Question

Rationalise the denominator of each of the following:
(i) $ \dfrac{3}{{\sqrt 5 }} $
(ii) $ \dfrac{3}{{2\sqrt 5 }} $
(iii) $ \dfrac{1}{{\sqrt {12} }} $

Answer 1

Hint: To rationalise the denominator of the given numbers, we will multiply and divide by conjugate of the denominator. Remember that conjugate of $ a + b $ is $ a - b $ . Also remember that conjugate of any real number is itself.

Complete step-by-step answer:
In the given problem, we have to rationalise the denominator of the given numbers. That is, we have to simplify the given mathematical expression such that it does not have an irrational number in its denominator.
(i) Here the given number is $ \dfrac{3}{{\sqrt 5 }} $ . We can see that the denominator is $ \sqrt 5 $ which is an irrational number. We know that $ \sqrt 5 $ is a real number and conjugate of any real number is itself. Hence, the conjugate of $ \sqrt 5 $ is $ \sqrt 5 $ . Let us multiply and divide by $ \sqrt 5 $ to given numbers. So, we can write
 $\Rightarrow \dfrac{3}{{\sqrt 5 }} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }} = \dfrac{{3\sqrt 5 }}{{\sqrt {25} }} = \dfrac{{3\sqrt 5 }}{5} $
Hence, we can say that after rationalising the denominator $ \dfrac{3}{{\sqrt 5 }} $ is equal to $ \dfrac{{3\sqrt 5 }}{5} $ .
(ii) Here the given number is $ \dfrac{3}{{2\sqrt 5 }} $ . We can see that in the denominator there is $ \sqrt 5 $ which is an irrational number. Let us multiply and divide by $ \sqrt 5 $ to given numbers. So, we can write
 $\Rightarrow \dfrac{3}{{2\sqrt 5 }} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }} = \dfrac{{3\sqrt 5 }}{{2\sqrt {25} }} = \dfrac{{3\sqrt 5 }}{{2 \times 5}} = \dfrac{{3\sqrt 5 }}{{10}} $
Hence, we can say that after rationalising the denominator $ \dfrac{3}{{2\sqrt 5 }} $ is equal to $ \dfrac{{3\sqrt 5 }}{{10}} $ .
(iii) Here the given number is $ \dfrac{1}{{\sqrt {12} }} $ . This number can be written as
$\Rightarrow \dfrac{1}{{\sqrt {12} }} = \dfrac{1}{{\sqrt {4 \times 3} }} = \dfrac{1}{{\sqrt 4 \times \sqrt 3 }} = \dfrac{1}{{2\sqrt 3 }} $ . Now we can see that in the denominator there is $ \sqrt 3 $ which is an irrational number. We know that $ \sqrt 3 $ is a real number and conjugate of any real number is itself. Hence, the conjugate of $ \sqrt 3 $ is $ \sqrt 3 $ . Let us multiply and divide by $ \sqrt 3 $ to give a number. So, we can write
$\Rightarrow \dfrac{1}{{\sqrt {12} }} = \dfrac{1}{{2\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{{2\sqrt 9 }} = \dfrac{{\sqrt 3 }}{{2 \times 3}} = \dfrac{{\sqrt 3 }}{6} $
Hence, we can say that after rationalising the denominator $ \dfrac{1}{{\sqrt {12} }} $ is equal to $ \dfrac{{\sqrt 3 }}{6} $ .

Note: In this type of problems, to rationalise the denominator we have to multiply and divide by conjugate of the denominator. Remember that if the denominator is of the form $ a + \sqrt b $ then its conjugate is $ a - \sqrt b $ . On multiplying $ a + \sqrt b $ and $ a - \sqrt b $ , we will get $ {\left( a \right)^2} - {\left( {\sqrt b } \right)^2} = {a^2} - b $