$V$ is a variety defined over $K$ if $I(V)\subset\bar{K}[x]$ s.t. $I(V)$ is generated by elements of $K[x]$. $I(V/K)=\{f\in K[x]\vert f(p)=0,\forall p\in V\}$
"If $V$ is defined over $K$, it is not enough to check $I(V/K)$ is prime in $K[x]$. For example, consider ideal $(x_1^2-2x_2^2)\subset Q[x,y]$."
$\textbf{Q:}$ I do not see why $(x_1^2-2x_2^2)\subset Q[x,y]$ is not prime here. $x_1^2-2x_2^2$ is irreducible and $Q[x,y]$ is UFD. So I get it is prime instead. What is implied in the above statement? I do not quite get the point. Or is the book trying to say $I(V/K)C[x,y]$ is not prime or $I(V/K)$ is not contraction of a prime?
The ideal can be prime over $K$ and fail to be prime over $\overline K$. In your case, the ideal is prime over $\mathbb Q$ but not over $\overline{\mathbb Q}$ and so the resulting algebraic set is not a variety. (It is not irreducible.)