If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$  
3) $a+b+c=0$ 
4) $c=0$ 
Update
I just tried to sub $c$ into both of the equations:
$c^2+cb+a=0$ and $c^2+ac+b=0$ which then gives us the equality
$c^2+cb+a=c^2+ac+b$
$ => cb+a=ac+b$
$=> b(c-1)=a(c-1)$
which gives me then a=b which is contradictory because the integers are supposed to be distinct.
Update #2
Ok it looks like 1) is true, 2) is true, 3) is true, and 4) is false .. right?

Express $$x^2+bx+a=(x-c)(x-d)$$
$$x^2+ax+b=(x-c)(x-f)$$ 
Matching the coefficients of the powers of $x$ for both equations, we have
$$c+d=-b$$
$$c+f=-a$$
$$cd=a$$
$$cf=b$$
Solving for $d$ and $f$ in terms of $a$ and $b$ we have
$$d=a,f=b$$
leading to $c$ having only one possible value, which is $1$.
Thus we have $$c+a=-b\Rightarrow a+b+c=0$$
as well as
$$a+b+1=0\Rightarrow a+b = -1$$
Also, $$c(a+c)=a+1=-b$$
Thus, options (1), (2) and (3) are true.
