Given the equations $x^2+kx+6=0$ and $x^2-kx+6=0$. If, when the roots of the equation are suitably listed, 
each root of the second equation is $5$ more than the corresponding root of the first equation, then $k$ equals:
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ -5 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ -7 \qquad \textbf{(E)}\ \text{none of these}$
Let the two roots of $x^2 + kx + 6$ be $r$ and $s$.  By [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas),
\[r+s=-k\]
Each root of $x^2 - kx + 6$ is five more than each root of the original, so using Vieta's Formula again yields
\[r+5+s+5 = k\]
\[r+s+10=k\]
Substitute $r+s$ to get
\[-k + 10 = k\]
\[10 = 2k\]
\[5 = k\]
The answer is $\boxed{\textbf{(A)}}$.