To solve this problem, we need to determine the domain of the function $f(x) = \frac{x^2 + 2x}{\sqrt{-2x - 6}}$.

Step 1: Analyze the Square Root Expression

The square root in the denominator, $\sqrt{-2x - 6}$, must be defined. This requires that the expression inside the square root must be non-negative:

$-2x - 6 \geq 0$

Step 2: Solve the Inequality

To find the values of $x$ that satisfy this inequality, we solve it as follows:

$-2x \geq 6$ $x \leq -3$

Step 3: Consider the Denominator

Since the square root is in the denominator, it cannot be zero. Therefore, we also need to ensure:

$\sqrt{-2x - 6} \neq 0$

This occurs when $-2x - 6 = 0$, which simplifies to:

$x = -3$

Since $x = -3$ would make the denominator zero, $x = -3$ must be excluded from the domain.

Final Domain

The domain of $f(x)$ consists of all $x$ such that:

$x < -3$

Conclusion

The correct answer is:

$\boxed{A}$

This corresponds to the domain $\{x \in \mathbb{R}, x < -3\}$.

Do you have any questions, or would you like further details? Here are some related questions:

1. How do you solve inequalities involving square roots?
2. What happens to a function if the denominator is zero?
3. Can a square root expression in the denominator ever be negative?
4. What is the difference between domain and range in a function?
5. How does excluding certain values from the domain affect the function?

Tip: Always check both the numerator and denominator when determining the domain of a function.