Question

if x = √3+√2/√3+√2 and y = √3-√2/√3+√2 , then find the value of x^2 + y^2

Answer 1

Ans.

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ = \frac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} ) } \\ = \frac{3 + 2 \sqrt{6} + 2}{3 - 2} \\ = 5 + 2 \sqrt{6} \\ \\ and \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{( \sqrt{3} - \sqrt{2} )( \sqrt{3} - \sqrt{2} )}{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} ) } \\ = \frac{3 - 2 \sqrt{6} + 2}{3 - 2} \\ = 5 - 2 \sqrt{6}$

So,
${x}^{2} + {y}^{2} \\ = {(5 + 2 \sqrt{6} )}^{2} + {(5 - 2 \sqrt{6} )}^{2} \\ = (25 + 20 \sqrt{6} + 24) + (25 - 20 \sqrt{6} + 24) \\ = 98$

I HOPE THAT THIS HELPS YOU.