The sum of all real values of $z$ satisfying the equation $(z^2 + 5z + 5)^{z^2+4z-60} = 1$ is
Step-by-Step Solution
Key Concept: For a cubic polynomial to be non-positive, we analyze conditions from evaluations at specific points and the discriminant to find the feasible range.
We need $f(a) \leq 0$. From the condition $f(2) \leq 0$, we get $a^2 + 7a + 1 \leq 0$, giving $\frac{-7-3\sqrt{5}}{2} \leq a \leq \frac{-7+3\sqrt{5}}{2}$. From $f(2) \leq 0$, we get $a^2 + 8a + 4 \leq 0$, or $-4-2\sqrt{3} \leq a \leq -4+2\sqrt{3}$. For $D > 0$: $a^2 - 4(a^2 + 6a) > 0$ gives $a^2 + 8a < 0$, or $-8 < a < 0$. Combining all three conditions: $\frac{-7-3\sqrt{5}}{2} \leq a \leq -4+2\sqrt{3}$. The integral values in this range are $-6, -5, -4, -3, -2, -1$.
Correct Answer: -6, -5, -4, -3, -2, -1