https://r2.erweima.ai/mathbot/image/1726158072313a79jzc4f.jpg

The equation in the image is:

\[
x \cdot (\sqrt{x} + 2) \cdot (x - 3) \cdot (x + 2) \cdot \left( \frac{1}{x} - 2 \right) = 0
\]

To find the solutions for \(x\), we need to solve each factor of the equation set to zero. Here are the factors:

1. \(x = 0\)
2. \(\sqrt{x} + 2 = 0\)
3. \(x - 3 = 0\)
4. \(x + 2 = 0\)
5. \(\frac{1}{x} - 2 = 0\)

Let's solve each of them one by one:

1. **\(x = 0\)**:
   \[
   x = 0
   \]

2. **\(\sqrt{x} + 2 = 0\)**:
   \[
   \sqrt{x} = -2
   \]
   Since the square root of a number cannot be negative, this equation has no real solutions.

3. **\(x - 3 = 0\)**:
   \[
   x = 3
   \]

4. **\(x + 2 = 0\)**:
   \[
   x = -2
   \]

5. **\(\frac{1}{x} - 2 = 0\)**:
   \[
   \frac{1}{x} = 2 \implies x = \frac{1}{2}
   \]

### Summary of Solutions:
The solutions for the equation are:

\[
x = 0, \, x = 3, \, x = -2, \, x = \frac{1}{2}
\]

Would you like more details or have any questions about this? 

### Related Questions:
1. How do you solve equations involving square roots?
2. What happens when an equation has no real solutions?
3. How do you solve rational equations like \(\frac{1}{x} - 2 = 0\)?
4. What are the conditions for a polynomial equation to have real roots?
5. Can the square root function produce a negative result?

### Tip:
Always check each solution to ensure it does not violate any conditions, such as division by zero or taking the square root of a negative number.

Solve the equation: x * (√x + 2) * (x - 3) * (x + 2) * (1/x - 2) = 0

Step-by-step solution to solve the polynomial equation x * (√x + 2) * (x - 3) * (x + 2) * (1/x - 2) = 0. Covers key algebraic concepts like square roots, rational equations, and the zero product property. Suitable for high school students in Grades 10-12.

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