# Consider $az^2+bz+c=0$ where $a,b,c$ are all complex numbers

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?

• You need it to be $a(z^2+b'z+c')$ where $z^2+b'z+c'$ has real coefficients and real roots. Sep 22 '20 at 18:03

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$$\dfrac{\,\overline b\,}{\overline a},\dfrac{\,\overline c\,}{\overline a}\in\Bbb R\quad\text{and}\quad\left(\dfrac{\,\overline b\,}{\overline a}\right)^2-4\dfrac{\,\overline c\,}{\overline a}\geqslant0.$$