jmc

We consider two cases, $x$ is nonnegative (so $|x|=x$), and $x$ is negative (so $|x|=-x$).

When $x\ge 0,$ the equation becomes $x^2-2x-1=0$. Applying the quadratic formula gives $x=1\pm \sqrt{2}.$ However, $x$ must be nonnegative in this case, so we have $x=1+\sqrt{2}$.

When $x<0,$ the equation becomes $x^2+2x+1=0$, so $(x+1{)}^2=0$ and $x=-1$.

Thus, the smallest value of $x$ is $x=\boxed{-1}.$