A quadratic polynomial is nonnegative if and only if the discriminant is nonpositive

Question

I want to 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?

Answer 1

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

Answer 2

Hint: the vertex $x = -\dfrac{b}{a} \Rightarrow f_{\text{min}} = f\left(-\frac{b}{a}\right) = \cdots$

Answer 3

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 $b^2−ac≤0$

Answer 4

A better statement would be the following

Thoerem. Let $Q(x)=ax^2+bx+c$ be a quadratic polynomial with $a \ne 0$. Then, $Q(x)\ge 0$ for all $x\in\mathbb{R}$ if and only if $a > 0$ and $\Delta = b^2 -4ac \leq 0$.

Proof. First note that by the technique of making squares we have

\begin{equation} Q(x) = a\big(x+\frac{b}{2a}\big)^2 - \frac{\Delta}{4a}. \end{equation}

From this it should be clear that if $a\ge 0$ and $\Delta \leq 0$ then for all $x$ we have $Q(x) \ge 0$. To prove the converse, suppose that $Q(x) \ge 0$ holds for all $x$. Letting $x=-\frac{b}{2a}$ leads to $Q(x) = -\frac{\Delta}{4a} \ge 0$ which implies that $a\, \Delta\leq 0$. But $a < 0$ cannot occur since if this holds then $\Delta \ge 0$. So, we have $\frac{\Delta}{4a^2} \ge 0$. Setting $x=-\frac{b}{2a}+\sqrt{\frac{\Delta}{4a^2} + y}$ with $y > 0$ leads to $Q(x) = a\,y < 0$ which is a contradiction.

Note. The intuition behind choosing the values for $x$ in the proof was the extremum of $Q(x)$ which satisfies $Q'x) = 0$ and some $x$ greater than the larger root of $Q(x)$ which is $x=-\frac{b}{2a}+\sqrt{\frac{\Delta}{4a^2}}$ when $\Delta \ge 0$.