<s>[INST] Solutions to a quadratic given 1 solution in form a+bi I was just really confused as to how I only ended up with 1 of 2 answers for the following question.
Given that $-2+bi$ is a solution of $x^2+ax+(3+a) $ find constants $a$ and $b$ given that they are real.
As soon as I saw that $-2+bi$ was a solution, I immediately jumped to $ -2-bi$  must also be a solution, by the fundamental theorem of algebra. By doing the sum and product of the solution a quadratic could be obtained
Sum $ (-2+bi)+(-2-bi)=-4$
Product $(-2-bi)(-2+bi)=(-2)^2-(-bi)^2=4+b^2$
Thus the quadratic $x^2+4x+(4+b^2)$ is obtained 
Equation both sides of the equation 
$\\x^2+4x+(4+b^2)=x^2+ax+(3+a)$ 
$a=4$
Therefore$ b={\sqrt 3}$   or $ b={-\sqrt 3}$ 
However, the solutions seem to suggest that an extra solution can be
$ b={0}, a=7$ 
Did I eliminate a solution by doing the sum and product of the solutions to find the quadratic? Or does it have to do with my working process
 [/INST] Let the second root be $c-ib$ (because $2+ib+c-ib=a$ must be real). Then
$$(x+2-ib)(x-c+ib)=x^2+(2-c)x-2c+b^2+i(bc+2b)=x^2+ax+a+3.$$
For the imaginary term to vanish, $$b=0\lor c=-2$$
gives us the three solutions
$$a=7,b=0,c=-5$$ and $$a=4,b=\pm\sqrt3,c=-2.$$
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