The origin and the roots of the equation `z^2 + pz + q = 0` form an equilateral triangle If -

Options

A. `p^(2)=q`
B. `p^(2)=3q`
C. `p^(2)=3p`
D. `q^(2)=p`

Explanation

To solve the problem, we need to establish the relationship between the coefficients of the quadratic equation \( z^2 + pz + q = 0 \) and the condition that the origin and the roots of the equation form an equilateral triangle.

### Step-by-Step Solution:

1. **Identify the Roots and Origin**:
Let the roots of the quadratic equation be \( z_1 \) and \( z_2 \). The origin is represented by \( z_3 = 0 \). Therefore, the vertices of the equilateral triangle are \( z_1, z_2, \) and \( z_3 \).

2. **Use Vieta's Formulas**:
According to Vieta's formulas for the quadratic equation \( z^2 + pz + q = 0 \):
- The sum of the roots \( z_1 + z_2 = -p \)
- The product of the roots \( z_1 z_2 = q \)

3. **Equilateral Triangle Property**:
For three points \( z_1, z_2, z_3 \) to form an equilateral triangle, the following condition must hold:
\[ z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \]
Substituting \( z_3 = 0 \):
\[ z_1^2 + z_2^2 + 0^2 = z_1 z_2 + z_2 \cdot 0 + 0 \cdot z_1 \]
This simplifies to:
\[ z_1^2 + z_2^2 = z_1 z_2 \]

4. **Express \( z_1^2 + z_2^2 \)**:
We can express \( z_1^2 + z_2^2 \) using the identity:
\[ z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1 z_2 \]
Substituting \( z_1 + z_2 = -p \) and \( z_1 z_2 = q \):
\[ z_1^2 + z_2^2 = (-p)^2 - 2q = p^2 - 2q \]

5. **Set Up the Equation**:
Now we have:
\[ p^2 - 2q = q \]
Rearranging gives:
\[ p^2 - 3q = 0 \]
or
\[ p^2 = 3q \]

6. **Conclusion**:
Therefore, the condition that the origin and the roots of the equation \( z^2 + pz + q = 0 \) form an equilateral triangle is given by:
\[ p^2 = 3q \]

### Final Answer:
The condition is \( p^2 = 3q \).