Table of Contents

Last modified on March 28th, 2023

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, ${\dfrac{2}{\sqrt{3}}}$, 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 ${\dfrac{4}{\sqrt{5}}}$**

Solution:

As we know, we need to multiply the sq. root term with itself,

${\therefore \dfrac{4}{\sqrt{5}}\times \dfrac{\sqrt{5}}{\sqrt{5}}}$

${=\dfrac{4\sqrt{5}}{\sqrt{5}}}$

Rationalize a denominator applying **QUOTIENT** **RULE of SQUARE ROOTS**

**Rationalize ${\sqrt{\dfrac{3}{5}}}$.**

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

${\dfrac{\sqrt{3}}{\sqrt{5}}}$

${=\dfrac{\sqrt{3}}{\sqrt{5}}\times \dfrac{\sqrt{5}}{\sqrt{5}}}$ (rationalizing)

${=\dfrac{\sqrt{3}\sqrt{5}}{\sqrt{5^{2}}}}$

${=\dfrac{\sqrt{3}\sqrt{5}}{5}}$

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 ${\dfrac{1}{\sqrt{5}+\sqrt{3}}}$, 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, ${\left( \sqrt{5}+\sqrt{3}\right) \times \left( \sqrt{5}-\sqrt{3}\right)}$

multiplying the denominator with its conjugate, we get:

${\left( \sqrt{5}+\sqrt{3}\right) \times \left( \sqrt{5}-\sqrt{3}\right)}$

${=\left( \sqrt{5}\right) ^{2}-\left( \sqrt{3}\right) ^{2}}$

=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 ${\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}}$**

Solution:

As we know, we need to multiply the denominator with its conjugate.

${\therefore \dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\times \dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}}$

${=\dfrac{7+5-2\times \sqrt{7}\times \sqrt{5}}{7-5}}$

${=\dfrac{12-2\sqrt{35}}{2}}$

${=6-\sqrt{35}}$

Rationalize a denominator using **ALGEBRAIC IDENTITIES**

**Rationalize ${\dfrac{2}{\sqrt{a}-\sqrt{b}}}$**

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

${\therefore \dfrac{2}{\sqrt{a}-\sqrt{b}}\times \dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}}$

${=\dfrac{2\left( \sqrt{a}+\sqrt{b}\right) }{\left( \sqrt{a}\right) ^{2}-\left( \sqrt{b}\right) ^{2}}}$

${=\dfrac{2\left( \sqrt{a}+\sqrt{b}\right) }{a-b}}$

Rationalize a denominator using **3 TERMS**

**Rationalize ${\dfrac{1}{1+\sqrt{5}-\sqrt{3}}}$**

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

${=\dfrac{1}{1+\sqrt{5}-\sqrt{3}}\times \dfrac{1+\sqrt{5}+\sqrt{3}}{1+\sqrt{5}+\sqrt{3}}}$

${=\dfrac{1+\sqrt{5}+\sqrt{3}}{\left( 1+\sqrt{5}\right) ^{2}-\left( \sqrt{3}\right) ^{2}}}$

${=\dfrac{1+\sqrt{5}+\sqrt{3}}{1+2\sqrt{5}+5-3}}$

${\dfrac{1+\sqrt{5}+\sqrt{3}}{3+2\sqrt{5}}}$

${=\dfrac{1+\sqrt{5}+\sqrt{3}}{3+2\sqrt{5}}\times \dfrac{3-2\sqrt{5}}{3-2\sqrt{5}}}$

${=\dfrac{-7+3\left( \sqrt{5}+\sqrt{3}\right) -2\left( \sqrt{5}+\sqrt{15}\right) }{-11}}$

Eliminating ‘-‘ sign,

${=\dfrac{-7+3\left( \sqrt{5}+\sqrt{3}\right) -2\left( \sqrt{5}+\sqrt{15}\right) }{-11}\times \dfrac{-1}{-1}}$

${=\dfrac{7-3\left( \sqrt{5}+\sqrt{3}\right) +2\left( \sqrt{5}+\sqrt{15}\right) }{11}}$

Last modified on March 28th, 2023