What is a condition for which a quadratic expression $ax^2+bx+c$ is greater than or equal to zero?
We know that the leading coefficient $a$ is positive for this case, but I'm confused on whether the discriminant $D$ is:
- $D \leqslant 0,\;\;$ or
- $D = 0$.
Tell me if you know with suitable logics/reasons.
First of all suppose that $a,b,c\in\mathbb R$, $a\neq 0$. Let $p(x)=ax^2+bx+x$. Let's work with complex solutions to avoid getting into unnecessary problems with the roots, then the roots of the equation are: $$ x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} $$ where $\Delta=b^2-4ac$. Now let's analyze a little the behavior of the discriminant. Now if we get into the discussion of real roots:
Case 1: If $\Delta> 0$, then we have that the graph of $p(x)$ intersects the $x-$axis at two points (turns out $x_1\neq x_2$ both in $\mathbb R$), therefore the graph has a negative and a positive part.
Case 2: If $\Delta= 0$, so the short graph intersects the $x-$axis, at a single point (turns out $x_1=x_2\in\mathbb R$). That is, there are only two possibilities:
Case 3: If $\Delta< 0$, so the graph never intersects the $x-$axis (we have square root of a negative number, therefore there is no real number that is root of $p(x)$.). Again we have two possibilities:
With which you can state the following:
Let $p(x)=ax^2+bx+c$ with $a,b,c\in\mathbb R$, $a\neq 0$. $$p(x)\geq 0 \quad,\forall x\in\mathbb R\Leftrightarrow a>0\text{ and }\Delta\leq 0$$
The proof is simple, but I won't do it since it's not really your question. But I tried to explain what would be happening behind this result.