Typos Give Rise To Problems

From Vieta's Formulas, $-a = a + b$ and $b = ab$ . We quickly get that either $b = 0 \Rightarrow a = 0$ or $a = 1 \Rightarrow b= -2$ .

This means that the sum of $a + ab + b$ is $0+0+0+1-2-2 = \boxed{-3.}$

Here is my proof, as well as links to the same problem, with different answers:

Links to the same problem: Just apply Vieta's and Troll Quadratic

Let $f(x)=x^2 + ax + b$ . We have

$\begin{aligned} f(a) &= 2a^2+b &= 0\\ f(b) &= b^2 + ab + b &= 0\\ \Rightarrow b&=-2a^2 \text{ and } b(b+a+1) &= 0\\ \Rightarrow \quad & -2a^2 (-2a^2 + a + 1) &= 0\\ \Rightarrow \quad & 2a^2(a-1)(2a+1) &= 0 \end{aligned}$

Thus, we have 3 solutions for $a$ , which are $a= - \frac {1}{2}, 0, 1$ . From this, we get all solutions are

$(a, b) = (-\dfrac{1}{2}, -\dfrac {1}{2}), (0, 0), (1, -2)$

However, we must have these roots to be integers, so the $a=b=-\frac{1}{2}$ case is nulled, leaving us with an answer of $\boxed{-3}$ .

$x^2 + ax + b \equiv (x-a)(x-b) = x^2 + (-a-b)x+ab$

therefore, $a=-a-b$ and $b=ab$ form these two equation , $(a,b)=(0,0)$ or, $(a,b)=(1,-2)$

$\sum_{}(a+ab+b)=(0+0+0)+(1+1\times(-2)+(-2))=\boxed{-3}$