Как решить уравнение?

Решите задачу:

$x^2+\frac{1}{x^2}+x+\frac{1}{x}=0\\\\t=x+\frac{1}{x}\; ;\; t^2=(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2\; \; \to \; \; x^2+\frac{1}{x^2}=t^2-2\\\\t^2+t-2=0\\\\t_1=1,\; t_2=-2\; \; \to \\\\1)\; \; x+\frac{1}{x}=1\\\\x+\frac{1}{x}-1=0\; \; \to \; \; \frac{x^2-x+1}{x}=\; \; \to \; \; \left \{ {{x^2-x+1=0} \atop {x\ne 0}} \right. \\\\x^2-x+1=0\; ;\; D=1-4=-3\ \textless \ 0\; \to \; net\; reshenij\\\\2)\; \; x+\frac{1}{x}=-2$

$x+\frac{1}{x}+2=0\; \; \to \; \; \frac{x^2+2x+1}{x}=0\; \; \to \; \; \left \{ {{x^2+2x+1=0} \atop {x\ne 0}} \right. \\\\x^2+2x+1=(x+1)^2=0\; \; \to \; \; x=-1\\\\Otvet:\; x=-1.$

$P.S.(x+\frac{1}{x})^2=x^2+2*x*\frac{1}{x}+(\frac{1}{x})^2=x^2+2*1+\frac{1^2}{x^2}=\\\\=x^2+2+\frac{1}{x^2}$