# Find roots of 3 degree polynomial such that they are in geometric progression

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 $$q\alpha = \sqrt{-1}$$. 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.

• See this question. Jun 14, 2019 at 15:00

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 $$\alpha$$.

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 $$m$$.

• You haven't rule out the complex cube-roots $a=e^{\pm i\pi/3}$ yet. Jun 14, 2019 at 15:05

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 $$q\neq 0$$