For how many values of $n$ in the range $n \in [5, 100]$ is $D$ a perfect square, where $D$ is the discriminant of $x^2 + 2x - n = 0$?
Step-by-Step Solution
Key Concept: The discriminant of a quadratic can be analyzed by setting it equal to a perfect square and solving for the parameter.
For the quadratic $x^2 + 2x - n = 0$, the discriminant is $D = 4 + 4n = 4(1 + n)$. For $D$ to be a perfect square, $1 + n$ must be a perfect square. Let $1 + n = k^2$ for some positive integer $k$, so $n = k^2 - 1$. With $n \in [5, 100]$, we have $5 \leq k^2 - 1 \leq 100$, giving $6 \leq k^2 \leq 101$. This yields $k \in \{3, 4, 5, 6, 7, 8, 9, 10\}$ (since $\sqrt{6} \approx 2.45$ and $\sqrt{101} \approx 10.05$). The corresponding values are $n \in \{8, 15, 24, 35, 48, 63, 80, 99\}$, giving 8 values.
Correct Answer: 8