For the equation $3 x^{2} + p x^{2} + 3 = 0$ , if one of the roots is the square of the other then $p = . . . . . . . . . . . . . . .$

- \frac{1}{3}

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- 1

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- 6

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\frac{2}{3}

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Solution

The correct option is B $- 6$
$α + α^{2} = \frac{- p}{3}$
$α^{3} = 1 \Rightarrow (α + 1) (α^{2} + α + 1) = 0$
$(α + 1) (\frac{- p}{3} + 1) = 0$
$α = + 1$
$p = - 6$

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