How to Rationalize the Denominator

Rationalizing a denominator means removing any radical expressions such as square or cube roots from the denominator of a fraction. We do it by multiplying the original fraction with a value such that the denominator no longer has any radicals.

If a fraction has a monomial denominator which is a radical, we rationalize the denominator by multiplying itself with both the top (numerator) and bottom (denominator) of a fraction.

For a fraction, ${\displaystyle \frac{2}{\sqrt{3}}}$, we rationalize the denominator by simply multiplying $\sqrt{3}$ with $\sqrt{3}$ to get a rational denominator, i.e. 3.

We will see how to rationalize the denominator through some more examples.

Rationalize ${\displaystyle \frac{4}{\sqrt{5}}}$

Solution:

As we know, we need to multiply the sq. root term with itself,
$\therefore {\displaystyle \frac{4}{\sqrt{5}}}\times {\displaystyle \frac{\sqrt{5}}{\sqrt{5}}}$
$={\displaystyle \frac{4\sqrt{5}}{\sqrt{5}}}$

Rationalize $\sqrt{{\displaystyle \frac{3}{5}}}$.

Solution:

Here we will apply the quotient rule of square roots to write the fraction with distinct radical symbols for both numerator and denominator.
So the fraction is
${\displaystyle \frac{\sqrt{3}}{\sqrt{5}}}$
$={\displaystyle \frac{\sqrt{3}}{\sqrt{5}}}\times {\displaystyle \frac{\sqrt{5}}{\sqrt{5}}}$ (rationalizing)
$={\displaystyle \frac{\sqrt{3}\sqrt{5}}{\sqrt{5^2}}}$
$={\displaystyle \frac{\sqrt{3}\sqrt{5}}{5}}$

Multiplying by the Conjugate

If a fraction has a binomial denominator with one or two radical terms, we rationalize the denominator by multiplying its conjugate with both the top and bottom of a fraction.

For a fraction ${\displaystyle \frac{1}{\sqrt{5}+\sqrt{3}}}$, we rationalize the denominator by simply multiplying the denominator with its conjugate (the same binomial with an opposite middle sign) i.e. $\sqrt{5}-\sqrt{3}$.

So, $(\sqrt{5}+\sqrt{3})\times (\sqrt{5}-\sqrt{3})$

multiplying the denominator with its conjugate, we get:
$(\sqrt{5}+\sqrt{3})\times (\sqrt{5}-\sqrt{3})$
$={\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2$
=5-3
=2

Make the figure no reduced, keep the denominator.

This method of rationalizing with the conjugate is also known as rationalizing the denominator with subtraction.

Rationalize ${\displaystyle \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}}$

Solution:

As we know, we need to multiply the denominator with its conjugate.
$\therefore {\displaystyle \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}}\times {\displaystyle \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}}$
$={\displaystyle \frac{7+5-2\times \sqrt{7}\times \sqrt{5}}{7-5}}$
$={\displaystyle \frac{12-2\sqrt{35}}{2}}$
$=6-\sqrt{35}$

Rationalize ${\displaystyle \frac{2}{\sqrt{a}-\sqrt{b}}}$

Solution:

To rationalize (√a -√b), we use the rationalizing factor (√a +√b).
To rationalize (√a + √b), we use the rationalizing factor is (√a − √b).
Here we will use the factor $\sqrt{a}+\sqrt{b}$ to implement algebraic identities
$\therefore {\displaystyle \frac{2}{\sqrt{a}-\sqrt{b}}}\times {\displaystyle \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}}$
$={\displaystyle \frac{2(\sqrt{a}+\sqrt{b})}{{\left(\sqrt{a}\right)}^2-{\left(\sqrt{b}\right)}^2}}$
$={\displaystyle \frac{2(\sqrt{a}+\sqrt{b})}{a-b}}$

Rationalize ${\displaystyle \frac{1}{1+\sqrt{5}-\sqrt{3}}}$

Solution:

Here we will consider the denominator as (a + b) – c, and multiply the fraction with (a + b) + c
Here (a + b) = 1 + √5, and c = √3
$={\displaystyle \frac{1}{1+\sqrt{5}-\sqrt{3}}}\times {\displaystyle \frac{1+\sqrt{5}+\sqrt{3}}{1+\sqrt{5}+\sqrt{3}}}$
$={\displaystyle \frac{1+\sqrt{5}+\sqrt{3}}{{(1+\sqrt{5})}^2-{\left(\sqrt{3}\right)}^2}}$
$={\displaystyle \frac{1+\sqrt{5}+\sqrt{3}}{1+2\sqrt{5}+5-3}}$
${\displaystyle \frac{1+\sqrt{5}+\sqrt{3}}{3+2\sqrt{5}}}$
$={\displaystyle \frac{1+\sqrt{5}+\sqrt{3}}{3+2\sqrt{5}}}\times {\displaystyle \frac{3-2\sqrt{5}}{3-2\sqrt{5}}}$
$={\displaystyle \frac{-7+3(\sqrt{5}+\sqrt{3})-2(\sqrt{5}+\sqrt{15})}{-11}}$
Eliminating ‘-‘ sign,
$={\displaystyle \frac{-7+3(\sqrt{5}+\sqrt{3})-2(\sqrt{5}+\sqrt{15})}{-11}}\times {\displaystyle \frac{-1}{-1}}$
$={\displaystyle \frac{7-3(\sqrt{5}+\sqrt{3})+2(\sqrt{5}+\sqrt{15})}{11}}$