How do you solve a quadratic equation containing a negative power?

Question

I was trying to factor a polynomial with Wolfram and I noticed a quadratic form I've never considered.

$$n^2-4+\frac{6}{n}=0$$ The purpose of it is not important but it made me wonder. How do you solve it? A normal quadratic $\quad ax^2+bx+c\quad$ is solved by the quadratic formula:

$$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

but this equation seems to take the form $\quad ax^2+0x-c+dx^{-1}=0.\quad$ How do you solve it? Do I multiply through by $n$ and then solve it as a cubic?

Answer 1

Hint Multiplying by $n$ you get the cubic equation $$n^3-4n+6=0$$

The Rational Root Test tells you that the only potential rational roots are $\pm1, \pm 2, \pm 3, \pm6$ and you can see that none of them works. This means that you cannot factor this equation easily; you need to use the Cubic Formula (or solve it as a cubic).

Answer 2

First, you have to multiply by $n$, obtaining: $$n^3-4n+6$$. Now you can use the cubic formula, obtaining the only real solution $x=2.5251022548143$.

Answer 3

You need to solve for $n$ the cubic equation $$n^3-4n+6=0$$

Just follow the steps given here with $a=1$, $b=0$, $c=-4$ and $d=6$. You will get $$\Delta=-716 \qquad p=-4 \qquad q=6$$ So, only one real root.

To get it, use the hyperbolic method and get $$n=-\frac{4 }{\sqrt{3}}\cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{9 \sqrt{3}}{8}\right)\right)=-2.52510225481432$$