Find the quadratic equation whose roots are 2+3i, 2-3i.

Answer

To find the quadratic equation whose roots are \(2 + 3i\) and \(2 - 3i\), we can follow these steps:

### Step 1: Identify the roots
The roots of the quadratic equation are given as:
- \(r_1 = 2 + 3i\)
- \(r_2 = 2 - 3i\)

### Step 2: Calculate the sum of the roots
The sum of the roots \(S\) can be calculated as:
\[ S = r_1 + r_2 = (2 + 3i) + (2 - 3i) \]
When we simplify this:
\[ S = 2 + 3i + 2 - 3i = 4 \]

### Step 3: Calculate the product of the roots
The product of the roots \(P\) can be calculated as:
\[ P = r_1 \times r_2 = (2 + 3i)(2 - 3i) \]
This expression can be simplified using the difference of squares formula:
\[ P = 2^2 - (3i)^2 = 4 - 9i^2 \]
Since \(i^2 = -1\), we substitute:
\[ P = 4 - 9(-1) = 4 + 9 = 13 \]

### Step 4: Form the quadratic equation
The standard form of a quadratic equation with roots \(r_1\) and \(r_2\) is given by:
\[ x^2 - Sx + P = 0 \]
Substituting the values of \(S\) and \(P\):
\[ x^2 - 4x + 13 = 0 \]

### Final Answer
Thus, the quadratic equation whose roots are \(2 + 3i\) and \(2 - 3i\) is:
\[ \boxed{x^2 - 4x + 13 = 0} \]