smc

We are given that the roots are real, so the discriminant is positive, which means $p^2-4(8)(1)>0$. This leads to $|p|>4\sqrt{2}$. By Vieta, the sum of the roots is $-p$, so we have $|-(r_1+r_2)|\ge 4\sqrt{2}$, or $|r_1+r_2|>4\sqrt{2}$, which is option $\boxed{|r_1+r_2|>4\sqrt{2}}$.