The values of $k$ for which the equation $2x^2-kx+x+8=0$ will have real and equal roots are: 
$\textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9}$
$b^2-4ac=0\implies b^2=4ac\implies (1-k)^2=4(2)(8)\implies (1-k)^2=8^2\implies 1-k=8\implies k=-7, 1-k=-8\implies k=9$, $\fbox{A}$