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

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Question

(C) Let the roots of the quadratic equation $3x^2 + px + 3 = 0$ be $\alpha$ and $\alpha^2$.From the properties of roots,the product of the roots is gi

Vedclass · Accepted Answer

$3$

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(C) Let the roots of the quadratic equation $3x^2 + px + 3 = 0$ be $\alpha$ and $\alpha^2$.From the properties of roots,the product of the roots is given by $\alpha \cdot \alpha^2 = \frac{c}{a} = \frac{3}{3} = 1$.This implies $\alpha^3 = 1$,so $\alpha = 1, \omega, 	ext{ or } \omega^2$,where $\omega$ is a complex cube root of unity.If $\alpha = 1$,then the sum of roots is $\alpha + \alpha^2 = 1 + 1 = 2 = -\frac{p}{3}$,which gives $p = -6$. This contradicts the condition $p > 0$.If $\alpha = \omega$ or $\alpha = \omega^2$,then the sum of roots is $\alpha + \alpha^2 = \omega + \omega^2 = -1$.Using the sum of roots formula,$\alpha + \alpha^2 = -\frac{p}{3}$,we get $-1 = -\frac{p}{3}$.Therefore,$p = 3$.

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

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