2. Given real numbers $a, b, c$ satisfy
$$
a+b+c=2, a b c=4 \text {. }
$$

Then the minimum value of $|a|+|b|+|c|$ is ( ).
(A) 2
(B) 3
(C) $3 \sqrt[3]{4}$
(D) 6
2.D.

Obviously, at least one of $a$, $b$, and $c$ is positive. Without loss of generality, let it be $a$.
From $b+c=2-a$ and $bc=\frac{4}{a}$, we get the quadratic equation in $x$
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$

which has two real roots.
Thus, $(2-a)^{2}-\frac{16}{a} \geqslant 0$.
Combining this with $a>0$, we get $a \geqslant 4$.
Then $|a|+|b|+|c|=a+|b+c|$
$$
=a+|2-a|=2a-2 \geqslant 6 \text{. }
$$