If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational? If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational ?
Is it a "If and only if " statement or "only if " statement ?
For $a,b,c \in \mathbb Q$ , I think it is a "if and only if" statement . Am I correct ?
I can prove that if $a+b+c=0$ and $a,b,c \in \mathbb Q$ , then roots are rational.
But I can not prove that if roots are rational and $a,b,c \in \mathbb Q$ , then $a+b+c=0$.
Any help ? or any conditions that $ax^2 +bx+c =0 $
 should satisfy in order to have rational roots ?
My clarifications :
1) What is a rational number ? It is a number which can be written in the form $\frac{p}{q} $ , where $q \neq 0$ and $p , q \in \mathbb Z$
2)
In this case : Let's take the quadratic equation $ax^2+bx+c=0$ where $a \neq 0$.
We all know that the roots are given by , $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
Case 1 : Suppose that $a,b,c \in \mathbb Q$.
Then $x$ is rational if and only if $b^2-4ac$ is a perfect square or zero  $(0,1,4,9,16,...)$.
Now we need to make $b^2-4ac$ a perfect square !
Now observe that if $a+b+c=0$ , then  $b^2-4ac = (-a-c)^2-4ac=(a-c)^2$
That is if $a,b,c \in \mathbb Q$ and $a+b+c=0$ then the solutions are rational.
Case 2 : Suppose that $a,b,c \in \mathbb Q$.
If $c=0$ , then all the roots are rational. (This is easy if all $a,b,c$ are rationals)
Case 3 : Suppose that $b,c \in \mathbb R- \mathbb Q$.(If $a$ is irrational we can divide by $a$ )
$a+b+c=0$ condition does not satisfy. 
Ex : $(1-\sqrt{2})x^2-2x+(1+\sqrt{2})=0$

If $c = -a-b$, then the discriminant is $$b^2-4ac = b^2 +4a(a+b) = b^2 +4ab +4a^2 = (b+2a)^2.$$
Since the discriminant is a perfect square, then the roots are always rational.
Their values are
$$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
   =\dfrac{-b \pm (b+2a)}{2a} 
   \in \left\{ 1, -\dfrac{a+b}{a} \right\}$$
Of course, now that you know this, you can say that
$$(x-1)(ax+a+b) = ax^2 + bx -(a+b) = ax^2+ bx + c$$
