Question

Rationalise the denominator of √2/√2+√3

Answer 1

The answer is given below :

Now,

$\frac{ \sqrt{2} }{ \sqrt{2} + \sqrt{3} } \\ \\ = \frac{ \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3 } - \sqrt{2} } \\ \\ = \frac{ \sqrt{2} ( \sqrt{3} - \sqrt{2} )}{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} )} \\ \\ = \frac{ \sqrt{2.3} - \sqrt{2.2} }{3 - 2} \\ \\ = \frac{ \sqrt{6} - 2 }{1} \\ \\ = \sqrt{6} - 2$

Thank you for your question.

Answer 2

Hi there !!

Given,

to rationalize the denominator in

$\frac{ \sqrt{2} }{ (\sqrt{2} + \sqrt{3}) }$

$= \frac{( \sqrt{2})( \sqrt{2} - \sqrt{3}) }{( \sqrt{2} + \sqrt{3})( \sqrt{2} - \sqrt{3} ) }$
Using the identity in the denominator ie, (a +b)(a - b) = a² - b²
we have,

$= \frac{( \sqrt{2}( \sqrt{2} ) - (\sqrt{2})( \sqrt{3} ) }{( \sqrt{2}) {}^{2} - (\sqrt{3}) {}^{2} }$

$= \frac{2 - \sqrt{6} }{2 - 3}$

$= \frac{2 - \sqrt{6} }{ - 1}$

$= - (2 - \sqrt{6} )$

= -2 + ✓6 is the answer

or

✓6 - 2