A quadratic polynomial is nonnegative for all $x$ if and only if the discriminant is nonpositive Show that if $a>0$ the inequality $ax^2+2bx+c\ge 0 $ for all values of $x$ if and only if $b^2-ac\le 0$.
I tried to prove it by: $ax^2+2bx+c≥ b^2-ac$.
Used partial derivatives with respect to $a$. Sorry, I am new to proofs. Can someone help me out?
 A: The quadratic formula is all you need to prove this. 
If you have a quadratic polynomial: 
$$ ax^2+bx+c=0$$
Then, to find the zeroes, you rearrange the above expression into the quadratic formula: 
$$ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
From here, you can tell the discriminant, $\Delta =b^2-4ac$ and from $\Delta $ you can tell how many zeroes (where the graph touches the x-axis) the function will have. If $\Delta =0$ the polynomial function will have 1 zero, since the plus or minus term becomes zero and there is only one value for x. If $\Delta\gt0 $ then the function will have two zeroes, since the plus or minus term afects the negative b term. If $\Delta\lt0 $ then the function will not have any real zeroes because the term inside the root will be negative and the square root of a negative number is an imaginary number. 
Therefore, when $b^2-4ac\le0$ the polynomial equation can only have a maximum of one zero, its square root term (the root of the determinant $\Delta $) will only either yield a zero or an imaginary number. If we also know that $a\gt0$, then we know that the parabola defined by the quadratic will open upward like a happy face. We also know that it will only have one zero, maximum. Thus, the parabola will always be above the x-axis, or it will only touch it once, making the above statement $a>0$,$ax2+2bx+c≥0$ for all values of x if,and only if $b^2−ac≤0$
A: Hint: the vertex $x = -\dfrac{b}{a} \Rightarrow f_{\text{min}} = f\left(-\frac{b}{a}\right) = \cdots$
A: The  discriminant of the quadratic $\;y=ax^2+2bx+c\;$ is 
$$\Delta=4b^2-4ac$$
The above parabola is "hoovering"above the $\;x$- axis (including being tangent to it) iff it has at most one real root (why?), and this happens iff is discriminant is non-positive. End the argument
