If $x$ is real and satisfies $x + 2 > \sqrt {x + 4} ,$ then
$x < - 2$
$x > 0$
$ - 3 < x < 0$
$ - 3 < x < 4$
The product of all real roots of the equation ${x^2} - |x| - \,6 = 0$ is
The solutions of the quadratic equation ${(3|x| - 3)^2} = |x| + 7$ which belongs to the domain of definition of the function $y = \sqrt {x(x - 3)} $ are given by
The number of solutions of $\frac{{\log 5 + \log ({x^2} + 1)}}{{\log (x - 2)}} = 2$ is
$\{ x \in R:|x - 2|\,\, = {x^2}\} = $
Suppose $a$ is a positive real number such that $a^5-a^3+a=2$. Then,