The roots of the equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$ are

The $x$-coordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3|x| + 2 = 0$ and the $y$-coordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$. Then the possible vertices of the square are:

Let $p(x)$ be a quadratic polynomial such that $p(0) = 1$. If $p(x)$ leaves a remainder of $4$ when divided by $x - 1$ and a remainder of $6$ when divided by $x + 1$,then:

If $f(x)$ is a second degree polynomial such that $f(x) \geq 0$ for all $x \in R$,$f(-3) = 0$,and $f(0) = 18$,then find $f(3)$.

Let $a_n, a_{n-1}, \ldots, a_1, a_0 \in \mathbb{C}$ and $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ be a polynomial. If the polynomial $f(x)$ is monic,then:

If the roots of the given equation $({m^2} + 1){x^2} + 2amx + {a^2} - {b^2} = 0$ are equal,then: