Show that $Ax^2+Bx+C>0 $ for all real $x$ if and only if $A>0$ and $B^2-4AC<0$.

Question

Show that

For all real $x$ , $Ax^2+Bx+C>0 $ if and only if $A>0$ and $B^2-4AC<0$.

Case 1 :

Suppose that $A>0$ and $B^2-4AC<0$.

Let
$y=Ax^2+Bx+C$.

Then $Ax^2+Bx+(C-y)=0$

Since $x \in \mathbb R $ , $$B^2 -4A(C-y) \geq 0$$

$$B^2 -4AC+4Ay \geq 0$$

$$y \geq \frac{4AC-B^2 }{4A}$$

Since $ 4AC-B^2>0 $ and $A>0$ , $$y \geq Some Positive Value$$

Thus $y>0$. DONE

Case 2: ?

How can I prove this side ?

Any help ?

Answer 1

For $A\neq0$ because $$Ax^2+Bx+C=A\left(x^2+\frac{B}{A}x+\frac{C}{A}\right)=A\left(x+\frac{B}{2A}\right)^2-\frac{B^2-4AC}{4A}.$$ Indeed, for $A<0$ we see that $Ax^2+Bx+x=x^2\left(A+\frac{B}{x}+\frac{C}{x^2}\right)\rightarrow-\infty$ for $x\rightarrow\infty$.

Thus, we need $A>0$ and from here we need $B^2-4AC<0$,

otherwise, $x=-\frac{B}{2A}$ will get a counterexample.

Also, there is the case $A=B=0$ and $C>0$.

Answer 2

Let $p(x)=Ax^2+Bx+C$. First multiply by $4A$ noting that if $A\lt 0$ this changes the sign.

We have $$4Ap(x)=4A^2x^2+4ABx+4AC=(2Ax+B)^2+(4AC-B^2)\ge 4AC-B^2$$

If $A\neq 0$ there is equality when $x=-\frac B{2A}$.

If $A$ is positive then $4Ap(x)$ is positive for all values of $x$ precisely when the minimum value $4AC-B^2\gt 0$, and this is the condition for $p(x)$ too.

If $A$ is negative, we can make $4Ap(x)$ as large and positive as we like since $4AC-B^2$ is constant and $(2Ax+B)^2$ is unbounded above [I will leave you to make this argument as precise and tight as you need it], so $p(x)$ is negative for large enough values of $|x|$.

I am assuming that $A\neq 0$ is implied in the question, but this case is easier to deal with.