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 $\dfrac{-1\pm\sqrt{17}}{8}$?
Best Answer
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.