If the inequality $x^2 + ax + a^2 + 6a < 0$ is satisfied for all $x \in (1,2)$, then the sum of all the integral values of $a$ must be equal to
Step-by-Step Solution
Key Concept: Combine conditions from function values and the discriminant to find the feasible range for parameter $a$, then identify and sum the integers in that range.
We evaluate $f(1) < 0 \Rightarrow a^2 + 7a + 1 < 0$, which gives $a \in \left(\frac{-7-3\sqrt{5}}{2}, \frac{-7+3\sqrt{5}}{2}\right)$. We evaluate $f(2) < 0 \Rightarrow a^2 + 8a + 4 < 0$, which gives $a \in \left(-4-2\sqrt{3}, -4+2\sqrt{3}\right)$. The discriminant condition $D > 0$ gives $a^2 - 4a > 0$, so $a \in (-\infty, 0) \cup (4, \infty)$. Taking the intersection of all three conditions yields $a \in \left(\frac{-7-3\sqrt{5}}{2}, -4+2\sqrt{3}\right)$. The integral values of $a$ in this interval are $\{-6, -5, -4, -3, -2, -1\}$, giving a sum of $-21$.
Correct Answer: 2