Why does $f(x)=ax^2 + bx + c \ge 0\ \forall x \in \mathbb R$ imply $f$ has at most one real distinct root and discriminant $D \le 0$? 
Why does $f(x)=ax^2 + bx + c \ge 0 \ \forall x \in \mathbb R$ imply $f$ has at most one real distinct root and discriminant $D \le 0$?

I've been wondering why the following result is true. Intuitively it seems to be true if you consider the graph of a polynomium of degree 2 that doesn't intersect the $x$-axis twice. However this is by no means rigorous and I think the answer should be simple, since the text is introductory.
 A: If $f$ had two distinct roots, these would be simple roots, hence at each of these the function changes sign. As the function is nowhere negative, this is impossible.
A: The roots are $$x=\frac{-b\pm \sqrt{D}}{2a}$$
If $D\lt0$, and assuming that the coefficients are real, the roots must be complex clearly, in which case we have $0$ real roots.
If $D=0$, $$x=-\frac{b}{2a}$$ is the only solution obviously.
The above is assuming $a\ne 0$, if it is equal to $0$, then $x=\frac{-c}{b}$ trivially.
A: Ok. Let's assume a is positive, otherwise we take -a
$$f(x) = a[(x + \frac{b}{2a})^2-\frac{b^2}{4a}+\frac{c}{a}]$$
$$f(x) = a[(x + \frac{b}{2a})^2+\frac{-b^2+4ac}{4a}](*)$$
I just tried to force a square, do you see how it is equal to the initial f? 
Now this quantity has a minimum for $x=-\frac{b}{2a}$ and as we know f is positive, therefore the quantity on the right in the main equation (*) is necessarily positive. It's $-\Delta >0$ the discriminant that is defined like $\Delta = b^2 - 4ac$. The discriminant is negative.
Note that $f(x) = 0$ on $x=-\frac{b}{2a}$ if and only if the discriminant is zero. That is what we call a double root.
EDIT : By the way, a has to be positive if the function stays positive. If you take the limit in $+\infty$ it behaves the same way as $ax^2$ so the positivity implies $a>0$
