Table of Contents
Last modified on August 3rd, 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 ${\dfrac{4}{\sqrt{5}}}$
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 ${\sqrt{\dfrac{3}{5}}}$.
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 ${\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}}$
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 ${\dfrac{2}{\sqrt{a}-\sqrt{b}}}$
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 ${\dfrac{1}{1+\sqrt{5}-\sqrt{3}}}$
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 August 3rd, 2023