If one of the roots of the equation $x^{2} - \sqrt{p}x + q = 0$ where $p, q \in R$ is $x = -\sqrt{p}$,then prove that $2p + q = 0$.

(N/A) Given the quadratic equation is $x^{2} - \sqrt{p}x + q = 0$.
Since $x = -\sqrt{p}$ is a root of the equation,it must satisfy the equation.
Substitute $x = -\sqrt{p}$ into the equation:
$(-\sqrt{p})^{2} - \sqrt{p}(-\sqrt{p}) + q = 0$
$p - (-\sqrt{p} \cdot \sqrt{p}) + q = 0$
$p - (-p) + q = 0$
$p + p + q = 0$
$2p + q = 0$
Hence,it is proved.