Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$

Question

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

$$x=?$$

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.

Answer 1

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.)

Answer 2

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 $3x ^ 2-x + 1>0$.

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To find complex roots we can refer to Quartic equation.