The roots of the quadratic equation $2x^2 - 6x + 3 = 0$ are $\alpha$ and $\beta$. Find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

1. 利用韦达定理求出两根之和与两根之积

对于一元二次方程 $2x^2 - 6x + 3 = 0$，各项系数分别为 $a = 2$, $b = -6$, $c = 3$。
由根与系数的关系（韦达定理）可得：
$\alpha + \beta = -\frac{b}{a} = -\frac{-6}{2} = 3$
$\alpha \beta = \frac{c}{a} = \frac{3}{2}$

2. 将目标代数式通分，代入根与系数的值求出结果

将代数式 $\frac{1}{\alpha} + \frac{1}{\beta}$ 通分得：
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$
代入 $\alpha + \beta = 3$ 和 $\alpha \beta = \frac{3}{2}$：
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{\frac{3}{2}} = 3 \times \frac{2}{3} = 2$