Write the quadratic equation with a lead coefficient of 1 whose roots are 3+2i and 3-2i

To write a quadratic equation with roots \(3 + 2i\) and \(3 - 2i\), we can use the fact that if \(a\) and \(b\) are roots of a quadratic equation, then the equation can be written as:

\[
(x - a)(x - b) = 0
\]

Given roots \(3 + 2i\) and \(3 - 2i\), we can substitute them as \(a\) and \(b\):

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

Expanding this expression:

1. Rewrite the terms as a difference of squares:

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

2. Use the identity \((A - B)(A + B) = A^2 - B^2\), where \(A = x - 3\) and \(B = 2i\):

   \[
= (x - 3)^2 - (2i)^2
\]

3. Now, expand \((x - 3)^2\) and \((2i)^2\):

   \[
= (x^2 - 6x + 9) - (4i^2)
\]

4. Since \(i^2 = -1\), \(4i^2 = 4(-1) = -4\):

   \[
= x^2 - 6x + 9 + 4
\]

5. Simplify:

   \[
= x^2 - 6x + 13
\]

Therefore, the quadratic equation with roots \(3 + 2i\) and \(3 - 2i\) is:

\[
x^2 - 6x + 13 = 0
\]

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

Here are some related questions to explore:
1. How can complex roots be identified in a quadratic equation?
2. What is the relationship between the coefficients and the roots of a quadratic equation?
3. Can a quadratic equation have more than two complex roots?
4. How do we derive the discriminant for equations with complex roots?
5. What real-world applications involve quadratic equations with complex roots?

**Tip:** For a quadratic equation with complex roots, the imaginary parts of the roots will always cancel each other out, leading to a quadratic equation with real coefficients.

Learn how to derive the quadratic equation with complex roots 3+2i and 3-2i. This step-by-step guide demonstrates the use of the difference of squares and complex conjugates to form a quadratic equation with real coefficients. Suitable for students in Grades 9-11.

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