If $x=\frac{-1}{\sqrt{3}}$ and $\sqrt{x=\frac{2}{\sqrt{3}}}$ are zeroes of polynomial $p ( x )=3 x ^{2}-1,$ then $p\left(\frac{-m}{l}\right)$

$p\left(\frac{-1}{\sqrt{3}}\right)$ and $p\left(\frac{2}{\sqrt{3}}\right)$ should be $0$

Here, $p\left(\frac{-1}{\sqrt{3}}\right)=3\left(\frac{-1}{\sqrt{3}}\right)^{2}-1=3\left(\frac{1}{3}\right)-1=1-1=0,$ and $p\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{2}{\sqrt{3}}\right)^{2}-1=3\left(\frac{4}{3}\right)-1=4-1=3$

Hence, $x=\frac{-1}{\sqrt{3}}$ is a zero of the given polynomial. However, $x=\frac{2}{\sqrt{3}}$ is not a zero of the given polynomial.