# Roots of polynomial equation in geometric progression

Find the relation between $a, b, c, d$ if the roots of $ax^3+bx^2+cx+d=0$ are in geometric progression.

By considering $(\alpha+\beta)(\beta+\gamma)(\alpha+\gamma)$ show that the above cubic equation has two roots equal in size but opposite in sign if and only if $ad=bc$.

I can do the second part if I am given some hints on the first. I can use $\beta$ as the middle root and make $\alpha=\beta/r$ and $\gamma=r \times \beta$, but haven't got anywhere so far.

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The monic polynomial with roots $\beta/r,\beta, r\beta$ is $$x^3-\beta\left(1+r+\frac1r\right) x^2+\beta^2\left(1+r+\frac1r\right)x-\beta^3$$ We find $\beta=-\frac cb$ and $\beta^3=-\frac da$, hence $$db^3=ac^3$$ as necessary condition. Why is it also sufficient? What additional condition(s) would we need if we wanted the roots to be real?

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If $\alpha, \beta, \gamma$ are the roots of $ax^3+bx^2+cx+d=0$

Using Vieta's Formulas, $$\alpha+\beta+\gamma=-\frac ba, \alpha\beta+\beta\gamma+\gamma\alpha=\frac ca,\alpha\beta\gamma=-\frac da$$

If $\alpha,\beta,\gamma$ are in Geometric Progression, $\frac\beta\alpha=\frac\gamma\beta$ is constant $=r$(say). Clearly, $\alpha r\ne 0$

$\implies \beta=\alpha r, \gamma=\beta r=\alpha r^2$

$$\implies \alpha+\alpha r+\alpha r^2=-\frac ba\implies \alpha(1+r+r^2)=-\frac ba$$

$$\alpha\alpha r+\alpha r\alpha r^2+\alpha r^2\alpha=\frac ca\implies \alpha^2 r(1+r+r^2)=\frac ca$$

$$\alpha\alpha r\alpha r^2=-\frac da\implies \alpha^3 r^3=-\frac da$$

So, $$\frac{\alpha^2 r(1+r+r^2)}{\alpha(1+r+r^2)}=\frac{\frac ca}{-\frac ba}\implies \alpha r=-\frac cb \text{ if }1+r+r^2\ne0$$

Then, $$\left(-\frac cb\right)^3=-\frac da\implies b^3d=c^3a$$

If $1+r+r^2=0,$

$b=c=0$ and $r=\frac{-1\pm\sqrt3}2$ i.e., r is a complex cube root of $1$

So, the equation reduces to $ax^3+d=0$ whose roots are $\left(-\frac da\right)^\frac13,\left(-\frac da\right)^\frac13r, \left(-\frac da\right)^\frac13r^2$.

So, we don't find any relationship among $a,b,c$ and $d$ here.

In the 2nd case, let us find the equation whose roots are $\alpha+\beta,\beta+\gamma, \gamma+\alpha$ using the Transformation of Equations.

If $y=\alpha+\beta=\alpha+\beta+\gamma-\gamma=-\frac ba-\gamma\implies \gamma=-\left(y+\frac ba\right)$

As $\gamma$ is a root of the given equation, $-a\left(y+\frac ba\right)^3+b\left(y+\frac ba\right)^2-c\left(y+\frac ba\right)+d=0$

On simplification, $$ay^3+(\cdots)y^2+(\cdots)y^2+\frac{bc-ad}a=0$$ whose roots are $\alpha+\beta,\beta+\gamma, \gamma+\alpha$

Using Vieta's Formulas again, $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=\frac{ad-bc}{a^2}$

If among $\alpha,\beta,\gamma$ two have same absolute value but opposite sign, $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=0\implies ad=bc$ (the condition is necessary)

Sufficiency: Again, if $ad=bc,$ $\frac ab=\frac cd=t$(say,) so $a=bt,c=dt$

Then the equation becomes $$btx^3+bx^2+dtx+d=0\implies bx^2(tx+1)+d(tx+1)=0$$

So, $$(tx+1)(bx^2+d)=0$$ One root is $-\frac1t$ What about the others?

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$x=\sqrt{-d/b}$ which is not really $1/t$, the thing we need? –  bbr4in Feb 1 '13 at 21:59
@user52187, I left it for you to find the two roots other than $-\frac1t=-\frac ba=-\frac dc$ to prove the proposition as it is after all a 'homework'. –  lab bhattacharjee Feb 2 '13 at 3:27
So is what I wrote incorrect? –  bbr4in Feb 3 '13 at 22:09
@user52187, no. You are right when you say $\frac1t\ne \pm\sqrt{-\frac db}$. But 'homework' is not supposed to be answered fully here as my last comment has pointed out. –  lab bhattacharjee Feb 4 '13 at 3:29