Understanding Primitive roots I am trying to find a single primitive root modulo $11$. The definition in my textbook says "Let  $a$  and  $n$  be relatively prime integers with  ($a \neq 0$)  and  $n$  positive. Then the least positive integer  $x$  such that  $a^x\equiv1\pmod{\! n}$  is called the  order of $a$ modulo $n$ and is denoted by $\text{ord}_{n}a$".
So what I don't understand is how I can find a single primitive root modulo $11$ if I am not also given $a$.
Or is it that maybe I understand things after all since $2$ is a primitive root modulo $11$ since $2^{10} \equiv \phi(11)\equiv 10\pmod{\! 11}$ and $2$ is a generator for the group $\mathbb{Z}/11\mathbb{Z}$?
In any case, I am confused since I need to find a second primitive root modulo $11$ and I'm not sure how to do that other than by guessing and checking $a^{10}$ for $a \in \{3,4,5,6,7,8,9,10\}$. Any help would be appreciated. 

You just need to verify the candidates to primitive roots one by one in order, the candidates are all the numbers $k\in[1,n-1]$, where $n$ is the "modulo n" in your question. 
The way of knowing which ones are primitive roots is by calculating the following congruences for each one of the candidates $k$: 
$k^1 \pmod{n}$
$k^2\pmod{n}$
...
$k^{n-1}\pmod{n}$
This is mod your number $n$, in this case $\pmod{11}$. 
If all the congruences of $k$ as above are not $0$ and different, so the total of non-zero congruences of $k$ are exactly $\varphi(n)$ and different between them, which is equivalent to say that the period of $k^j \pmod n$ is exactly $\varphi(n)$, then $k$ is a primitive root modulo $n$.
So in your case, verify each number in $[1,10]$, for instance $2$, and then calculate $2^1,2^2..2^{10}$ and for each value calculate mod $11$, if no one of those congruences is $0$ and all of them are different then $2$ is a primitive root.
Indeed, the primitive roots modulo $11$ are $\{2,6,7,8\}$. Here is an online calculator.
