Let $\alpha$ and $\beta$ be the real roots of the equation $x^2 - x(k - 2) + \left(k^2 + 3k + 5\right) = 0$. The maximum value of $\alpha^2 + \beta^2$ is:
Step-by-Step Solution
Key Concept: The maximum of a sum of squares subject to discriminant constraints occurs at the boundary where the discriminant condition is tight.
From the discriminant condition $D \geq 0$, we find $K \in [-4, -\frac{4}{3}]$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab = 19 - (K+5)^2$, the maximum occurs when $(K+5)^2$ is minimized, giving maximum $a^2 + b^2 = 19 - (-4+5)^2 = 18$.
Correct Answer: 1