Which quadratic has `2+3i and 2-3i` as its solutions?

Options

A. `x^(2)+4x-13=0`
B. `x^(2)-4x+13=0`
C. `x^(2)+13x-4=0`
D. `x^(2)-13x-4=0`

Answer

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

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

### Step 2: Use the factored form of a quadratic
The general form of a quadratic equation with roots \(a\) and \(b\) is given by:
\[ (x - a)(x - b) = 0 \]
In our case, the roots are \(2 + 3i\) and \(2 - 3i\). Therefore, we can write:
\[ (x - (2 + 3i))(x - (2 - 3i)) = 0 \]

### Step 3: Simplify the expression
We can rewrite the expression:
\[ (x - 2 - 3i)(x - 2 + 3i) = 0 \]
This is a difference of squares, which can be expressed as:
\[ ((x - 2) - 3i)((x - 2) + 3i) = (x - 2)^2 - (3i)^2 \]

### Step 4: Calculate \((x - 2)^2\) and \((3i)^2\)
First, we expand \((x - 2)^2\):
\[ (x - 2)^2 = x^2 - 4x + 4 \]
Next, we calculate \((3i)^2\):
\[ (3i)^2 = 9i^2 = 9(-1) = -9 \]

### Step 5: Substitute back into the equation
Now we substitute back into our expression:
\[ (x - 2)^2 - (3i)^2 = (x^2 - 4x + 4) - (-9) \]
This simplifies to:
\[ x^2 - 4x + 4 + 9 = x^2 - 4x + 13 \]

### Step 6: Write the final quadratic equation
Thus, the quadratic equation with roots \(2 + 3i\) and \(2 - 3i\) is:
\[ x^2 - 4x + 13 = 0 \]

### Conclusion
The quadratic equation that has \(2 + 3i\) and \(2 - 3i\) as its solutions is:
\[ \boxed{x^2 - 4x + 13 = 0} \]