Question

If $a=\frac{2-\sqrt{5}}{2+\sqrt{5}}$ and $b=\frac{2+\sqrt{5}}{2-\sqrt{5}}$, then find the value of $a^2-b^2$.

Answer 1

Let, $a=\frac{2-\sqrt{5}}{2+\sqrt{5}}$ and $b=\frac{2+\sqrt{5}}{2-\sqrt{5}}$.

Now

$a+b$

$=\frac{2-\sqrt{5}}{2+\sqrt{5}}$$+\frac{2+\sqrt{5}}{2-\sqrt{5}}$

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

$=\frac{2(2^2+5)}{(-1)}$

$=-\left(18\right)$......(1).

Again,

$a-b$

$=\frac{2-\sqrt{5}}{2+\sqrt{5}}$$-\frac{2+\sqrt{5}}{2-\sqrt{5}}$

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

$=\frac{-4\left(2.\sqrt{5}\right)}{(-1)}$

$=\left(8\sqrt{5}\right)$......(2).

Now

$a^2-b^2$

$=(a+b)(a-b)$

$=-144\sqrt{5}$ [ Using (1) and (2)]