If three of the roots of $x^4+ax^2+bx+c=0$ are $1$, $2$, and $3$ then the value of $a+c$ is:
$\text{(A) } 35 \quad \text{(B) } 24 \quad \text{(C) } -12 \quad \text{(D) } -61 \quad \text{(E) } -63$
Since this is a quartic equation, there are going to be $4$ solutions. 
By Vieta's formulas, since the $x^3$ term is $0$, the sum of the roots is also $0$. Therefore, the 4th root of this polynomial is $-6$. 
Lastly, by Vieta's Formulas, $a+c=-(1\cdot2+1\cdot3+1\cdot-6+2\cdot3+2\cdot-6+3\cdot-6)+1\cdot2\cdot3\cdot-6=-25-36=-61= \fbox{D}$