An ideal that is radical but not prime.

Why don't you use a simple example ... Suppose the ring to be $R=\mathbf{Z}$ and take the ideal $6\mathbf{Z}$. Since the $\sqrt{m\mathbf{Z}}=r\mathbf{Z}$ where $r=\Pi_{p|m} p$ and $p$ is a prime number. Here $\sqrt{6\mathbf{Z}}=6\mathbf{Z}$ but $6\mathbf{Z}$ is not a prime ideal since 6 is not prime.

$x(x-1)$ is certainly not prime. Suppose $f^{n}\in \langle x(x-1) \rangle$. Since $f^{n}\in \langle x \rangle$, and $\langle x \rangle$ is prime, for some $m<n,f^{m}\in \langle x \rangle$. Proceed this way we can prove $f$ must be divisible by $x$, and similarly for $x-1$. This gives $f$ is divisible by $x(x-1)$. So this ideal is its own radical.