HMMT February

Note that $x=0$ is clearly not a solution, so we can divide the equation by $x^{2}$ to get $\left(x^{2}+2+\frac{1}{x^{2}}\right)+3\left(x+\frac{1}{x}\right)+1=0$. Letting $y=x+\frac{1}{x}$, we get that $y^{2}+3 y+1=0$, so $y=x+\frac{1}{x}=\frac{-3 \pm \sqrt{5}}{2}$. Since $\frac{-3+\sqrt{5}}{2}$ has absolute value less than 2, the associated $x$ are on the unit circle, and thus the two solutions for $x$ in this case each have magnitude 1. For $\frac{-3-\sqrt{5}}{2}$, the roots are negative reals that are reciprocals of each other. Thus, the sum of their absolute values is the absolute value of their sum, which is $\frac{3+\sqrt{5}}{2}$. Thus, the sum of the magnitudes of the four solutions are $1+1+\frac{3+\sqrt{5}}{2}=\frac{7+\sqrt{5}}{2}$.