Question

If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1},y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$ Find $x^2+xy+y^2$.

Answer 1

We have,

$x=\frac{\sqrt{5}+1}{\sqrt{5}-1},y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$

Since,

$=x^2+xy+y^2$

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

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

$=\left(\frac{6+2\sqrt{5}}{6-2\sqrt{5}}\right)+\left(\frac{6-2\sqrt{5}}{6+2\sqrt{5}}\right)+1$

$=\frac{3+\sqrt{5}}{3-\sqrt{5}}+\frac{3-\sqrt{5}}{3+\sqrt{5}}+1$

$=\frac{{(3+\sqrt{5})}^2+{(3-\sqrt{5})}^2}{(3-\sqrt{5})(3+\sqrt{5})}+1$

$=\frac{9+5+6\sqrt{5}+9+5-6\sqrt{5}}{9-5}+1$

$=\frac{28}{4}+1$

$=7+1$

$=8$

Hence, this is the answer.