Complex roots meaning the quadratic equation is always positive?

Question

Does it mean that if you have complex roots for a quadratic equation, i can take in any real number as its domain? Then would the formula for range, ie,[-D/4a,inf) where D is the discriminant, still be applicable here? Im really confused here as it does not seem intuitive to me that a complex root quadratic equation can take in any real number as its domain.

Answer 1

The domain and range are fixed entities, any polynomial will have domain = $\mathbb{R}$. If $D \lt 0$, the range will be $[-\frac{D}{4a}, \infty)$ if $a \gt 0$ and $(-\infty, -\frac{D}{4a}]$ if $a \lt 0$.

Answer 2

It's not correct to say that if a polynomial has complex (non real) roots, then the quadratic equation is always positive. Take for example $x^2+1=0$ this has complex roots and it's always positive, while: $$-1-x^2=0$$ has no real roots but $-1-x^2<0 \;\;\forall x \in R$.

Answer 3

Any function can be defined over any domain. In particular the quadratic $$f(x)=ax^2+bx+c,$$ where $a,b,c\in \mathrm R$ may be defined over complex or real values or even operators, etc. But if we take $x$ to be real, then obviously $f(x)$ is always real. It is not compulsory for $f(x)$ to vanish. When it does, it always vanishes at some point in the domain, here some real value of $x.$

However, if $f(x)$ never vanishes, then it means that there is no real number $x$ that makes it disappear. Thus, if it will vanish, then the values $x$ that will make it vanish cannot be real. Hence they may be complex. I hope this clears your confusion.