Question

Rationalize the Denominator of $\Big (\dfrac{1 + \sqrt 2}{\sqrt 5 + \sqrt 3} \Big ) + \Big (\dfrac{1 - \sqrt 2}{\sqrt 5 - \sqrt 3}\Big )$

Answer 1

Given to Rationalize the denominator :-

Rationalize the Denominator of $\Big (\dfrac{1 + \sqrt 2}{\sqrt 5 + \sqrt 3} \Big ) + \Big (\dfrac{1 - \sqrt 2}{\sqrt 5 - \sqrt 3}\Big )$

Solution :-

We will do separately take the first term and we shall solve it

$\dfrac{1 + \sqrt{2} }{ \sqrt{5} + \sqrt{3} }$

In order to rationalize the denominator we have to multiply and divide with its Rationalizing factor

Rationalizing factor for

$\sqrt{5} + \sqrt{3} \: \: is \: \: \sqrt{5} - \sqrt{3}$

$\dfrac{1 + \sqrt{2} }{ \sqrt{5} + \sqrt{3} } \times \dfrac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

$\dfrac{(1 + \sqrt{2})( \sqrt{5} - \sqrt{3} ) }{( \sqrt{5} + \sqrt{3})( \sqrt{5} - \sqrt{3}) }$

$\dfrac{ \sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} }{( \sqrt{5}) {}^{2} - ( \sqrt{3}) {}^{2} }$

$\dfrac{ \sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} }{5 - 3}$

$\dfrac{ \sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} }{2}$

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We shall do rationalization for second term

$\dfrac{1 - \sqrt{2} }{ \sqrt{5} - \sqrt{3} }$

In order to rationalize so multiply and divide with its Rationalizing factor

Rationalizing factor for

$\sqrt{5} - \sqrt{3 \: } \: is \: \: \sqrt{5} + \sqrt{3}$

$\dfrac{1 - \sqrt{2} }{ \sqrt{5} - \sqrt{3} } \times \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

$\dfrac{(1 - \sqrt{2})( \sqrt{5} + \sqrt{3} ) }{( \sqrt{5} - \sqrt{3})( \sqrt{5} + \sqrt{3} ) }$

$\dfrac{ \sqrt{5} + \sqrt{3} - \sqrt{10} - \sqrt{6} }{( \sqrt{5}) {}^{2} - ( \sqrt{3} ) {}^{2} }$

$\dfrac{ \sqrt{5} + \sqrt{3} - \sqrt{10} - \sqrt{6} }{5 - 3}$

$\dfrac{ \sqrt{5} + \sqrt{3} - \sqrt{10} - \sqrt{6} }{2}$

Now, adding the both terms

$\Big (\dfrac{1 + \sqrt 2}{\sqrt 5 + \sqrt 3} \Big ) + \Big (\dfrac{1 - \sqrt 2}{\sqrt 5 - \sqrt 3}\Big )$

$\dfrac{ \sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} }{2} + \dfrac{ \sqrt{5} + \sqrt{3} - \sqrt{10} - \sqrt{6} }{2}$

$\dfrac{ \sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} + \sqrt{5} + \sqrt{3} - \sqrt{10} - \sqrt{6} }{2}$

$\dfrac{2 \sqrt{5} - 2 \sqrt{6} }{2}$

$(\sqrt{5} - \sqrt{6} ) \div 1$

So, $\Big (\dfrac{1 + \sqrt 2}{\sqrt 5 + \sqrt 3} \Big ) + \Big (\dfrac{1 - \sqrt 2}{\sqrt 5 - \sqrt 3}\Big )$ =

$\sqrt{5} - \sqrt{6}$

Hence denomiantor rationalized