Question

If $a=\frac{2+\sqrt{5}}{2-\sqrt{5}}$, $b=\frac{2-\sqrt{5}}{2+\sqrt{5}}$, find $a^2-b^2$.

Answer 1

As $a=\frac{2+\sqrt{5}}{2-\sqrt{5}}$, so,

$a^2=\frac{{\left(2+\sqrt{5}\right)}^2}{{\left(2-\sqrt{5}\right)}^2}$

$=\frac{9+4\sqrt{5}}{9-4\sqrt{5}}$

And as $b=\frac{2-\sqrt{5}}{2+\sqrt{5}}$, then,

$b^2=\frac{{\left(2-\sqrt{5}\right)}^2}{{\left(2+\sqrt{5}\right)}^2}$

$=\frac{9-4\sqrt{5}}{9+4\sqrt{5}}$

Now,

$a^2-b^2=\frac{9+4\sqrt{5}}{9-4\sqrt{5}}-\frac{9-4\sqrt{5}}{9+4\sqrt{5}}$

$=\frac{{\left(9+4\sqrt{5}\right)}^2-{\left(9-4\sqrt{5}\right)}^2}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}$

$=\frac{81+80+72\sqrt{5}-81-80+72\sqrt{5}}{81-80}$

$=144\sqrt{5}$