Real roots of $z^2+\alpha z + \beta=0$ Question:-
If equation $z^2+\alpha z + \beta=0$ has a real root, prove that $$(\alpha\bar{\beta}-\beta\bar{\alpha})(\bar{\alpha}-\alpha)=(\beta-\bar{\beta})^2$$

I tried goofing around with the discriminant but was unable to come with anything good. Just a hint towards a solution, might work.
 A: Eliminate $r$ between
$$r^2+r\alpha+\beta=0$$ and $$r^2+r\bar\alpha+\bar\beta=0.$$
By Cramer,
$$r^2=-\frac{\left|\begin{matrix}\beta&\alpha\\\bar\beta&\bar\alpha\end{matrix}\right|}{\left|\begin{matrix}1&\alpha\\1&\bar\alpha\end{matrix}\right|},$$
$$r=-\frac{\left|\begin{matrix}1&\beta\\1&\bar\beta\end{matrix}\right|}{\left|\begin{matrix}1&\alpha\\1&\bar\alpha\end{matrix}\right|}.$$
A: Let the roots be $-x,-y$ with $x$ real. Then $\alpha=x+y$ and $\beta=xy$, hence
$$
\alpha\bar{\beta}-\beta\bar{\alpha}=(x+y)x\bar{y}-xy(x+\bar{y})=x^2(\bar{y}-y)   \\
(\bar{\alpha}-\alpha)=\bar{y}-y         \\
(\beta-\bar{\beta})^2 =x^2 (y-\bar{y})^2  \\
$$
It is easy to see that the product of the first two is the last.
A: Let $x$ be a real root of the given equation. Then we have
\begin{align*}
x^2+\alpha x+\beta&=0\\
x^2+\overline\alpha x+\overline\beta&=0
\end{align*}
and after subtracting we get
\begin{gather*}
(\alpha-\overline\alpha)x+(\beta-\overline\beta)=0\\
x=-\frac{\beta-\overline\beta}{\alpha-\overline\alpha}
\end{gather*}
Now let us plug this into the original equation
\begin{align*}
x^2&=-\alpha x -\beta\\
\frac{(\beta-\overline\beta)^2}{(\alpha-\overline\alpha)^2} &= \frac{\alpha(\beta-\overline\beta)}{\alpha-\overline\alpha} - \beta\\
\frac{(\beta-\overline\beta)^2}{(\alpha-\overline\alpha)^2} &= \frac{\alpha(\beta-\overline\beta)}{\alpha-\overline\alpha} - \frac{\beta(\alpha-\overline\alpha)}{\alpha-\overline\alpha}\\
\frac{(\beta-\overline\beta)^2}{(\alpha-\overline\alpha)^2} &= \frac{\beta\overline\alpha-\alpha\overline\beta}{\alpha-\overline\alpha}   \\
(\beta-\overline\beta)^2 &= (\beta\overline\alpha-\alpha\overline\beta)(\alpha-\overline\alpha)
\end{align*}
This is basically what we wanted to prove. (In the given problem, the sign in both brackets is changed to opposite, which does not change the expression.)
