If one root of the quadratic equation $x^2 + px + (1 - p) = 0$ is $(1 - p)$,then what are its roots?

If $a, b, c$ are distinct real numbers and $a^3 + b^3 + c^3 = 3abc$,then the equation $ax^2 + bx + c = 0$ has two roots,out of which one root is

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:

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then the set of possible values of $a$ is

The roots of the cubic equation $3x^3+4x^2-5x-2=0$ are diminished by $h$,and a cubic equation with these diminished roots is formed. If the transformed equation does not contain the $x^2$ term,then the roots of the transformed equation are

The roots of the equation $|x - 2|^2 + |x - 2| - 6 = 0$ are: