Rationalize the denominator and simplify $ \dfrac{3}{{\sqrt 5 - \sqrt 3 }} $ .

Hint: Rationalizing the denominator means getting rid of any square roots or cube roots by multiplying and dividing the given expression by the conjugate of the denominator. Conjugate is nothing but the same expression but with different sign between the terms like $ a + b $ is the conjugate of $ a - b $ . So here the denominator has two square roots terms subtracted from each other. So to rationalize the given expression, we have to multiply and divide the expression by the denominator’s conjugate. The conjugate of the denominator is $ \sqrt 5 + \sqrt 3 $ . Using this information, we have to simplify the given expression.

Complete step-by-step answer:
We are given to rationalize and simplify $ \dfrac{3}{{\sqrt 5 - \sqrt 3 }} $
So we have to first multiply and divide the expression by the conjugate of the denominator which is $ \sqrt 5 + \sqrt 3 $ .
The expression now becomes, $ \dfrac{3}{{\sqrt 5 - \sqrt 3 }} \times \dfrac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} = \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}} $
Considering $ \sqrt 5 $ as ‘a’ and $ \sqrt 3 $ as ‘b’, $ \left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right) $ becomes $ \left( {a - b} \right)\left( {a + b} \right) $
We already know that $ \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} $
Therefore, $ \left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right) $ becomes $ {\left( {\sqrt 5 } \right)^2} - {\left( {\sqrt 3 } \right)^2} $
On substituting the above obtained value in $ \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}} $ , we get
 $ \Rightarrow \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} $
The value of $ {\left( {\sqrt 5 } \right)^2} $ is 5 because the square root gets cancelled by the square and in the same way $ {\left( {\sqrt 3 } \right)^2} $ is 3
 $ \Rightarrow \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} $
Subtracting 3 from 5 results 2
 $ \Rightarrow \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{2} $
Now the denominator is rationalized, which means it has no square root or cube root terms.
Therefore, the expression $ \dfrac{3}{{\sqrt 5 - \sqrt 3 }} $ after rationalizing is $ \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{2} $
So, the correct answer is “ $ \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{2} $ ”.

Note: Instead of leaving the result like $ \dfrac{{3\left( {\sqrt 5 + \sqrt 3 } \right)}}{2} $ , we can expand the numerator by sending 3 inside the brackets like $ \dfrac{{3\sqrt 5 + 3\sqrt 3 }}{2} $ , and sending the 3 inside the roots like $ \dfrac{{\sqrt {5 \times {3^2}} + \sqrt {3 \times {3^2}} }}{2} = \dfrac{{\sqrt {45} + \sqrt {27} }}{2} $ . Whenever we send a variable or a number inside a square root, the number becomes its square such as 3 becomes 9, 5 becomes 25 etc.