Let $X$ be a random variable. The variance ($\sigma_X^2$) of $X$ is calculated by the formula:
$$\sigma_X^2=\Bbb E[X^2]-(\Bbb E[X])^2$$
Is there two operators acting on $X$ that outputs $\sqrt{\Bbb E[X^2]}+\Bbb E[X]$ for the first ($T_1$) and $\sqrt{\Bbb E[X^2]}-\Bbb E[X]$ for the second ($T_2$)? Satisfying the relationship:
$$T_1(X)T_2(X)=\sigma_X^2 $$
And if there is not, but there is some kind of relation between difference of squares and the variance?