6. Let $a, b$ be integers, and the quadratic equation
$$
x^{2}+(2 a+b+3) x+\left(a^{2}+a b+6\right)=0
$$

has a double root $\alpha$, and the quadratic equation
$$
2 a x^{2}+(4 a-2 b-2) x+(2 a-2 b-1)=0
$$

has a double root $\beta$. Then the quadratic equation with integer coefficients and roots $\alpha, \beta$ is ( ).
(A) $2 x^{2}+7 x+6=0$
(B) $2 x^{2}+x-6=0$
(C) $x^{2}+4 x+4=0$
(D) $x^{2}+(a+b) x+a b=0$
6. A.

From the discriminants of the two equations being 0, we have
$$
\begin{array}{l}
(2 a+b+3)^{2}-4\left(a^{2}+a b+6\right)=0, \\
(4 a-2 b-2)^{2}-8 a(2 a-2 b-1)=0,
\end{array}
$$

which simplifies to
$$
\begin{array}{l}
(b+3)^{2}=12(2-a), \\
(b+1)^{2}=2 a .
\end{array}
$$

Eliminating \(a\) gives \(7 b^{2}+18 b-9=0\).
The integer root is \(b=-3\).
Thus, \(a=2\).
Therefore, the two equations are
$$
x^{2}+4 x+4=0 \text { and } 4 x^{2}+12 x+9=0 \text {. }
$$

The equal roots of the first equation are \(\alpha=-2\), and the equal roots of the second equation are \(\beta=-\frac{3}{2}\).

It is easy to see that the quadratic equation with integer coefficients and roots \(\alpha, \beta\) is \(2 x^{2}+7 x+6=0\).