In Dummit & Foote, the definition of primary ideal says:
A proper ideal of a commutative ring is called primary if whenever $ab \in Q$ and $a \notin Q$, then $b \in {\rm rad}(Q)$.
Suppose $I$ is an ideal such that ${\rm rad}(I)$ is prime.
Suppose $ab \in I$. Then $ab \in {\rm rad}(I)$, hence $a \in {\rm rad}(I)$ or $b \in {\rm rad}(I)$.
Case 1: $a \notin I$, $a \notin {\rm rad}(I)$. Then $b \in {\rm rad}(I)$. So, $I$ is primary by definition.
Case 2: $a \notin I$, but $a, b \in {\rm rad}(I)$. Then $I$ is primary by definition.
Case 3: $a \notin I$, $b \notin {\rm rad}(I)$. Then, $a \in {\rm rad}(I)$. In this case, since $b \notin {\rm rad}(I)$, then $b \notin I$. Then we have $b \notin I$ but $a \in {\rm rad}(I)$. So, $I$ is primary.
What is wrong with this argument?
Indeed Case 3 is incorrect. The issue is that you ended up with the wrong conclusion! You showed $b \notin I$ and $a \in \text{rad}(I)$, but you needed to conclude that $a \notin I $ and $b \in \text{rad}(I)$! In other words, in total you proved the following:
If $ab \in I$ and $a \notin I$, then either $b \in \text{rad}(I)$ or ($b \notin I$ and $a \in \text{rad}(I)$)
But this is not the definition of a primary ideal!