Question

What is the quadratic equation that has the roots 3+i and 3−i ?(1 point) Responses x2−6x+8=0 x squared minus 6 x plus 8 equals 0 x2+6x+10=0 x squared plus 6 x plus 10 equals 0 x2−6x+10=0 x squared minus 6 x plus 10 equals 0 x2+6x−10=0

Answer 1

To find the quadratic equation that has the given roots \(3+i\) and \(3-i\), we can utilize the fact that if a polynomial has roots \(r_1\) and \(r_2\), it can be expressed in the form:

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

Substituting the roots \(r_1 = 3+i\) and \(r_2 = 3-i\):

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

This can be simplified as follows:

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

Using the difference of squares, we have:

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

Since \(i^2 = -1\), this becomes:

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

Now, expanding \((x - 3)^2\):

\[
= x^2 - 6x + 9 + 1
\]

Thus:

\[
= x^2 - 6x + 10
\]

This means the quadratic equation with roots \(3+i\) and \(3-i\) is:

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

So, the correct response is:

**x²−6x+10=0** (or x squared minus 6 x plus 10 equals 0).