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Where $a\in \textbf{Z}[i] $ and $a \not\in \textbf{Z}$, suppose for the quadratic formula, $ b^2-4ac = 0 \Rightarrow b^2 = 4 ac \Rightarrow c= \frac{b^2}{4a} $ and $ b=a $, so that $\displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\frac{-b}{2a}=\frac{-b}{2b}=-\frac{1}{2}$. More generally, suppose $d \in \textbf{Z}$ and $b=da$. Then the quadratic formula will have a root of $ \displaystyle -\frac{db}{2b}=- \frac{d}{2}$, thus any real number could be a root.

I was going to ask this question, but then answered it in trying to show that I had made an attempt...did I do so correctly? Could a complex quadratic polynomial have more than 1 real root?

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  • $\begingroup$ $ix^2 - i$ has the roots $\pm 1$. Is this the type of thing you're looking for? $\endgroup$ – MartianInvader Nov 6 '13 at 16:22
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If $f$ is a quadratic polynomial with more than 1, i.e. 2 real roots, it can be written as $\alpha(x-a)(x-b)$, where $a,b$ are the roots and $\alpha$ can be `anything'. If you allow $\alpha$ to be complex, then yes, a complex quadratic polynomial may have 2 real roots. However, if you require the polynomial to be monic (having leading coefficient 1), $\alpha$ has to be 1 and this cannot be the case anymore.

I do have to admit that I don't quite follow what you're doing, though.

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