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The given quadratic equation is:

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

We are told that the equation has equal roots, and \( p \) is a non-zero constant. To find the value of \( p \), we will use the condition for equal roots in a quadratic equation.

### Step 1: Discriminant Condition for Equal Roots
For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant is given by:

\[
\Delta = b^2 - 4ac
\]

If the equation has equal roots, the discriminant must be zero, i.e.,

\[
\Delta = 0
\]

For the equation \( x^2 + 3px + p = 0 \), we have:
- \( a = 1 \)
- \( b = 3p \)
- \( c = p \)

### Step 2: Compute the Discriminant
The discriminant is:

\[
\Delta = (3p)^2 - 4(1)(p)
\]

Simplifying this expression:

\[
\Delta = 9p^2 - 4p
\]

### Step 3: Set the Discriminant to Zero
Since the equation has equal roots, we set \( \Delta = 0 \):

\[
9p^2 - 4p = 0
\]

### Step 4: Solve for \( p \)
We can factor this equation as:

\[
p(9p - 4) = 0
\]

This gives two possible solutions:

1. \( p = 0 \)
2. \( 9p - 4 = 0 \), which simplifies to \( p = \frac{4}{9} \)

However, the problem specifies that \( p \) is non-zero, so we discard \( p = 0 \). Thus, the solution is:

\[
p = \frac{4}{9}
\]

The value of \( p \) is \( \frac{4}{9} \).

Would you like more details, or do you have any other questions? 

Here are 5 related questions:
1. What are the conditions for distinct roots of a quadratic equation?
2. How does the discriminant help classify the nature of roots?
3. What is the general form of a quadratic equation?
4. How can you factor quadratic equations like the one solved here?
5. What does it mean when a quadratic equation has complex roots?

**Tip:** Always check for special conditions (like non-zero values) when solving quadratic equations!

The equation x^2 + 3px + p = 0, where p is a non-zero constant, has equal roots. Find the value of p.

Learn how to solve the quadratic equation x^2 + 3px + p = 0, where p is a non-zero constant, using the discriminant condition for equal roots. This problem covers key algebraic concepts and is suitable for Grades 9-11 students.

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