Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula: 

$$4+\frac{1}{x}-\frac{1}{x^2}=0$$

The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
Here's my working:
Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$
Rearranging into standard form:
$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$
Multiply by $-1$ to get a positive leading coefficient $a$:
$$\frac{1}{x^2}-\frac{1}{x}-4=0$$
I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.
Plugging into quadratic function:
$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
I find this challenging due to the coefficients $a$ and $b$ being fractions.
How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?
 A: Note that$$4+\frac1x-\frac1{x^2}=\frac1{x^2}\left(4x^2+x-1\right).$$So, solve the equation $4x^2+x-1=0$.
A: Multiplying both sides by $x^2$ will result in 
$$4x^2+x-1=0$$
Now, with $a=4, b=1, c=-1$ use the quadratic formula and let us know what you get.
A: Other answers followed my suggestion in the comments.  
Here's an alternative:
Let $z=\dfrac1x.$  Then we have $-z^2+z+4=0$, so, using the quadratic formula, $z=\dfrac{-1\pm\sqrt{17}}{-2}.$ 
Therefore $x=\dfrac1z=\dfrac{-2}{-1\pm\sqrt{17}}=\dfrac{2(-1\mp\sqrt{17})}{16}.$
A: From
$$4+\left(\frac1x\right)-\left(\frac1x\right)^2=0$$
using the standard formula blindly,
$$\frac1x=\frac{1\pm\sqrt{17}}2$$
and obviously
$$x=\frac2{1\pm\sqrt{17}}.$$

Though this seems to contradict the expected answer, consider
$$\frac2{1\pm\sqrt{17}}=\frac{2(1\mp\sqrt{17})}{1-(\sqrt{17})^2}=\frac{-1\pm\sqrt{17}}8.$$
A: Note that $$4+\frac1x-\frac1{x^2}=4+\frac1x-\left(\frac1x\right)^2,$$ so making the substitution $y=\frac1x,$ we obtain the quadratic $$-y^2+y+4=0,$$ which should be more familiar. Solve for $y,$ and since neither solution for $y$ should be equal to $0,$ use $x=\frac1y$ to solve for $x.$
A: None of the other answers so far has really addressed the error that you made in your attempt. After rearranging the original equation into $$\frac{1}{x^2}-\frac{1}{x}-4=0,$$ you then decided that $a=1/x^2$, $b=-1/x$ and $c=-4$. Substituting these names for the corresponding values in this equation gives you $$a+b+c=0.$$ This is no longer a quadratic equation—it doesn’t even have an unknown to solve for!  
Instead, as other answers explain, you need to either multiply by $x^2$ to eliminate $x$ from all of the denominators, or introduce a new variable such as $y=1/x$. Either approach will give you an equation that looks more like one you’re used to. Remember to reject $x=0$ if it comes up as a solution to the modified equation (it won’t).
