Proof that a non-negative quadratic equation has at most one solution without using calculus I'm working through Lax and Terrell's 'Calculus with Applications'. One of the first exercises is to work through the proof of the Cauchy-Schwarz inequality (without calculus). I'm all good for my proof except for the first part: i.e., how do I prove that a non-negative quadratic equation has at most one solution?
I can 'explain' this geometrically, i.e., a non-negative quadratic equation can only have a solution where the vertex touches the X axis. However, this doesn't feel very 'rigorous'.
Is there a better, algebraic, proof I could use here?
Thank you!
 A: Suppose that $f(x):= ax^2 + bx + c \geq 0$ for every $x \in \mathbb{R}$ and that there exist $x_1,x_2, x_1 \ne x_2$ s.t. $f(x_1) = f(x_2) = 0$. Then, by polynomial long division, $f(x) = (x-x_1)q(x) + r(x)$ where $r(x)$ has degree $0$ and $q(x)$ has degree $1$. But then $0= 0f(x_1) = 0q(x) + r(x_1) = r(x_1)$. So $r(x_1) = 0$, but $r(x) = d$ for some $d \in \mathbb{R}$ (since it has degree $0$), so $d = 0$. Thus $f(x) = (x-x_1)q(x)$. Then $q(x) = ux-v$ for some $u,v \in \mathbb{R}$. Also $0=f(x_2)= (x_2 - x_1)q(x_2)$. Then $x_2 - x_1 \ne 0$, by assumption, so $q(x_2) = 0$. Thus $ux_2 - v= 0$, or $v = ux_2$. But then
$$f(x) = (x-x_1)(ux+v) = ux^2 +(v-ux_1)x -x_1v = ax^2 + bx+c,$$
so $u=a$, so  $v= ax_2$. Thus, putting it together,
$$f(x) = (x-x_1)(ax-ax_2) = a(x-x_1)(x-x_2).$$
Now let $x_3$ between $x_1$ and $x_2$ (we can without loss of generality suppose $x_1 <x_2$, so $x_1 < x_3 < x_2$). Then $x_3 - x_1 > 0$ and $x_3 - x_2 <0$, so $k:= (x_3-x_1)(x_3-x_2)<0$. Thus
$$f(x_3) = a(x_3 -x_1)(x_3-x_2) = ak$$
so $f(x_3)$ has the opposite sign of $a$ (since $k$ is strictly negative). Then by assumption, for all $x$, $f(x) \geq 0$, so $f(x_3)\geq 0$. Thus $a \leq 0$. Then let $x_4 >x_1, x_2$. Then $x_4-x_1, x_4 - x_2>0$, so $k' := (x_4-x_1)(x_4-x_2) >0$. So
$$f(x_4) = a(x_4-x_1)(x_4 -x_2) = ak'.$$
Then $a \leq 0$ and $k' \geq 0$, so $f(x_4) \leq 0$. But also, by assumption $f(x_4) \geq 0$. Thus $f(x_4) =0$. But $k' \ne 0$, so $a = 0$.
Thus
$$f(x) = bx +c,$$
but it is easy to see that this is nonnegative for all $x$ iff $b =0$. Thus $f(x) = c$, but $f(x_1) =0$, so $0 = f(x_1) = c$, so $f(x) = 0$. But we assumed that $f$ has only two roots, but $f(x) = 0$ has infinitely many roots, so we have a contradiction. Thus a nonnegative quadratic cannot have two roots. But any quadratic also has at most two roots, so a nonnegative quadratic has at most one root.
A: Consider $Ax^2+Bx+C\ge 0$ for all $x,$ with $A\ne 0.$

*

*Take $x_0\ne 0$ where $|x_0|$ is large enough that $|B/Ax_0|<1/2$ and $|C/Ax_0^2|<1/2.$ Then $1+B/Ax_0+C/Ax_0^2\ge 1-|B/Ax_0|-|C/Ax_0^2|>0.$ Therefore $0\le Ax_0^2+Bx_0+C= Ax_0^2(1+B/Ax_0 +C/Ax_0^2)\implies Ax_0^2\ge 0\implies$ $\implies A\ge 0\implies A>0.$


*Let $x_1=-B/2A$. Then $0\le Ax_1^2+Bx_1+C=A(x_1+B/2A)^2 +(4AC-B^2)/4A=(4AC-B^2)/4A,$ and since $A>0,$ therefore $0\le 4AC-B^2.$


*The Quadratic Formula:   $Ax^2+Bx+C =0\iff x=\frac {-B\pm \sqrt {B^2-4AC}}{2A}.$ But $B^2-4AC\le 0 $ so $\sqrt {B^2-4AC} \not\in \Bbb R$ unless $B^2-4AC=0.$ So IF there exists $x_2$ with $Ax_2^2+Bx_2+C= 0$ THEN $B^2-4AC=0$ and $x_2=\frac {-B\pm \sqrt {B^2-4AC}}{2A}=\frac {-B\pm \sqrt {0}}{2A}=\frac {-B}{2A}.$
BTW. By the methods of 1., if $p$ is a non-zero polynomial and $p(x)\ge 0$ for all $x$ then deg($p$) is even and the leading co-efficient of $p$ is positive.
