How to prove that $f(x)=ax^2+bx+c=0$ has a unique solution, provided that $\forall x :f(x) \geq 0$?

Question

Let $f$ be a real function $f(x)=ax^2+bx+c:a\neq0$ such that $\forall x \in \mathbb R:f(x) \ge0$.

I've proved that if $\sqrt{\Delta}=\sqrt{b^2-4ac}$ exists, then $x=\dfrac{-b+\sqrt{\Delta}}{2a}$ or $x=\dfrac{-b-\sqrt{\Delta}}{2a}$. Supposing that there exists $x:f(x)=0$, how to prove that $\sqrt{\Delta}$ exists and that it's equal to $0$?

Answer 1

If there exists $\;x_1\;$ s.t. $\;f(x)=0\;$ , then it is given by

$$x_1=\frac{-b\pm\sqrt\Delta}{2a}\;,\;\;\text{with}\;\;\;\Delta:=b^2-4ac$$

It thus must be that $\;\Delta\ge0\;$ , otherwise its square root is not a real number. but if $\;\Delta>0\;$ then there are two possibilities above because of the $\;\pm\;$ signs, and we thus get a different solution $\;x_2\;$ for $\;f(x)=0\;$. So we can say that

$$x_1=\frac{-b+\sqrt\Delta}{2a}\;,\;\;x_2=\frac{-b-\sqrt\Delta}{2a}$$

and because $\;\sqrt\Delta\neq0\;$ , we then get $\;x_1\neq x_2\;$ . Suppose $\;x_1<x_2\;$ , then:

$$f(x)=a(x-x_1)(x-x_2)$$

but then, for any $\;x_1<d<x_2\;$ , we get $\;f(d)<0\;$ (since clearly $\;a>0\;$ ...why?) . Contradiction. Finish now the argument.

Answer 2

If $f(x)=ax^2+bx+c=0$ has two roots, $x_1< x_2$, then we can write it like $f(x)=a(x-x_1)(x-x_2)$ and then we have two cases:

$1)$ If $a>0$ then take a number $p$ such that $x_1<p<x_2$ and then we will have $f(p)<0$ what is a contradiction;

$2)$ If $a<0$ then take a number $p$ such that $p< x_1$ or $p>x_2$ and then we will have $f(p)<0$ what is a contradiction;

Otherwise, if $x_1=x_2$ then we can write $f(x)=a(x-x_1)^2$, with $a>0$.

Answer 3

First note that $a > 0$ otherwise the polynomial must became negative for some large $x$ or it is equal to the null polynomial and your solution is not unique.

Then for all $x$ you have : $$ ax^2+bx+c = a\left( x+\frac{b}{2a} \right)^2- \frac{b^2-4ac}{4a} =a\left( x+\frac{b}{2a} \right)^2- \frac{\Delta}{4a} $$ So your condition implies : $$a\left( x+\frac{b}{2a} \right)^2- \frac{\Delta}{4a} \ge 0 \text{ so } a\left( x+\frac{b}{2a} \right)^2\ge \frac{\Delta}{4a} $$ If we take the specific case $x=\frac{-b}{2a} $ we find $0\ge \Delta $. Now remember that any real solution requires that $\Delta \ge 0 $, and conclude that whenever a solution exists, it must be unique - and equals to $\frac{-b}{2a} $.