Let S be the set of all non-zero real numbers | vedclass

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(D) For the quadratic equation $\alpha x^2 - x + \alpha = 0$ to have two distinct real roots,the discriminant $D > 0$.$D = (-1)^2 - 4(\alpha)(\alpha)

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$(A, D)$

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(D) For the quadratic equation $\alpha x^2 - x + \alpha = 0$ to have two distinct real roots,the discriminant $D > 0$.$D = (-1)^2 - 4(\alpha)(\alpha) = 1 - 4\alpha^2 > 0$ $\Rightarrow \alpha^2 Given $|x_1 - x_2| Using $(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$,we get $\left(\frac{1}{\alpha}\right)^2 - 4(1) $\frac{1}{\alpha^2} - 4  \frac{1}{5}$.So,$|\alpha| > \frac{1}{\sqrt{5}}$,which means $\alpha \in \left(-\infty, -\frac{1}{\sqrt{5}}\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$.Combining the conditions,$\alpha \in \left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right) \cup \left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$.Thus,the intervals $(A)$ and $(D)$ are subsets of $S$.

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