Quadratic equation property

Question

It is known that if $x_1$ and $x_2$ are the roots of the equation $f(x)=ax^2+bx+c=0\:$ and $x_1 < k < x_2$, $\;\sqrt{b^2-4ac}<0\;$ and $\;af(k)<0$. Why is that so?

Answer 1

$\sqrt{b^2-4ac}$, by definition, denotes the nonnegative square root of the nonnegative number $b^2-ac$, so the assertion that $\sqrt{b^2-4ac}<0$ is absurd.

What is true is this:

If the equation $f(x)=ax^2+bx+c=0\:$ has two real roots $x_1<x_2$ and if a number $k$ separates these roots, i.e. $x_1<k<x_2$ then the discriminant of the equation $\Delta=b^2-4ac>0$ and $af(k)<0$, i.e. the leading coefficient $a$ and $f(k)$ have opposite signs.

The reason for the last assertion is that the existence of two roots implies a factorisation $f(x)$ as $\;f(x)=a(x-x_1)(x-x_2)$, so that $$af(k)=a^2(k-x_1)(k-x_2)<0,$$ and $k-x_1>0\;$ whereas $\;k-x_2<0$.