Question

God of mathematics solve it..........

Answer 1

$\red \bigstar \bold \answer \\ \\ \frac{ - 1}{ \sqrt{3} } is \: the \: zero \: of \: the \: given \: polunomial \\ \\ \green \bigstar solution \\ p = \binom{ - 1}{ \sqrt{3} } = 3 \binom{ - 1}{ \sqrt{3 {}^{} } } {}^{2} - 1 = 3 \binom{1}{3} - 1 = 0 \\ \frac{1}{ \sqrt{ - 3} } is \: the \: zero \: of \: p(x) = 3 {x}^{2} - 1 \\ p \binom{2}{ \sqrt{3} } = 3 \binom{2}{ - 3} {}^{2} - 1 = 3 \binom{4}{3} - 1 = 3 \\ hence \: verified \: \frac{ - 1}{ \sqrt{3} } is \: the \: zero \: of \: polynomial \: but \\ \frac{2}{ \sqrt{3} } is \: not$