In the worst-case scenario, it requires
six tests
to locate the radioactive rods. Several answers already describe strategies for locating the radioactive rods. I will give another.
Testing strategy: Start by testing two rods. If neither of these rods is radioactive, use the five remaining tests on five of the six remaining rods, one at a time. If two of these rods are radioactive, we are done. If only one is radioactive, then the single untested rod must also be radioactive.
If the first test indicates at least one of the two rods is radioactive, then divide the remaining six rods into two groups of three, and test both groups. If neither group contains a radioactive rod, then the initial two rods tested must both be radioactive. If one of the two groups of three is radiactive, we will use our final three test on a single rod each, one rod from the initial group of two, and two rods from the group of three containing a radioactive rod. Each of these groups has exactly one radioactive rod, so testing all but one of the rods (one at a time) is enough to determine which rod in each group is radioactive.
Proof that this cannot be done with fewer tests:
Suppose we have a strategy that is guaranteed to find the radioactive rods with five tests. The first test will involve $k$ rods. Initially, there are ${8\choose 2}=28$ possibilities for the pair of radioactive rods. After the first measurement, there will be either $28-{8-k\choose 2}$ possibilities (if at least one of the tested rods is radioactive), or ${8-k\choose 2}$ possibilities (if none of the tested rods is radioactive). Only four tests will then remain, so we can only distinguish between $2^4=16$ possibilities. This means we need $28-{8-k\choose 2}\leq 16$ and ${8-k\choose 2}\leq 16$. Direct computation shows that this is only satisfied for $k=2$, so our strategy must start by testing exactly two rods.
Suppose that neither of the two rods we test first is radioactive. Our next test will involve $j$ of the remaining rods. At this point, there are ${6\choose 2}=15$ possibilities for the pair of radioactive rods. After the second test, the number of possibilities will either be $15-{6-j\choose 2}$ or ${6-j\choose 2}$. After this test, we will only have three tests remaining, so we need $15-{6-j\choose 2}\leq 8$ and ${6-j\choose 2}\leq 8$. There is no value of $j$ for which both of these inequalities hold.