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.

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.
