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.

Just to give a different approach, note that any solution to $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ must lie in $(-1,0)\cup(1,\infty)$ because the left hand side is negative on $(-\infty,-1)\cup(0,1)$ whereas the right hand side, being a square root, is always non-negative, and the expression is undefined at $x=0$ and $x=\pm1$. Now if $x\lt0$, then $3x^4-x^3+2x^2-2x+1$ is a sum of positive terms (i.e., $x^3$ and $2x$ are negative, so $-x^3$ and $-2x$ are positive), while if $x\gt1$, then
$$3x^4-x^3+2x^2-2x+1\gt3x^4-x^4+2x^2-2x^2+1=2x^4+1\gt0$$
so $3x^4-x^3+2x^2-2x+1$ has no roots in the feasible solution set for $x$. (In fact, as user's answer shows, it has no roots in $(0,1)$ either.)
