Solution:
Let $x_{0}$ be a root of the cubic, then $x^{3}+p x+q=(x-x_{0})(x^{2}+a x+b)=x^{3}+(a-x_{0}) x^{2}+(b-a x_{0}) x-b x_{0}$. So $a=x_{0}$, $p=b-a x_{0}=b-x_{0}^{2}$, $-q=b x_{0}$. Hence $p^{2}=b^{2}-2 b x_{0}^{2}+x_{0}^{4}$. Also $4 x_{0} q=-4 x_{0}^{2} b$. So $p^{2}-4 x_{0} q=b^{2}+2 b x_{0}^{2}+x_{0}^{4}=(b+x_{0}^{2})^{2} \geq 0$.

Solution:
As the equation $x_{0} x^{2}+p x+q=0$ has a root $(x=x_{0})$, we must have $D \geq 0 \Leftrightarrow p^{2}-4 q x_{0} \geq 0$. (Also the equation $x^{2}+p x+q x_{0}=0$ having the root $x=x_{0}^{2}$ can be considered.)