Question

2 root 6 / root 3 + root 2

Answer 1

Solution:-

given 》 2 root 6 / root 3 + root 2

》
$\frac{2 \sqrt{6} }{ \sqrt{3} } + \sqrt{2} \\ \frac{2 \sqrt{6} + \sqrt{6} }{ \sqrt{3} } \\ \frac{3 \sqrt{6} }{ \sqrt{3} } \\ \sqrt{3 \times 6} \\ 3 \sqrt{2}$

Answer 2

Hey friend,

First of all,
Sorry for late solving in comments but I didn't got the notification that you have commented

But as soon as I Saw I have answered there and here elaborated

Here is the answer you were looking for:
Here we go =D
$\frac{2 \sqrt{6} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator we get,

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

Using the identity :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

Putting x = √3
and y = √2

$= \frac{2 \sqrt{6} ( \sqrt{3} - \sqrt{2} )}{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{2 \sqrt{6} \times \sqrt{3} - 2 \sqrt{6} \times \sqrt{2} }{3 - 2} \\ \\ = 2 \sqrt{18} - 2 \sqrt{12}$

Splitting √18 and √12 we get

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

We can also write it as

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

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺