To find the quadratic equation with the roots \(3+i\) and \(3-i\), we can use the fact that if a quadratic equation has roots \(r_1\) and \(r_2\), it can be expressed as:
\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]
In this case, the roots are \(r_1 = 3+i\) and \(r_2 = 3-i\).
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Sum of the roots: \[ r_1 + r_2 = (3+i) + (3-i) = 3 + 3 + i - i = 6 \]
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Product of the roots: \[ r_1 \cdot r_2 = (3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 9 + 1 = 10 \]
Now we can substitute these values back into the quadratic equation format:
\[ x^2 - (6)x + 10 = 0 \]
Thus, the quadratic equation is:
\[ x^2 - 6x + 10 = 0 \]
So the correct response is:
x² - 6x + 10 = 0.