Question

If √3+√2 by √3-√2=a+b√6
Then find the values of a+b

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ = 5 + 2 \sqrt{6} \\ \\ 5 + 2 \sqrt{6} = a + b \sqrt{6} \\ \\ a = 5 \\ \\ b = 2$

Hope this helps!!!

@Mahak24

Thanks....
☺