For a given value of $k$ the product of the roots of $x^2-3kx+2k^2-1=0$
is $7$. The roots may be characterized as:
$\textbf{(A) }\text{integral and positive} \qquad\textbf{(B) }\text{integral and negative} \qquad \\ \textbf{(C) }\text{rational, but not integral} \qquad\textbf{(D) }\text{irrational} \qquad\textbf{(E) } \text{imaginary}$
If the product of the roots are $7$, then by [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), 
\[2k^2-1=7\]
Solve for $k$ in the resulting equation to get
\[2k^2=8\]
\[k^2=4\]
\[k=\pm 2\]
That means the two quadratics are $x^2-6x+7=0$ and $x^2+6x+7=0$.  Since $b^2$, $a$, and $c$ are the same, the [discriminant](https://artofproblemsolving.com/wiki/index.php/Discriminant) of both is
$36-(4 \cdot 1 \cdot 7) = 8$.  Because $8$ is not a perfect square, the roots for both are irrational, so the answer is $\boxed{\textbf{(D)}}$.