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 $$\left| {{x_1} - {x_2}} \right| < 1.$$ Which of the following intervals is (are) $$a$$ subset(s) os $$S$$?

If $${3^x}\, = \,{4^{x - 1}},$$ then $$x\, = $$

If $$\alpha $$ and $$\beta $$ are the roots of $${x^2}$$+ px + q = 0 and $${\alpha ^4},{\beta ^4}$$ are the roots of $$\,{x^2} - rx + s = 0$$, then the equation $${x^2} - 4qx + 2{q^2} - r = 0$$ has always

Let a, b, c be real numbers, $$a \ne 0$$. If $$\alpha \,$$ is a root of $${a^2}{x^2} + bx + c = 0$$. $$\beta \,$$ is the root of $${a^2}{x^2} - bx - c = 0$$ and $$0 < \alpha \, < \,\beta $$, then the equation $${a^2}{x^2} + 2bx + 2c = 0$$ has a root $$\gamma $$ that always satisfies