Question

can answer this math question about rationalization ​

Answer 1

$a = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

Rationalizing the denominator we get:

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

Similarly,

$b = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \bf 5 - 2 \sqrt{6}$

Now,

L.H.S.

a² + b²

$= {(5 + 2 \sqrt{6} )}^{2} + {(5 - 2 \sqrt{6} )}^{2} \\ \\ = (25 + 24 + 20 \sqrt{6} ) + (25 + 24 - 20 \sqrt{6} ) \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ = \bf 98$

L.H.S = R.H.S

Hence proved