Find the necessary and sufficient conditions on $a$, $b$ so that $ax^2 + b = 0$ has a real solution. This question is really confusing me, and I'd love some help but not the answer. :D
Is it asking: What values of $a$ and $b$ result in a real solution for the equation $ax^2 + b = 0$? $a = b = 0$ would obviously work, but how does $x$ come into play? There'd be infinitely many solutions if $x$ can vary as well ($a = 1$, $x = 1$, $b = -1$, etc.).
I understand how necessary and sufficient conditions work in general, but how would it apply here? I know it takes the form of "If $p$ then $q$" but I don't see how I could apply that to the question. Is "If $ax^2 + b = 0$ has a real solution, then $a$ and $b =$ ..." it?

If $ax^2 + b=0$, then $x^2=\dfrac{-b}{a}$.  The solutions are therefore $x=\pm\sqrt{-b/a}$.  For the equation to have a real solution, $-b/a$ must be non-negative.  A fraction is non-negative if


*

*the numerator and denominator are positive, or

*the numerator and denominator are negative, or

*the numerator is $0$ and the denominator is not $0$.

