Question

Rationalize the denominator
√2÷ root 2+root 3
$\sqrt{2 } \div \sqrt{2} + \sqrt{3}$
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Answer 1

$= > \frac{ \sqrt{2} }{ \sqrt{2} + \sqrt{3} }$

multiply and divide with √2 - √3

$= > \frac{ \sqrt{2} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} }$

• (a + b) (a - b) = a² - b²

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

$= > \frac{ \sqrt{4} - \sqrt{6} }{2 - 3}$

$= > \frac{ \sqrt{4 } - \sqrt{6} }{ - 1} \\ = > \frac{2 - \sqrt{3 \times 2} }{ - 1}$