The quadratic equation whose one root is $\frac{1}{2 + \sqrt{5}}$ will be

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 $S$?
$(A) \left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right)$
$(B) \left(-\frac{1}{\sqrt{5}}, 0\right)$
$(C) \left(0, \frac{1}{\sqrt{5}}\right)$
$(D) \left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$

Let $x_1, x_2, x_3 \in \mathbb{R} \setminus \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for:

Let $a \neq 0$ and $p(x)$ be a polynomial of degree greater than $2$. If $p(x)$ leaves remainders $a$ and $-a$ when divided respectively by $x+a$ and $x-a$,then the remainder when $p(x)$ is divided by $x^2-a^2$ is:

Solve the equation $3x^{2} - 4x + \frac{20}{3} = 0$.

If $1-i$ is a root of the equation $x^2+ax+b=0$ where $a$ and $b$ are real numbers,then $b$ is equal to