Real roots of $z^2+\alpha z + \beta=0$

Question

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

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I tried goofing around with the discriminant but was unable to come with anything good. Just a hint towards a solution, might work.

Answer 1

Eliminate $r$ between

$$r^2+r\alpha+\beta=0$$ and $$r^2+r\bar\alpha+\bar\beta=0.$$

By Cramer,

$$r^2=-\frac{\left|\begin{matrix}\beta&\alpha\\\bar\beta&\bar\alpha\end{matrix}\right|}{\left|\begin{matrix}1&\alpha\\1&\bar\alpha\end{matrix}\right|},$$ $$r=-\frac{\left|\begin{matrix}1&\beta\\1&\bar\beta\end{matrix}\right|}{\left|\begin{matrix}1&\alpha\\1&\bar\alpha\end{matrix}\right|}.$$

Answer 2

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.