Yes; we claim that the root $x=-1$ must always be achieved. In the first quadratic, evaluating $x$ at $-1$ gives $11$ and in the second quadratic, evaluating $x$ at $-1$ gives $-9$. Therefore, somewhere in the journey, evaluating $x$ at $-1$ would give $0$, implying integer roots. This makes sense; if the quadratic was in the form $(x+1)(x+(1-a))$, $a \in \mathbb{Z}$, then it would be equivalent to $x^2+(2-a)x+(1-a)$. This is the same thing as saying that the linear term would have to be exactly one more than the constant term. This is obviously true, since at some instance in our journey, we will have a quadratic in the form $x^2+bx+b$, $b \in \mathbb{Z}$. We have two cases. In the first one, we increase the linear term by $1$, giving the above form. In the second one, we decrease the constant term by $1$, which also gives us the above form. Either of these cases must happen at least once because to get from $10x$ to $20x$, we need to add to the linear term, and to get from $20$ to $10$, we need to decrease the constant term.