Question

If $x=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y=\frac{\sqrt{3}-1}{\sqrt{3}+1}$ find value of $x^2+y^2$

Answer 1

From the picture we have $x=2+\sqrt{3}$ and $y=2-\sqrt{3}$.

Then,

$x^2+y^2$

$=\{4+3+2\sqrt{3}\}+\{4+3-2\sqrt{3}\}$

$=7+7$

$=14$.