<s>[INST] Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ $$\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$$
$$x=?$$
I solved this but the equation $ (2x + 1) (3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1) = 0 $ is formed I answer $ x =- \frac {1} {2} $ I know there is, but I couldn't do the next expression. I need help with that, or someone will solve it in a better way. I'd be happy with that.
 [/INST] Your way is fine with a solution for $x=-\frac12$, now we have that
$$3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1=x^2(3x ^ 2-x  + 1)+(x-1)^2 > 0$$
since $3x ^ 2-x  + 1>0$.

To find complex roots we can refer to Quartic equation.
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