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?

  • $\begingroup$ $ix^2 - i$ has the roots $\pm 1$. Is this the type of thing you're looking for? $\endgroup$ – MartianInvader Nov 6 '13 at 16:22

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.