<s>[INST] For the given quadratic equation find the value of p 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?
 [/INST] 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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