2. For the equation in $y$
$$
4 y^{2}+4(2-k) y-k^{2}=0
$$

the two real roots $y_{1}, y_{2}$ satisfy $\left|y_{1}\right|=2+\left|y_{2}\right|$. Then the value of $k$ is ( ).
(A) 1 or 4
(B) 0 or 1
(C) 0 or 4
(D) 0 or 1 or 4
2. C.
$$
\begin{array}{l}
\text { Given }\left|y_{1}\right|=2+\left|y_{2}\right| \\
\Rightarrow\left(y_{1}+y_{2}\right)^{2}-2 y_{1} y_{2}-2\left|y_{1} y_{2}\right|=4 \text {. }
\end{array}
$$

By Vieta's formulas, we have
$$
y_{1}+y_{2}=k-2, y_{1} y_{2}=-\frac{k^{2}}{4} \text {. }
$$

Thus, $(k-2)^{2}=4 \Rightarrow k=0$ or 4.