** **For the equation \[3{x^2} + px + 3 = 0,\,\,p > 0\], if one of the root is square of the other, then *p* is equal to

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For this quadratic equation,

$\text { Let } \alpha, \alpha^{2} \text { be the roots of } 3 x^{2}+p x+3=0$

$\text { Now, } \quad S=\alpha+\alpha^{2}=-p / 3 \text {, }$

$P=\alpha^{3}=1$

$\Rightarrow \quad \alpha=1, \omega, \omega^{2}$

$\text { Now, } \quad \alpha+\alpha^{2}=-p / 3$

$\Rightarrow \omega+\omega^{2}=-p / 3$

$\Rightarrow -1=-p / 3$

$\Rightarrow p=3$

Hence **p = 3.**

Last updated date: 24th Sep 2023

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