Let S be the set of all non - zero real numbers α such that the...

\alpha{x}^{2} - x + \alpha = 0$<br > \left|{x}_{1} - {x}_{2}\right| < 0$
\left|\dfrac{\sqrt{1 - 4{\alpha}^{2}}}{\alpha}\right| < 1$<br >$\sqrt{1 - 4{\alpha}^{2}} < | \alpha |$<br >$1 - 4{\alpha}^{2} < {\alpha}^{2}$<br >${\alpha}^{2} - \dfrac{1}{5} > 0$<br >$\alpha \in \left(- \infty, - \dfrac{1}{\sqrt{5}}\right) \cup \left(\dfrac{1}{\sqrt{5}}, \infty\right)$<br >Also,$1 - 4{\alpha}^{2} > 0$<br >${\alpha}^{2} - \dfrac{1}{4} < 0$<br > \left(\alpha - \dfrac{1}{2}\right)\left(\alpha + \dfrac{1}{2}\right) < 0$
$\alpha \in \left(- \dfrac{1}{2}, \dfrac{1}{2}\right)$
Taking intersection
$\alpha \in \left(- \dfrac{1}{2}, - \dfrac{1}{\sqrt{5}}\right) \cup \left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{2}\right)$