Let the roots of $x^2-3x+1=0$ be $r$ and $s$. Then the expression $r^2+s^2$ is: $\textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}\qquad\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number}$
You may recognize that $r^2+s^2$ can be written as $(r+s)^2-2rs$. Then, by [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), \[r+s=-(-3)=3\]and \[rs=1.\]Therefore, plugging in the values for $r+s$ and $rs$, \[(r+s)^2-2rs=(3)^2-2(1)=9-2=7.\]Hence, we can say that the expression $r^2+s^2$ is $\boxed{\text{(A)}\ \text{a positive integer.}}$