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

Solving the equation, i get the value of p as -6 but the question states that p>0. Is there another way of solving this or is the question wrong?

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  • $\begingroup$ I agree with you; I think the question has a mistake. $\endgroup$ – user84413 Sep 14 '14 at 21:39
  • $\begingroup$ @user84413: You can consider complex roots and then $p>0$ is possible. $\endgroup$ – gammatester Sep 15 '14 at 7:52
  • $\begingroup$ @gammatester Thanks - I should have realized that. $\endgroup$ – user84413 Sep 15 '14 at 22:01
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The value is $p=3$.

Proof: Let $a, a^2$ be the roots of the equation. Then you get $$3x^2+px+3 = 3(x-a)(x-a^2) = 3x^2-3(a^2+a)x+a^3,$$ So you first have to solve $a^3=1$. This gives the three solutions (the three cube-roots of unity) $$a_1=1,\quad a_2=-\frac{1}{2}+\frac{i}{2}\sqrt{3},\quad a_3=-\frac{1}{2}-\frac{i}{2}\sqrt{3}$$ which in turn give the possible values for $p$ $$p_1 = -3(a_1^2+a_1) = -6$$ $$p_2 = -3(a_2^2+a_2) = 3$$ $$p_3 = -3(a_3^2+a_3) = 3$$ $p_1\;$ is your negative value, but $p_2=p_3=3\;$ is positive and therefore $p=3\;$ is the answer to the question.

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  • $\begingroup$ Wow!Thanks a lot gammatester :) $\endgroup$ – Mohit Singh Sep 15 '14 at 14:02

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