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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Answer
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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 \\
$
Hence, option C is correct.
Note: In this problem first we will find the sum and product of the roots then use cube roots of unity and its application to reach the required answer.
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 \\
$
Hence, option C is correct.
Note: In this problem first we will find the sum and product of the roots then use cube roots of unity and its application to reach the required answer.
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