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.
Rationalize the Denominator
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 QUOTIENTRULE 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) ${=\dfrac{\sqrt{3}\sqrt{5}}{\sqrt{5^{2}}}}$ ${=\dfrac{\sqrt{3}\sqrt{5}}{5}}$
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 ${\dfrac{1}{\sqrt{5}+\sqrt{3}}}$, 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.
Rationalize the Denominator Using Conjugate
This method of rationalizing with the conjugate is also known as rationalizing the denominator with subtraction.
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, ${=\dfrac{-7+3\left( \sqrt{5}+\sqrt{3}\right) -2\left( \sqrt{5}+\sqrt{15}\right) }{-11}\times \dfrac{-1}{-1}}$ ${=\dfrac{7-3\left( \sqrt{5}+\sqrt{3}\right) +2\left( \sqrt{5}+\sqrt{15}\right) }{11}}$
