The appearance of complex numbers when factoring quadratics

Question

Why does $i$ only get involved in factoring quadratics like $x^2+9$, but not $x^2-9$? Why does the $+$ sign lead to complex numbers?

Answer 1

I think, you may want to ask when $i$ appears when we factor $ax^2+bx+c$, where $a,b,c\in\mathbb{R}$. $ax^2+bx+c$ can always be factored as $(x-\alpha)(x-\beta)$, where $\alpha$ and $\beta$ are solution of the equation $ax^2+bx+c=0$. In general, \begin{equation} x=\frac{-b+\sqrt{b^2-4ac}}{2a} \end{equation} Therefore, $i$ appears when $b^2 - 4ac <0$. Suppose that $b^2-4ac < 0$. Since $b^2$ is nonnegative, $ac > 0$. Then we can think two possiblities:

1. $a>0$ and $c>0$. (example: $x^2+9$, $x^2+2x+2$, ...)
2. $a<0$ and $c<0$. (example: $-x^2-9$, ...)