\frac{\sqrt{5}}{\sqrt{3}}$

The new expression has a denominator with a square root of 3, so we multiply both the numerator and denominator by this radical:

$latex \frac{\sqrt{5}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{5}\sqrt{3}}{\sqrt{9}}$

$latex =\frac{\sqrt{5}\sqrt{3}}{3}$

$latex =\frac{\sqrt{15}}{3}$

**EXAMPLE 4**

Simplify by rationalizing the denominator of $latex \frac{3 \sqrt{6}}{\sqrt{3}}$.

##### Solution

We multiply both the numerator and denominator by $latex \sqrt{3}$ to eliminate the radical from the denominator:

$latex \frac{3\sqrt{6}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{18}}{\sqrt{9}}$

$latex =\frac{3\sqrt{18}}{3}$

$latex =\sqrt{18}$

We can rewrite 18 as the product of 9 and 2 to simplify to the expression:

$latex \sqrt{18}=\sqrt{9\times 2}$

$latex =\sqrt{9}\sqrt{2}$

$latex =3\sqrt{2}$

**EXAMPLE 5**

Rationalize and simplify the expression $latex \frac{4- \sqrt{6}}{\sqrt{2}}$.

##### Solution

This expression contains a binomial in the numerator, but the process is the same since we still have a monomial in the denominator. We multiply both the numerator and denominator by $latex \sqrt{2}$:

$latex \frac{4-\sqrt{6}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}(4-\sqrt{6})}{\sqrt{4}}$

$latex =\frac{4\sqrt{2}-\sqrt{12}}{2}$

We can simplify the expression by recognizing that we can write 12 as the product of 4 and 3:

$latex \frac{4\sqrt{2}-\sqrt{12}}{2}=\frac{4\sqrt{2}-\sqrt{4\times 3}}{2}$

$latex =\frac{4\sqrt{2}-\sqrt{4}\sqrt{3}}{2}$

$latex =\frac{4\sqrt{2}-2\sqrt{3}}{2}$

$latex =2\sqrt{2}-\sqrt{3}$

**EXAMPLE 6**

Rationalize the expression $latex \frac{4}{2+\sqrt{2}}$.

##### Solution

This exercise is a bit different since we have two terms in the denominator, that is, a binomial. To eliminate the radical from the denominator, we have to multiply both the numerator and the denominator by the conjugate of the binomial.

To find the conjugate of a binomial, we simply have to change the sign of the middle. In this case, we have to multiply by $latex \frac{2- \sqrt{2}}{2- \sqrt{2}}$:

$latex \frac{4}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{4(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$

$latex =\frac{8-4\sqrt{2}}{4-2\sqrt{2}+2\sqrt{2}-\sqrt{4}}$

$latex =\frac{8-4\sqrt{2}}{4-2}$

$latex =\frac{8-4\sqrt{2}}{2}$

$latex =4-2\sqrt{2}$

**EXAMPLE 7**

Rationalize the expression $latex \frac{5}{6- \sqrt{3}}$.

##### Solution

Similar to the previous problem, we have to multiply both the numerator and the denominator by the conjugate of the binomial. In this case, we have to multiply by $latex \frac{6+\sqrt{3}}{6+\sqrt{3}}$:

$latex \frac{5}{6-\sqrt{3}}\times \frac{6+\sqrt{3}}{6+\sqrt{3}}=\frac{5(6+\sqrt{3})}{(6-\sqrt{3})(6+\sqrt{3})}$

$latex =\frac{30+5\sqrt{3}}{36-6\sqrt{3}+6\sqrt{3}-\sqrt{9}}$

$latex =\frac{30+5\sqrt{3}}{36-3}$

$latex =\frac{30+5\sqrt{3}}{33}$

**EXAMPLE 8**

Rationalize the expression $latex \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.

##### Solution

We multiply both the denominator and the numerator by the conjugate of the binomial. In this case, the conjugate is $latex \sqrt{3} – \sqrt{2}$:

$latex \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$latex =\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

$latex =\frac{\sqrt{9}-\sqrt{6}-\sqrt{6}+\sqrt{4}}{\sqrt{9}-\sqrt{6}+\sqrt{6}-\sqrt{4}}$

$latex =\frac{3-2\sqrt{6}+2}{3-2}$

$latex =3-2\sqrt{6}+2$

$latex =5-2\sqrt{6}$

## Rationalizing the denominator – Practice problems

Use what you have learned about rationalization to solve the following problems. Choose an answer and check it to make sure you selected the correct one.

## See also

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