If $|{x^2} - x - 6| = x + 2$, then the values of $x$ are
$-2, 2, -4$
$-2, 2, 4$
$3, 2, -2$
$4, 4, 3$
Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 1$ is
The number of pairs of reals $(x, y)$ such that $x=x^2+y^2$ and $y=2 x y$ is
Suppose $a, b, c$ are three distinct real numbers, let $P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$ When simplified, $P(x)$ becomes
The number of roots of the equation $|x{|^2} - 7|x| + 12 = 0$ is
If $3$ distinct real number $a$,$b$,$c$ satisfy $a^2(a + p) = b^2 (b + p) = c^2 (c + p)$ where $p \in R$, then value of $bc + ca + ab$ is