Question

If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ , then find the value of $x^2+xy-y^2$.

Answer 1

We have,

$x^2+xy-y^2\Rightarrow {\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)}^2+\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}-{\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right)}^2\Rightarrow \frac{3+2+2\sqrt{6}}{3+2-2\sqrt{6}}+1-\frac{3+2-2\sqrt{6}}{3+2+2\sqrt{6}}\Rightarrow \frac{5+2\sqrt{6}}{5-2\sqrt{6}}+1-\frac{5-2\sqrt{6}}{5+2\sqrt{6}}\Rightarrow \frac{{\left(5+2\sqrt{6}\right)}^2+\left(5-2\sqrt{6}\right)(5+2\sqrt{6})-{\left(5-2\sqrt{6}\right)}^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\Rightarrow \frac{25+24+20\sqrt{6}+25-24-25-24+20\sqrt{6}}{25-24}\Rightarrow \left(1+20\sqrt{6}\right)$