The equation \((x - 5)^2 = -9\) has no real solutions because the left side, \((x - 5)^2\), represents a squared term, which is always non-negative for real numbers. Squaring any real number results in zero or a positive number, so it cannot equal a negative number like \(-9\).

To confirm, let's analyze this:

1. Expanding or taking the square root of both sides doesn't change the fact that \((x - 5)^2 \geq 0\).
2. Since there’s no real value for \(x\) such that \((x - 5)^2 = -9\), we conclude that there are no real solutions.

However, if we consider complex numbers, we can proceed by taking the square root of both sides:

\[
x - 5 = \pm \sqrt{-9}
\]

Since \(\sqrt{-9} = 3i\) (where \(i\) is the imaginary unit), we have:

\[
x - 5 = \pm 3i
\]

Solving for \(x\), we find:

\[
x = 5 \pm 3i
\]

### Solution
The solutions in the complex number system are:
\[
x = 5 + 3i \quad \text{and} \quad x = 5 - 3i
\]

Would you like further details, or have any questions?

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Here are five related questions to expand on this concept:

1. How do you solve equations involving imaginary numbers?
2. What is the process for expanding and simplifying \((x - 5)^2\)?
3. Why is it impossible for real numbers to have a square root of a negative value?
4. How can complex solutions be graphed in the complex plane?
5. What are the real-world applications of complex numbers?

**Tip:** When you encounter a negative square, consider if complex numbers may be needed for solutions.

Explore the solution to the equation (x - 5)^2 = -9, where no real solutions exist due to the negative result under the square. Learn how to solve for complex solutions using imaginary numbers and understand why the square root of a negative value introduces complex solutions. Suitable for high school students in Grades 10-12.

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