Suppose $a^2-4b$ $\neq 0$. Let $\alpha, \beta$ be the (distinct) roots of the polynomial $x^2+ax+b$. Then there is a real number $c$ s.t. Claim: Let $a,b$$\in$$\mathbb{R}$ and suppose $a^2-4b$$\neq$$0$. Let $\alpha$ and $\beta$ be the (distinct) roots of the polynomial $x^2+ax+b$. Then there is a real number c such that either $\alpha$ $-$ $\beta$ $=$ $c$ or $\alpha$ $-$ $\beta$ $=$ $ci$.
My attempt
Proof. Suppose $a^2-4b$$\neq$ $0$ with $a,b$$\in$$\mathbb{R}$. Because $a^2-4b$$\neq$ $0$, it follows that either $a^2-4b>0$ or $a^2-4b<0$.
Case 1: If $a^2-4b>0$, then $\alpha$ and $\beta$ are two distinct real roots such that p($\alpha$) $=$ $0$ and p($\beta$) $=$ $0$. Thus $\alpha$ $-$ $\beta$ $=$ $c$ with $c$ $\in$$\mathbb{R}$.
Case 2: If $a^2-4b<0$, then $\alpha$ and $\beta$ are two distinct complex roots such that p($\alpha$) $=$ $0$ and p($\beta$) $=$ $0$. Thus $\alpha$ $-$ $\beta$ $=$ $ci$ with $c$ $\in$$\mathbb{R}$.
Hence there is a real number $c$ such that either $\alpha$ $-$ $\beta$ $=$ $c$ or $\alpha$ $-$ $\beta$ $=$ $ci$, as claimed.
I'm not sure whether my approach here is correct.
 A: Your proof of case 2 is clearly incomplete. Note that, in that case,$$\alpha=\frac{-a\pm\sqrt{a^2-4b}}2\quad\text{and}\quad\beta=\frac{-a\mp\sqrt{a^2-4b}}2.$$Since $a^2-4b<0$, this means that$$\alpha=\frac{-a\pm\sqrt{4b-a^2}\,i}2\quad\text{and that}\quad\beta=\frac{-a\mp\sqrt{4b-a^2}\,i}2.$$So,$$\alpha-\beta=\pm\sqrt{4b-a^2}\,i.$$
A: $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=a^2-4b\in\mathbb{R}$.
If $a^2-4b>0$, then $\alpha-\beta=c$ where $c^2=a^2-4b$.
If $a^2-4b<0$, then $\alpha-\beta=ci$, where $c^2=-(a^2-4b)$.
A: By looking at the complex conjugate $$\overline{x^2+ax+b}={\bar x}^2+a\bar{x}+b$$ you can note that: $z$ is a root of this quadratic if and only if so does $\bar z.$ In fact, this is the case with any polynomial with real coefficients. Hence, if $\alpha$ is real, then so does $\beta$ and $\alpha-\beta.$ On the other hand, if $\alpha$ is not real, then $\beta$ must be the complex conjugate of it, thus $\alpha-\beta$ is purely imaginary.
If $a^2-4b=0,$ then you can choose $c=0,$ and hence that data is irrelevant to the question.
