https://file.aiquickdraw.com/mathbot/file/1726113747603byfqplvs.jpg

The given equation is:

\[
(x + 6)^2 + (y + 1)^2 = 0
\]

This is a sum of two squared terms set equal to zero.

### Key Insights:
- Both terms \((x + 6)^2\) and \((y + 1)^2\) represent squares of real numbers.
- The square of any real number is **non-negative** (i.e., \(\geq 0\)).
- For the sum of two squares to be equal to zero, **each individual square must also be zero**. This means:

\[
(x + 6)^2 = 0 \quad \text{and} \quad (y + 1)^2 = 0
\]

### Solving for \(x\) and \(y\):
- \((x + 6)^2 = 0\) implies \(x + 6 = 0\), so \(x = -6\).
- \((y + 1)^2 = 0\) implies \(y + 1 = 0\), so \(y = -1\).

### Solution Set:
The only solution is the point \((x, y) = (-6, -1)\).

### Final Answer:
\[
\{(-6, -1)\}
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How does the equation of a circle relate to this problem?
2. What happens if the equation had a different constant on the right-hand side, like \( (x + 6)^2 + (y + 1)^2 = r^2 \)?
3. Can the sum of squares of real numbers ever be negative?
4. How can this problem be visualized geometrically?
5. How would the solution change if the exponents were odd rather than even?

**Tip:** Whenever you see an equation involving squared terms equal to zero, it's a strong indication that each term individually must be zero.

Learn how to solve the equation (x + 6)^2 + (y + 1)^2 = 0. Understand key algebraic concepts, including the non-negative property of squares and the geometric interpretation of circles. Suitable for high school students.

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