Question

What is the quadratic equation that has the roots 3+i and 3−i?

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 in the form:

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

Substituting the given roots:

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

Now, we can simplify this expression. First, we rewrite it as:

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

Next, we can recognize this as a difference of squares:

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

Since \(i^2 = -1\), we have:

\[
= (x - 3)^2 - (-1) = (x - 3)^2 + 1
\]

Now we expand \((x - 3)^2\):

\[
(x - 3)^2 = x^2 - 6x + 9
\]

Thus,

\[
(x - 3)^2 + 1 = x^2 - 6x + 9 + 1 = x^2 - 6x + 10
\]

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

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