The given quadratic equation is:

\[
x^2 - 3px + 2p^2 = 0
\]

To solve this quadratic equation for \(x\), we can use the **quadratic formula**:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where the equation is in the form \(ax^2 + bx + c = 0\). In our case:

- \(a = 1\)
- \(b = -3p\)
- \(c = 2p^2\)

Substitute these values into the quadratic formula:

\[
x = \frac{-(-3p) \pm \sqrt{(-3p)^2 - 4(1)(2p^2)}}{2(1)}
\]

Simplifying the terms:

\[
x = \frac{3p \pm \sqrt{9p^2 - 8p^2}}{2}
\]

\[
x = \frac{3p \pm \sqrt{p^2}}{2}
\]

\[
x = \frac{3p \pm p}{2}
\]

Now we have two possible solutions:

1. \( x = \frac{3p + p}{2} = \frac{4p}{2} = 2p \)
2. \( x = \frac{3p - p}{2} = \frac{2p}{2} = p \)

Thus, the two solutions are:

\[
x = 2p \quad \text{or} \quad x = p
\]

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you derive the quadratic formula?
2. What are the conditions for a quadratic equation to have real roots?
3. How does the discriminant affect the type of roots in a quadratic equation?
4. Can a quadratic equation have two equal roots? If so, under what conditions?
5. How does the value of \(p\) affect the solutions of the equation?

**Tip:** Always check the discriminant (\(b^2 - 4ac\)) before solving a quadratic equation to know the nature of the roots (real, equal, or complex).

Learn how to solve the quadratic equation x^2 − 3px + 2p^2 = 0 step-by-step. This algebra problem utilizes the quadratic formula and explores key concepts in quadratic equations. Suitable for students in Grades 9-10.

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