If $\alpha$ and $\beta$ are algebraic integers then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer. An algebraic integer is a complex number that is a root of a monic polynomial with coefficients in $\mathbb{Z}$. 

Let $\alpha$ and $\beta$ be algebraic integers. Then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.

What we have so far is that $2x+\alpha$ is an algebraic integer but since the set of algebraic integers is a ring I can't divide by 2 so I can't have an algebraic integer and also the polynomial wouldn't be monic. 
I appreciate the help.
 A: If I well remember, there's a characterization of integral elements over a ring R (in our case $R=\mathbb{Z}$), in particular $x$ is integral over $R$ if  and only if it is contained in a ring $ C$ such that $R \subset C$ and $C$ is a finite algebra over $R$.
In our case a rooth of $x^2+\alpha x+\beta$ is integral over $\mathbb{Z}[\alpha,\beta]$ and so there exist a ring $C$ with the property above.
Now $\mathbb{Z}[\alpha,\beta]$ is a finite $\mathbb{Z}$-module, because $\alpha$ and $\beta$ are algebraic integers. Then $C$ is a finite $\mathbb{Z}$-module and $x$ is an algebraic integer.
A: A standard trick is to take the norm. If you compute the polynomial
$$ g(x) = \prod_i \prod_j (x^2 + \alpha_i x + \beta_j)$$
where the $\alpha_i$ are the conjugates of $\alpha$ and the $\beta_j$ are the conjugates of $\beta$, then you'll find $g(x)$ has integer coefficients.
Another way to calculate the same thing is with resultants: if $m_\gamma$ is the minimal polynomial of $\gamma$, then
$$ g(x) = \mathop{\mathrm{Res}}_z(\mathop{\mathrm{Res}}_y(x^2 + yx + z, m_\alpha(y)), m_\beta(z)) $$
