Question

rationalize
$\sqrt{3 } + 1 \div 2 \sqrt{2} - \sqrt{3}$

Answer 1

Heya friend,
Here is the answer you were looking for:
$\frac{ \sqrt{3} + 1}{2 \sqrt{2} - \sqrt{3} }$

On rationalizing the denominator we get,

$= \frac{ \sqrt{3} + 1}{2 \sqrt{2} - \sqrt{3} } \times \frac{2 \sqrt{2} + \sqrt{3} }{2 \sqrt{2} + \sqrt{3} }$

Using the identity :

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

$= \frac{ \sqrt{3} (2 \sqrt{2} + \sqrt{3} ) + 1(2 \sqrt{2} + \sqrt{3}) }{ {(2 \sqrt{2} )}^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{2 \sqrt{6} + 3 + 2 \sqrt{2} + \sqrt{3} }{8 - 3} \\ \\ = \frac{3 + 2 \sqrt{2} + \sqrt{3} + 2 \sqrt{6} }{5}$

Hope this helps!!!

@Mahak24

Thanks...
☺