GS. 2 Let $\alpha, \beta, \gamma$ be real numbers satisfying $\alpha+\beta+\gamma=2$ and $\alpha \beta \gamma=4$. Let $v$ be the minimum value of $|\alpha|+|\beta|+|\gamma|$. Find the value of $v$.
If at least one of $\alpha, \beta, \gamma=0$, then $\alpha \beta \gamma \neq 4 \Rightarrow \alpha, \beta, \gamma \neq 0$
If $\alpha, \beta, \gamma>0$, then
$$
\begin{array}{l}
\frac{\alpha+\beta+\gamma}{3} \geq \sqrt[3]{\alpha \beta \gamma} \quad \text { (A.M. } \geq \text { G.M.) } \\
\frac{2}{3} \geq \sqrt[3]{4} \\
2^{3} \geq 27 \times 4=108, \text { which is a contradiction }
\end{array}
$$

If $\beta0$, WLOG let $\gamma0$
$$
\begin{array}{l}
\alpha=2-\beta-\gamma>2 \\
|\alpha|+|\beta|+|\gamma|=\alpha-(\beta+\gamma)=2+2(-\beta-\gamma) \geq 2+4 \sqrt{(-\beta)(-\gamma)}, \text { equality holds when } \beta=\gamma \\
4=(2-2 \beta) \beta^{2} \\
\beta^{3}-\beta^{2}+2=0 \\
(\beta+1)\left(\beta^{2}-2 \beta+2\right)=0 \\
\beta=-1 \text { (For the } 2^{\text {nd }} \text { equation, } \Delta=-4<0 \text {, no real solution) } \\
\gamma=-1, \alpha=4 \\
|\alpha|+|\beta|+|\gamma|=1+1+4=6 \\
v=\min \text {. of }|\alpha|+|\beta|+|\gamma|=6
\end{array}
$$