Solve x^2-x+1/x=fx

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

Solve for f

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

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$x^{2}-x+1=fxx$$

Multiply x and x to get x^{2}.

$$x^{2}-x+1=fx^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$fx^{2}=x^{2}-x+1$$

The equation is in standard form.

$$x^{2}f=x^{2}-x+1$$

Divide both sides by x^{2}.

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

Dividing by x^{2} undoes the multiplication by x^{2}.

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

Divide x^{2}-x+1 by x^{2}.

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{4f-3}+1}{2\left(1-f\right)}\text{; }x=\frac{-\sqrt{4f-3}+1}{2\left(1-f\right)}\text{, }&f\neq 1\\x=1\text{, }&f=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{4f-3}+1}{2\left(1-f\right)}\text{; }x=\frac{-\sqrt{4f-3}+1}{2\left(1-f\right)}\text{, }&f\neq 1\text{ and }f\geq \frac{3}{4}\\x=1\text{, }&f=1\end{matrix}\right.$$