Since you can also take the radical inside the ring $K[x_1,\ldots,x_n]$ this boils down to the following question: if we have a radical ideal $I$ in $K[x_1,\ldots x_n]$, will its extension in $\bar{K}[x_1,\ldots x_n]$ still be radical. 
In characteristic $p$ this is problematic, take for example $n=1$, take $t\in K$ to be an element without a $p$-th root of unity, and take $I$ to be the ideal generated by the polynomial $x^p-t$, which is radical, but its extension in the algebraic closure will not be radical. For characteristic zero the situation is different: if your ideal $I$ is radical, then it will be the intersection of all maximal ideals containing it in $K[x_1,\ldots x_n]$. Now, if $m$ is such a maximal ideal, then the extension of $m$ in $\bar{K}[x_1,\ldots x_n]$ will be an intersection of maximal ideals there: this is because in characteristic zero extensions are always separable. It follows that the extension of $I$ is also the intersection of maximal ideals in $\bar{K}[x_1,\ldots x_n]$, and therefore it is radical as well.