If $m$ and $n$ are the roots of $x^2+mx+n=0 ,m \ne 0,n \ne 0$, then the sum of the roots is:
$\text{(A) } -\frac{1}{2}\quad \text{(B) } -1\quad \text{(C) } \frac{1}{2}\quad \text{(D) } 1\quad \text{(E) } \text{undetermined}$
By Vieta's Theorem, $mn = n$ and $-(m + n) = m$. Dividing the first equation by $n$ gives $m = 1$. Multiplying the 2nd by -1 gives $m + n = -m$. The RHS is -1, so the answer is $\fbox{B}$