Question

If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$ find the value of $x^2+y^2+2xy$

Answer 1

$\begin{array}{c}x=\frac{\sqrt{5}+1}{\sqrt{5}-1},y=\frac{\sqrt{5}-1}{\sqrt{5}+1}\\ x^2+y^2+xy={\left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)}^2+{\left(\frac{\sqrt{5}-1}{\sqrt{5}+1}\right)}^2+2\left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)\left(\frac{\sqrt{5}-1}{\sqrt{5}+1}\right)\\ =\frac{6+2\sqrt{5}}{6-2\sqrt{5}}+\frac{6-2\sqrt{5}}{6+2\sqrt{5}}+2\\ =\frac{{\left(6+2\sqrt{5}\right)}^2+{\left(6-2\sqrt{5}\right)}^2}{6^2-{\left(2\sqrt{5}\right)}^2}+2\\ =\frac{36+20+24\sqrt{5}+36+20-24\sqrt{5}}{36-20}+2\\ =\frac{112}{16}+2\\ =7+2\\ =9\\ Whichistherequiredanswer.\end{array}$