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# For the equation $3{x^2} + px + 3 = 0;(p > 0),$ if one of the roots is square of the other. Then P is equal to:          A $\dfrac{1}{3}$          B 1          C 3          D $\dfrac{2}{3}$

Last updated date: 23rd Mar 2023
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Hint: If h, k are roots of the quadratic equation $a{x^2} + bx + c = 0$ then $h + k = - \dfrac{b}{a},hk = \dfrac{c}{a}$ .
The given quadratic equation is $3{x^2} + px + 3 = 0$ . Let $\alpha$ and $\beta$ be the two roots of the above equation. Given that $\beta = {\alpha ^2}$ . We know if the quadratic equation is in the form $a{x^2} + bx + c = 0$ and $h,k$ are its roots then $h + k = - \dfrac{b}{a},hk = \dfrac{c}{a}$ . According to our question,$\alpha + \beta = \alpha + {\alpha ^2} = - \dfrac{p}{3}{\text{ - - - - - - - - }}(1) \\ \alpha \beta = \alpha .{\alpha ^2} = {\alpha ^3} = \dfrac{3}{3} = 1{\text{ - - - - - - - }}(2) \\$
From equation (2), ${\alpha ^3} = 1$ then cube roots of unity are $1,\omega ,{\omega ^2}$ . Since, p is positive, 1 can’t be the root of the given equation. On substituting $\omega$ in place of $x$ in equation (1).
$\omega + {\omega ^2} = - \dfrac{p}{3}{\text{ }}[\omega = \dfrac{{ - 1 + \sqrt {3i} }}{2}] \\ \Rightarrow - 1 = - \dfrac{p}{3} \\ \Rightarrow p = 3 \\$