Question

$x=\frac{\sqrt{5}-2}{\sqrt{5}+2}$ and $y=\frac{\sqrt{5}+2}{\sqrt{5}-2}$.

Find the value of $x^2+y^2+xy$.

Answer 1

We rationalize the denominator,

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

$x=\frac{5+4-3\sqrt{5}}{5-4}$

$x=\frac{9-4\sqrt{5}}{1}$.

Then, $x^2=(9-4\sqrt{5})(9-4\sqrt{5})$

$=81+16\times 5-72\sqrt{5}$

$=161-72\sqrt{5}$ .

We rationalize the denominator,

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

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

$y=\frac{9+4\sqrt{5}}{1}$.

Then, $y^2=(9+4\sqrt{5})(9+4\sqrt{5})$

$=81+16\times 5+72\sqrt{5}$

$=161+72\sqrt{5}$ .

Now, $x=\frac{\sqrt{5}+2}{\sqrt{5}-2}$

and $y=\frac{\sqrt{5}-2}{\sqrt{5}+2}$.

Then, $xy=\frac{\sqrt{5}+2}{\sqrt{5}-2}\times \frac{\sqrt{5}-2}{\sqrt{5}+2}$

$xy=1$ .

Therefore, $x^2+y^2+xy=161-72\sqrt{5}+161+72\sqrt{5}+1=323$.