Let $I$ be an ideal in $K[t_1,\dots,t_n]$. Is it true that if the quotient $K[t_1,\dots,t_n]/I$ is reduced then $I$ is radical?
We say that a ring $R$ is reduced if $x^2 = 0$ implies $x=0$ for all $x \in R$. If $a \in \sqrt{I}$ then $\exists k \in \mathbb{N}$ such that $a^k \in I$, so $\overline{a^k} = \overline{a}^k = 0$ in the quotient. Now I don't know how to use the reducedness, which tells me that $\overline{a}^2 = 0$ implies $\overline{a} = 0$.
Here's a useful lemma to get you started:
Lemma: Let $R$ be a ring. Suppose that $R$ is reduced, i.e., that $x^2 = 0 \implies x=0$ for all $x \in R$. Then $R$ has no nonzero nilpotent elements.
Proof: Suppose $x^k = 0$. We claim $x = 0$. If $k = 2^m$ is a power of $2$, then the claim follows easily by induction: $0 = x^{2^m} = \left(x^{2^{m-1}}\right)^2$ implies $x^{2^{m-1}} = 0$. (For example, if $k = 4$, then $$ 0 = x^4 = (x^2)^2 \implies x^2 = 0 \implies x=0 $$ and this works similarly for any power of $2$.)
For $k$ not (necessarily) a power of $2$, choose $m$ large enough so that $2^m > k$. Multiplying the equation $0 = x^k$ by $x^{2^m - k}$, we have $0 = x^{2^m}$. Then $x = 0$ by the special case considered above. $\square$
Okay, now that we see that reduced $\implies$ no nonzero nilpotents, do you see how to complete the problem?