Consider $az^2+bz+c=0$ where $a,b,c$ are all complex numbers. What is the condition (ie the relation between $a,b,c$) for which the given quadratic has both real roots?
I took the conjugate of the given equation to get $$\overline a\overline z^2+\overline b\overline z + \overline c=0$$ Now since we need $z$ to be real, $z=\overline z$. $$\Rightarrow \overline az^2+\overline bz + \overline c=0$$
So now I thought of comparing this equation with the original and notice that they both must be identical hence the ratio of the respective coefficients must be equal. $$\frac{a}{\overline a}=\frac{b}{\overline b}=\frac{c}{\overline c}$$
(This condition I obtained actually matched the answer I have however the source isn't very trustworthy)
But here I got confused as to why we are applying condition for both roots common when what we want is for the roots to be real. Roots can be common but unreal. I don't even know how much of the results of quadratic with real coefficients carries over here. I can't even imagine a graph of this function so I'm just really confused as to what's even going on. Can someone help me in understanding this?
Best Answer
I will assume that $a\ne0$. Then, if $\beta =\dfrac{\,\overline b\,}{\overline a}$ and $\gamma=\dfrac{\,\overline c\,}{\overline a}$, the roots of $az^2+bz+c$ are the roots of $z^2+\beta z+\gamma$. If these roots are $r,s\in\Bbb R$, then\begin{align}z^2+\beta z+\gamma&=(z-r)(z-s)\\&=z^2-(r+s)z+rs\end{align}and therefore $\beta=-(r+s)$ and $\gamma=rs$. In particular, $\beta,\gamma\in\Bbb R$.
Now, note that\begin{align}(r-s)^2&=\bigl(-(r+s)\bigr)^2-4rs\\&=\beta^2-4\gamma\end{align}and therefore we must have $\beta^2-4\gamma\geqslant0$. On the other hand, if indeed $\beta,\gamma\in\Bbb R$ and if $\beta^2-4\gamma\geqslant0$, it follows from the quadratic formula that the roots of $z^2+\beta z+\gamma$ are indeed real.
So, the roots of $az^2+bz+c$ are real if and only if$$\dfrac{\,\overline b\,}{\overline a},\dfrac{\,\overline c\,}{\overline a}\in\Bbb R\quad\text{and}\quad\left(\dfrac{\,\overline b\,}{\overline a}\right)^2-4\dfrac{\,\overline c\,}{\overline a}\geqslant0.$$