Question

If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, find $x^2+y^2$.

Answer 1

We have, $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$\Rightarrow x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
[Rationalising the denominator]
$\Rightarrow x=\frac{{(\sqrt{3}+\sqrt{2})}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\frac{3+2++2\sqrt{3}\sqrt{2}}{3-2}=5+2\sqrt{6}$
Similarly , $y=5-2\sqrt{6}$
Now, $xy=(5+2\sqrt{6})(5-2\sqrt{6})=5^2-{\left(2\sqrt{6}\right)}^2$
$=25-24=1$
and, $x+y=5+2\sqrt{6}+5-2\sqrt{6}=10$
$\therefore {(x+y)}^2={10}^2$
$\Rightarrow x^2+y^2+2xy=100$
$\Rightarrow x^2+y^2+2\times 1=100$
$\Rightarrow x^2+y^2=98$