When we know the roots are r and s we can just multiply (x-r)(x-s) and set it to zero to recover the original quadratic equation. We can then scale to get rid of fractions or common factors if we like.
[tex]0 = (x - (-1 + i))(x - (-1 -i)) = x^2 - ( (-1+i)+(-1-i) ) x + (-1 + i)(-1 -i)[/tex]
[tex]0 = x^2 + 2x + 2[/tex]
The linear term coefficient is negative the sum of the roots, which in the case of conjugates is (negative) twice the real part, coefficient +2 here. The constant term is the product of the roots, which in the case of conjugates is the squared magnitude, here [tex]1^2+1^2[/tex].
We got a 2x in the middle so we get
Answer: 2x
