Question

Rationalize the denominator of the following-
$\dfrac{{\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }}$

Answer 1

Hint: In this question multiply both the numerator and the denominator part of the given expression with $\sqrt 2 - \sqrt 3 $. Then use the algebraic identity of $(a + b)(a - b) = {a^2} - {b^2}$ in the denominator part to get the answer.

Complete step-by-step answer:
Given expression
$\dfrac{{\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }}$
Now we have to rationalize the denominator.
To rationalize the denominator, you must multiply both the numerator and the denominator by the conjugate of the denominator.
Remember to find the conjugate all you have to do is change the sign between the two terms.
So the conjugate of the denominator is = $\sqrt 2 - \sqrt 3 $
So the rationalization of the given expression is
$ \Rightarrow \dfrac{{\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }} \times \dfrac{{\sqrt 2 - \sqrt 3 }}{{\sqrt 2 - \sqrt 3 }}$
Now simplify the above equation, in denominator use the property $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ so we have,
$ \Rightarrow \dfrac{{\sqrt 6 \sqrt 2 - \sqrt 6 \sqrt 3 }}{{{{\left( {\sqrt 2 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} = \dfrac{{\sqrt {12} - \sqrt {18} }}{{2 - 3}} = \dfrac{{\sqrt {12} - \sqrt {18} }}{{ - 1}} = \sqrt {18} - \sqrt {12} $
So this is the required rationalization of the given expression.
So this is the required answer.

Note: There is a very standard procedure to rationalize any given expression, we simply need to multiply both the numerator and the denominator by the conjugate of the denominator. Use of basic algebraic identities always helps saving a lot of time so it is advised to remember them.