Why, in the rules regarding the discriminant determining whether roots are real or not, does it matter that the quadratic has real coefficients?
For the roots to be real we have$$\Delta=b^2-4ac\ge 0 \implies b^2 \ge 4ac$$
If $b$ was imaginary $b^2$ would end up being negative. If the product $4ac < b^2$ we can still have $b^2 - 4ac < 0 $. In this scenario the rules regarding the discriminant does not hold.
$\endgroup$ 71 Answer
$\begingroup$Consider the quadratic $\,az^2+bz+c\,$ with complex coefficients $\,a,b,c \in \mathbb C\,$, $\,a \ne 0\,$.
The roots are still given by the same formula as in the real case $\,z_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}\,$ where $\,\Delta =b^2-4ac\,$ is the discriminant. In particular, $\,\Delta = 0\,$ iff the quadratic has a double root, and in that case $\,z_1=z_2=-\frac{b}{2a}\,$. This can be easily proved the same way as in the real case, for example by completing the square.
The discriminant $\,\Delta\,$ is complex in general, and complex numbers are not ordered, so $\,\Delta\,$ has no "sign" to speak of, and does not indicate the nature of the roots. Besides, unlike the real case, a quadratic with complex coefficients can have one real and one non-real complex root.
It is possible to determine the nature of the roots of a complex quadratic by reducing it to the case of two real quadratics. To simplify the calculations, let $\,z=x+iy\,$, $\,\frac{b}{a}=b'+ib''$, $\,\frac{c}{a}=c'+ic''$, then dividing the equation by $\,a \ne 0\,$ and isolating the real and imaginary parts gives:
$$ (x+iy)^2+(b'+ib'')(x+iy)+c'+ic'' = 0 $$
$$ \iff\;\;\;\; \begin{cases} \begin{align} x^2 - y^2+b'x-b''y + c' &= 0 \\ 2xy+b'y+b''x+c'' &= 0 \end{align} \end{cases} $$
For the equation to have a real root, the system must have a solution with $\,y=0\,$:
$$ \begin{cases} \begin{align} x^2 +b'x + c' &= 0 \\ b''x+c'' &= 0 \end{align} \end{cases} $$
If $\,b''=0\,$ the second equation requires $\,c''=0\,$, which is the case where $\frac{b}{a}, \frac{c}{a} \in \mathbb R$, so the original quadratic is equivalent to one with real coefficients. The roots will be either both complex, or both real, depending on the sign of $\,\Delta'=b'^{\,2}-4c'\,$.
If $\,b'' \ne 0\,$ then the second equation gives $\,x=-\frac{c''}{b''}\,$, and substituting back into the first equation $\,c''^{\,2}+b'b''c''+b''^{\,2}c'=0\,$, which is the condition for at least one real root.
