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?
 A: Hint: Mechanically "solve" the equation. What could go wrong?
As to what the question is asking, suppose that we are given fixed numbers $a$ and $b$. For example, let $a=-3$ and $b=17$. Does the equation have a real solution $x$? What about if $a=-3$ and $b=-5$? (And don't forget about the possibility that one or both of $a$ and $b$ might be $0$.) Can we come up with simple conditions on $a$ and $b$ so that just glancing at them tells us whether or not there is a solution?
A: I assume the question is "find conditions that are necessary and sufficient to guarantee solutions" rather than "find necessary conditions and also find sufficient sufficient conditions for a solution." If the former is the case, then you're asked for constraints on $a$ and $b$ such that (1) if the conditions are met then $ax^2+b$ is zero for some $x$ and (2) if the conditions are not met then $ax^2+b$ isn't zero for any $x$.
So, when will $ax^2+b$ have some value $x$ for which the expression is zero? Well, as André suggested, try to solve $ax^2+b=0$ "mechanically". By subtracting we have the equivalent equation $ax^2 = -b$ and we'd like to divide by $a$ to get $x^2=-b/a$. Unfortunately, we can't do that if $a=0$, so we need to consider two cases


*

*$a=0$ 

*$a\ne 0$


In the first case, our equation becomes $0\cdot x^2+b=0$, namely $b=0$ and if $b=0$ (and, of course $a=0$) then any $x$ will satisfy this. In other words, there's a solution (actually infinitely many solutions) if $a=b=0$, as you've already noted.
Now, in case (2) we can safely divide by $a$ to get $x^2=-b/a$. When does this have a solution? We know that $x^2\ge 0$ no matter what $x$ is, so when will our new equation have a solution? You said you don't want the full answer, so I'll denote the answer you discover by $C$, which will be of the form "some condition on $a\text{ and }b$". Once you've done that, your full answer will be:

$ax^2+b=0$ has a solution if and only if
  
  
*
  
*$a=b=0$, or
  
*Your condition $C$.
  

These are sufficient, since either guarantees a solution, and necessary, since if neither is true, then there won't be a solution (since we exhausted all possibilities for solutions).
A: 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 $0$.

