Hmm, I have been wondering about this when I went to solve the following equation: $$\frac{-x^2+2x}{5x-4} = 6$$
How come the above equation has two solutions, $-14 + 2\sqrt{55}$ and $-14 - 2\sqrt{55}$? I know when I simplify it, it turns to a quadratic equation, but how come it gets there? Also, does this apply to any rational equation in this form?
On
$$ \frac{-x^2+2x}{5x-4} = 6 \iff \\ -x^2+2x = 6 (5x-4) \wedge 5x-4 \ne 0 \iff \\ x^2 + 28 x - 24 = 0 \wedge x \ne 4/5 $$ And a polynomial of order 2 has up to two real solutions.
On
In general, if $p(x)$ is a polynomial of degree $m$, $q(x)$ a polynomial of degree $n \ne m$ such that $p(x)$ and $q(x)$ are coprime as polynomials, and $c$ a nonzero constant, $\dfrac{p(x)}{q(x)} = c$ is equivalent to $p(x) - c q(x) = 0$, and since $p(x) - c q(x)$ has degree $\max(n,m)$, that is the number of solutions in $\mathbb C$ (counted by multiplicity).
The equation can be rewritten as follows:
$$\frac{-x^2+2x-6(5x-4)}{5x-4} = 0 $$
Simplifying:
$$-x^2-28x+24=0$$
And solving:
$$x=-14\pm\sqrt{55}$$