Solve for $x$ in $6x^2-25x+12+\frac{25}{x}+\frac{6}{x^2}=0$

Question

I simplified to get $6x^4-25x^3+12x^2+25x+6=0$

Or $6 (x^2+1)^2+25x(1-x^2)=0$

Then I substituted $z=x^2+1$, to get $6z^2+25\sqrt{(z-1)}(2-z)=0$

I can't find a way to proceed further.

Answer 1

Substitute $y:=x-\frac{1}{x}$. Thus $y^2=x^2-2+\frac{1}{x^2}$.

Your original equation can be written as $6y^2-25y+24=0$.

Thus for y we get $\frac{25\pm\sqrt{625-96}}{12}=\frac{25\pm23}{12}=\{\frac{1}{6};4\}$

For $y=\frac{1}{6}$ we get for $x^2-\frac{1}{6}x+1=0$, which hasn't any real solution.

For $y=4$ we get $x^2-4x+1=0$, which has the solutions $\frac{4\pm\sqrt{16-4}}{2}=\underline{\underline{2\pm\sqrt{3}}}$.

Answer 2

Divide the whole equation by $x^2$ and simplifing:

$$6(x^2+\frac{1}{x^2})-25(x-\frac{1}{x})+12=6((x-\frac{1}{x})^2+2)-25(x-\frac{1}{x})+12=0$$

Put $x-\frac{1}{x}=y$ to solve the resulting equation:

$6y^2-25y+24=0$

Answer 3

Just use the rational root theorem after you made it into a polynomial

Candidate roots: $\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{1}{6}$.

Testing them you find that al four roots are rational:

$x\in\{2,3,-\frac{1}{2},-\frac{1}{3}\}$