Question

if
$x = \frac{ \sqrt{2 + 1} }{ \sqrt{2 + 1} }$
$y = \frac{ \sqrt{2 - 1} }{\sqrt{2 + 1} }$
then find the value of x²+y²-34​

Answer 1

We have,

$x = \frac{ \sqrt{2 + 1} }{ \sqrt{2 + 1} } \\ \therefore \: {x}^{2} = \frac{2 + 1}{2 + 1} = \frac{3}{3} = 1 \\ Also, \\ y = \frac{ \sqrt{2 - 1} }{ \sqrt{2 + 1} } \\ \therefore \: {y}^{2} = \frac{2 - 1}{2 + 1} = \frac{1}{3} \\ \\ So \: now, \: \\ {x}^{2} + {y}^{2} - 34 = 1 + \frac{1}{3} - 34 = \frac{4}{3} - 34 \\ \\ \implies \: {x}^{2} + {y}^{2} - 34 = \frac{ - 98}{3}$