Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^{2} - x + \alpha = 0$ has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $|x_1 - x_2| < 1$. Which of the following intervals is(are) a subset(s) of $S$?

$\alpha x^2 - x + \alpha = 0,\quad \alpha \neq 0$ $x_1, x_2 \in \mathbb{R},\ x_1 \neq x_2,\quad |x_1 - x_2| < 1$ $\text{To find: which given intervals are subsets of } S.$ $\text{Plan: } 1.\ D>0;\quad 2.\ |x_1-x_2|<1;\quad 3.\ \text{combine} \Rightarrow S$ $D = (-1)^2 - 4(\alpha)(\alpha) = 1 - 4\alpha^2 > 0$ $4\alpha^2 < 1 \Rightarrow \alpha^2 < \dfrac{1}{4} \Rightarrow |\alpha| < \dfrac{1}{2}$ $x_1 + x_2 = \dfrac{1}{\alpha},\quad x_1 x_2 = 1$ $(x_1 - x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 = \dfrac{1}{\alpha^2} - 4$ $\dfrac{1}{\alpha^2} - 4 < 1 \Rightarrow \dfrac{1}{\alpha^2} < 5 \Rightarrow \alpha^2 > \dfrac{1}{5}$ $\Rightarrow |\alpha| > \dfrac{1}{\sqrt{5}}$ $\dfrac{1}{\sqrt{5}} < |\alpha| < \dfrac{1}{2}$ $S = \left(-\dfrac{1}{2}, -\dfrac{1}{\sqrt{5}}\right) \cup \left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{2}\right)$ $(1)\ \left(-\dfrac{1}{2}, -\dfrac{1}{\sqrt{5}}\right) \subset S\ \checkmark$ $(4)\ \left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{2}\right) \subset S\ \checkmark$ $(2),\ (3)\ \not\subset S\ \times$ $\boxed{(1)\ \text{and}\ (4)}$