To solve the inequality \(x^2 - 3x > 0\), we follow these steps:

1. **Find the critical points** by solving the equation \(x^2 - 3x = 0\):
   \[
   x(x - 3) = 0
   \]
   So, the critical points are \(x = 0\) and \(x = 3\).

2. **Determine the intervals** to test:
   - \((-\infty, 0)\)
   - \((0, 3)\)
   - \((3, \infty)\)

3. **Test points in each interval** to see where the inequality \(x^2 - 3x > 0\) holds:
   - For \(x \in (-\infty, 0)\), choose \(x = -1\):
     \[
     (-1)^2 - 3(-1) = 1 + 3 = 4 > 0
     \]
     So, \(x^2 - 3x > 0\) in \((-∞, 0)\).

   - For \(x \in (0, 3)\), choose \(x = 1\):
     \[
     1^2 - 3(1) = 1 - 3 = -2 < 0
     \]
     So, \(x^2 - 3x < 0\) in \((0, 3)\).

   - For \(x \in (3, \infty)\), choose \(x = 4\):
     \[
     4^2 - 3(4) = 16 - 12 = 4 > 0
     \]
     So, \(x^2 - 3x > 0\) in \((3, \infty)\).

4. **Combine the intervals** where the inequality holds:
   \[
   x \in (-\infty, 0) \cup (3, \infty)
   \]

Therefore, the solution to the inequality \(x^2 - 3x > 0\) is:
\[
x \in (-\infty, 0) \cup (3, \infty)
\]

Would you like more details or have any questions? Here are some related problems you might find interesting:

1. Solve the inequality \(x^2 + 2x - 3 < 0\).
2. Find the solution set for \(2x^2 - 5x \leq 3\).
3. Determine the critical points of the function \(f(x) = x^2 - 4x + 4\).
4. Solve the inequality \(x^2 - 4 < 0\).
5. Solve \(x^3 - x > 0\).

**Tip:** When solving inequalities, always test points in each interval determined by the critical points to find where the inequality holds true.

Learn how to solve the inequality x^2 - 3x > 0 with a detailed step-by-step solution. This problem covers key algebraic concepts, including quadratic inequalities and critical point analysis. Suitable for students in Grades 10-12.

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