For the equation 3x^2 + px + 3 = 0, p 0,if | vedclass

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Question

(C) Let the roots be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients for $3x^2 + px + 3 = 0$:Sum of roots: $\alpha + \alpha^2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let the roots be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients for $3x^2 + px + 3 = 0$:Sum of roots: $\alpha + \alpha^2 = -p/3$Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = 3/3 = 1$Since $\alpha^3 = 1$,the possible values for $\alpha$ are $1, \omega, \omega^2$,where $\omega$ is a complex cube root of unity.If $\alpha = 1$,then $\alpha + \alpha^2 = 1 + 1 = 2$,so $2 = -p/3 \implies p = -6$. But we are given $p > 0$,so this case is rejected.If $\alpha = \omega$ or $\alpha = \omega^2$,then $\alpha + \alpha^2 = \omega + \omega^2 = -1$.Substituting this into the sum of roots equation: $-1 = -p/3 \implies p = 3$.Thus,$p = 3$.

For the equation $3x^2 + px + 3 = 0, p > 0$,if one of the roots is the square of the other,then $p$ is equal to:

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