Alg2.1.15

Find $x+\frac{1}{x}, x^2+\frac{1}{x^2}\, if x=\frac{\sqrt{5}-2}{\sqrt{5}+2}\,$

Substituting the value of x in the first one

$x+\frac{1}{x}=(\frac{\sqrt{5}-2}{\sqrt{5}+2})+(\frac{\sqrt{5}+2}{\sqrt{5}-2})\,$

Simplifying

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

$x+\frac{1}{x}=\frac{2(5+4)}{5-4}\,$

Hence

$x+\frac{1}{x}=18\,$

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

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

Simplifying $x^2+\frac{1}{x^2}=\frac{(9-4\sqrt{5})^2+(9+4\sqrt{5})^2}{(9-4(\sqrt{5})(9+4\sqrt{5})}\,$

$x^2+\frac{1}{x^2}=\frac{(9-4\sqrt{5})^2+(9+4\sqrt{5})^2}{(9-4(\sqrt{5})(9+4\sqrt{5})}\,$

$x^2+\frac{1}{x^2}=\frac{2(81+50)}{(81-80)}\,$

Final solution is

$x^2+\frac{1}{x^2}=262\,$