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I have the polynomial $P(x) = x^3 + mx^2-3x+1, m\in\mathbb{R}$. I need to find $m$ such that the roots of that polynomial are in geometric progression.

My attemp of solving this was to use Vieta's formulas. So
$x_1+x_2+x_3 = -m, x_1x_2+x_1x_3+x_2x_3 = -3, x_1x_2x_3 = -1$. If $x_1, x_2, x_3$ are in geometric progression then let $x_1 = \alpha, x_2 = q\alpha, x_3 = q^2\alpha$, where $q\in \mathbb{R}$ is the ratio of the geometric progression.
From first Vieta's formulas I get $\alpha q^2+\alpha q+(1+m) = 0$ and from third Vieta's formulas I get $q\alpha = \sqrt[3]{-1}$. From here I stuck. I don't know if my way of working this out is the right way. If it is could you please help me complete the solution, and if not I would very much appreciated If you would provide me a solution for this exercise.

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Assume the roots of the cubic are:

$$\frac{a}{r}, a, ar$$

Because it is given the roots need to be in GP. Product of the roots is given to be $-1$

$$\implies a^3 = -1$$ $$\implies a = -1$$

Also is given sum of products taken two at a time to be $-3$.

$$\implies \frac{1}{r} + r + 1 = -3$$

$$\implies r^2 +4r +1 = 0$$

Solve for $r$. And then you'll be able to solve for $m$.

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  • $\begingroup$ You haven't rule out the complex cube-roots $a=e^{\pm i\pi/3}$ yet. $\endgroup$ – user10354138 Jun 14 '19 at 15:05
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So you have \begin{align*} \alpha(1+q+q^2)&=-m\\ \alpha^2q(1+q+q^2)&=-3\\ (\alpha q)^3&=-1 \end{align*} So $\alpha q, 1+q+q^2\neq 0$. Thus dividing the second by the first gives $\alpha q=\frac3m\in\mathbb{R}$, so $\alpha q=-1$. Then the second equation gives $$ 1+q+q^2=-3q $$ which you can solve for $q=(2+\sqrt3)^{\pm 1}$ and hence $\alpha$.

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I would write for the three members of a geometric progression: $$a_1,a_1q,a_1q^2$$ then you will get $$a_1^3+ma_1^2-3a_1+1=0$$ $$a_1q^3+ma_1^2q^2-3a_1q+1=0$$ $$a_1^3q^6+ma_1^2q^4-3a_1q+1=0$$ and $$-(a_1+a_1q+a_1q^2)=m$$ $$a_1^2q+a_1^2q^2+a_1^2q^3=-3$$ $$-a_1^3q^3=1$$ Eliminating $a_1$ we get $$q+q^2+q^3=-3q$$ and $$1+q+q^2=mq$$ for $q\neq 0$

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