If $(\sqrt{3}+i{)}^{100}=2^{99}(p+iq),$ then $p$ and $q$ are roots of the equation

The correct option is A $x^2-(\sqrt{3}-1)x-\sqrt{3}=0$
$\because (\sqrt{3}+i)100=2^{99}(p+iq)$
${\left(2e^{i\frac{\pi }{6}}\right)}^{100}=2^{99}(p+iq)$
$2e^{i\frac{50\pi }{3}}=p+iq$
$\Rightarrow 2e^{i(16\pi +\frac{2\pi }{3})}=p+iq$
$\Rightarrow 2(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3})=p+iq$
$\therefore p=-1,q=\sqrt{3}$
Equation with roots $-1$ and $\sqrt{3}$ is
$x^2-(\sqrt{3}-1)x-\sqrt{3}=0$