What is $E(X^2)$ mean in literal terms? In my probability and statistics class we learned about expected value, or $E(X)$. We also did some work about finding expected values of functions and such, like $E(g(x))$. And in the case of finding the variance, one of the steps involve finding $E(X^2)$. Does this mean anything in real-life terms? 
Using a geometric distribution as an example, is $E(X^2)$ the expected number of times a trial needs to happen until the event $X$ happens twice in a row?

Here's an example.  Suppose $X$ is a random variable that represents the outcome of a roll of a die numbered $1$ to $6$ inclusive.  No assumption is made about the fairness of the die.  Then $X^2$ is a random variable that represents the outcome of the square of the roll; whereas $$X \in \{1, 2, 3, 4, 5, 6\},$$ we have $$X^2 \in \{1, 4, 9, 16, 25, 36\}.$$
Now, if we suppose that this die is indeed fair, then we can easily compute the expected value of $X$:  $$\operatorname{E}[X] = \sum_{x=1}^6 x \Pr[X = x] = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) = \frac{7}{2}.$$  The expectation of $X^2$ is:  $$\operatorname{E}[X^2] = \sum_{x=1}^6 x^2 \Pr[X = x] = \frac{1}{6}(1 + 4 + 9 + 16 + 25 + 36) = \frac{91}{6}.$$  This should give you some intuition behind the meaning of $X$ versus $X^2$ and their corresponding expectations.
