Solving the equation $X^2 + (3 +i)X + 1 +i =0$ I have to find the roots of the equation
$$X^2 + (3 +i)X + 1 +i =0.$$
The first step is to find the discrimant which is $4 + 2i$. Then, I assume that the square root of the discriminant is in the form of $a+bi$, so we have to solve the below system
$$a^2-b^2=4\qquad\text{ and }\qquad ab=1.$$
How can I solve the above system ? Please do not provide another solution (e.g using trigonometric form)

You're work so far is good, and indeed you want to find $a,b\in\Bbb{R}$ such that $(a+bi)^2=4+2i$, i.e.
$$a^2-b^2=4\quad\text{ and }\qquad 2ab=2.$$
The latter shows you that $a$ and $b$ are nonzero, and that $b=a^{-1}$. Plugging this into the former yields
$$a^2-a^{-2}=4,$$
and multiplying everything by $a^2$ and rearranging shows that
$$a^4-4a^2-1=0.$$
This is a quadratic in $a^2$, so by the quadratic formula $a^2=2\pm\sqrt{5}$. Because $a$ is real
$$a=\pm\sqrt{2+\sqrt{5}}.$$
