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The problem asks to find a quadratic equation that has the given solutions: \( 2 + 3i \) and \( 2 - 3i \).

These are complex conjugates, so the quadratic equation can be formed using the fact that for any quadratic equation \( ax^2 + bx + c = 0 \), the solutions can be found using the factored form:

\[
(x - (2 + 3i))(x - (2 - 3i)) = 0
\]

Now let's expand this product:

1. First, apply the difference of squares formula:
   \[
   (x - (2 + 3i))(x - (2 - 3i)) = (x - 2 - 3i)(x - 2 + 3i)
   \]
   This is of the form \( (a - b)(a + b) = a^2 - b^2 \), where \( a = x - 2 \) and \( b = 3i \).

2. Using the difference of squares:
   \[
   (x - 2)^2 - (3i)^2
   \]

3. Simplify \( (3i)^2 \):
   \[
   (3i)^2 = -9
   \]

4. So, the equation becomes:
   \[
   (x - 2)^2 + 9 = 0
   \]

5. Now expand \( (x - 2)^2 \):
   \[
   (x - 2)^2 = x^2 - 4x + 4
   \]

6. Substitute this back into the equation:
   \[
   x^2 - 4x + 4 + 9 = 0
   \]

7. Simplify:
   \[
   x^2 - 4x + 13 = 0
   \]

Thus, the quadratic equation is:

\[
x^2 - 4x + 13 = 0
\]

Would you like more details or have any questions about the solution? Here are some related questions:

1. How can we generalize this process for other complex conjugate solutions?
2. What is the discriminant of this quadratic equation, and what does it indicate about the nature of the roots?
3. How would the quadratic equation change if one of the solutions was purely real?
4. Can you use a different method to derive the same quadratic equation?
5. How does the sum and product of the roots relate to the coefficients of the quadratic equation?

**Tip**: When you have complex conjugate roots, their sum and product will always be real, which ensures the quadratic equation has real coefficients.

Determina la ecuación cuadrática que tenga las soluciones dadas: 2 + 3i y 2 - 3i

Discover how to determine a quadratic equation with given complex conjugate solutions, 2 + 3i and 2 - 3i. This problem explores the difference of squares formula and properties of complex numbers, making it ideal for Grades 9-12 students learning advanced algebra.

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