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

A brute force approach is to substitute $x=-2+bi$ into $x^2+ax+(3+a)$. Then
\begin{align}
x^2+ax+(3+a)
&= (-2+bi)^2+a(-2+bi)+(3+a) \\ 
&=(7-a-b^2)+b(a-4)i \\
\end{align}
where I have organized the last expression into real and imaginary parts (since $a,b$ are assumed real). If this last expression is to vanish as desired, then both real and imaginary parts are zero. From the latter, we deduce that $b=0$ or $a=4$. If $b=0$, then the condition $7-a-b^2=0$ yields $a=7$. If $a=4$, then this same condition implies $b^2=7-a=3$ and therefore $b=\pm \sqrt{3}$. This gives all three solutions.
