If $x=1$ is a solution of the equation $x^{2}+a x+1=0$, then the value of $a$ is
(A) 3
(B) -1
(C) 1
(D) 2
(E) -2
Since $x=1$ is a solution of the equation $x^{2}+a x+1=0$, then $1^{2}+a(1)+1=0$ or $2+a=0$ and so $a=-2$.

ANSWeR: (E)