smc

Note that if $y=-x^2+bx-8$ has its vertex on the $x$-axis, then $-y=x^2-bx+8$ will have its vertex on the x-axis as well. To find the location of the vertex of the parabola, we desire to put it in vertex form, where $-y=(x-h{)}^2+k$, and $(h,k)$ is the location of the vertex. However, we know that $k=0$, because the vertex is on the $x$-axis. Thus, we know that $x^2-bx+8$ must be the square of a linear term. Thus, $b=\pm 2\cdot 1\cdot \sqrt{8}=\pm 4\sqrt{2}$, which are both irrational. Thus, our answer is $\boxed{\text{a positive or a negative irrational number}}$.