Question

What is the quadratic equation that has the roots 3+i and 3-i? (1 point)

Answer 1

To find the quadratic equation with roots \(3+i\) and \(3-i\), we can use the fact that if \(r_1\) and \(r_2\) are the roots of a quadratic equation, then the equation can be expressed as:

\[
(x - r_1)(x - r_2) = 0
\]

In this case, \(r_1 = 3 + i\) and \(r_2 = 3 - i\). Therefore, the equation becomes:

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

Now, we simplify this product:

1. First, we can rewrite the terms:
\[
(x - (3+i)) = (x - 3 - i) \quad \text{and} \quad (x - (3-i)) = (x - 3 + i)
\]

2. Now, we can multiply the two expressions:
\[
(x - 3 - i)(x - 3 + i)
\]

This is a difference of squares, which will simplify to:
\[
[(x - 3)^2 - (i)^2]
\]

3. We know that \(i^2 = -1\), so:
\[
(x - 3)^2 - (-1) = (x - 3)^2 + 1
\]

4. Next, we expand \((x - 3)^2\):
\[
(x - 3)^2 = x^2 - 6x + 9
\]

So our equation becomes:
\[
x^2 - 6x + 9 + 1 = x^2 - 6x + 10
\]

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

\[
\boxed{x^2 - 6x + 10 = 0}
\]