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).