Question

Calculate the value of $x^2+x+\frac{1}{x}+\frac{1}{x^2}$. Given $x^2-3x+1=0$

Answer 1

$\Rightarrow x^2+\frac{1}{x^2}+2+\frac{1}{x}+x-2$

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

given $x^2-3x+1=0$

$x=\frac{3\pm \sqrt{9-4}}{2}$

$x=\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$

if $x=\frac{3+\sqrt{5}}{2}$

$x+\frac{1}{x}=\frac{3+\sqrt{5}}{2}+\frac{2}{3+\sqrt{5}}$

$x+\frac{1}{x}=3$

$(x+\frac{1}{x})+{(x+\frac{1}{x})}^2-2=3+3^2-2=10$

if $x=\frac{3-\sqrt{5}}{2}$

$x+\frac{1}{x}=\frac{3-\sqrt{5}}{2}+\frac{2}{3-\sqrt{5}}=3$

$(x+\frac{1}{x})+{(x+\frac{1}{x})}^2-2=3+3^2-2=10$