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.
Best Answer
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 $\alpha$.