For the equation 3x^ 2 + px + 3 = 0, p > 0 , if one of the roots is...

Let $\alpha, {\alpha}^{2}$ be the roots of $3x^{2} + px + 3$

$∴$ Product of the roots,&

$\alpha \cdot {\alpha}^{2} = 1 ⇒ {\alpha}^{3} = 1$

Again, $\alpha + {\alpha}^{2} = - \dfrac{p}{3}$

$⇒ 1 + 1 = - \dfrac{p}{3}$ & (if $\alpha = 1$)

$⇒ p = - 6$

But $p > 0$

$∴ \alpha = 1$ is not possible.

If $\alpha = \omega$, then $\alpha + {\alpha}^{2} = \omega + {ω}^{2} = - 1$

$∴ - 1 = - \dfrac{p}{3} ⇒ p = 3$

Again, if $\alpha = {ω}^{2}$, then

$\alpha + {\alpha}^{2} = {ω}^{2} + {ω}^{4} = {ω}^{2} + \omega = - 1$

$∴ - 1 = - \dfrac{p}{3}$

$⇒ p = 3$