For what value of $a$ does the equation $(a^2 - a - 2)x^2 + (a^2 - 4)x + a^2 - 3a + 2 = 0$ have more than two solutions?

The number of ordered pairs $(a, b)$ of integers such that $1 \leq a, b \leq 2021$ and the equations $x^2 - ax + b = 0$ and $x^3 - ax^2 + bx + a - b = 0$ have a common real root is

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:

If $x$ is real,then the minimum value of ${x^2} - 8x + 17$ is

The quadratic equation $2x^{2} - (a^{3} + 8a - 1)x + a^{2} - 4a = 0$ has roots of opposite signs. Then,

If the set of all $a \in R - \{1\}$,for which the roots of the equation $(1-a)x^2 + 2(a-3)x + 9 = 0$ are positive,is $(-\infty, -\alpha] \cup [\beta, \gamma)$,then $2\alpha + \beta + \gamma$ is equal to . . . . . . .