The cubic $x^3+ax^2+bx+c$ has three distinct zeros in GP and the reciprocal of these zeros are in AP then prove that $2b^2+3ac=0$

Question

I tried to solve this question: first assuming the zeros to be $m$, $mr$ and $mr^2$ in G.P. so that $\frac{1}{m}$, $\frac{1}{mr}$ and $\frac{1}{mr^2}$ in A.P.; then by solving it like $\frac{1}{mr}-\frac{1}{m}=\frac{1}{mr^2}-\frac{1}{mr}.$ Simplifying it we obtain $r=1$.

Then $$a= -(m+mr+mr^2)=-3m$$ $$b= m^2r+m^2r^2+m^2r^3=3m^2$$ $$c= -m^3r^3=-m^3$$ Then I put this value in the equation $2b^2+3ac$: $$2(3m^2)^2+3(-3m)(-m^3)=18m^4+9m^4=27m^4$$

So I am thinking it's a misprint and question was supposed to be: prove that $2b^2-3ac=0$.

Answer 1

As commented by @SouravGosh, since the zeros are assumed to be distinct, $r$ cannot be $1.$ So, "think outside the box": the central term of the arithmetic progression must not correspond to the central term of your geometric progression $m,mr,mr^2,$ i.e. what we have is not $\frac2{mr}=\frac1m+\frac1{mr^2}$ but (wlog): $$\frac2{mr^2}=\frac1m+\frac1{mr}.$$ The solution $\ne1$ is $$r=-2,$$ and $$2b^2+3ac=2(-6m^2)^2+3(-3m)(8m^3)=0.$$

Answer 2

Since $1/(1/x) = 1$, we can rephrase the problem to let the roots be in AP: $m - k, m, m + k$ and their reciprocals $\frac{1}{m - k}, \frac{1}{m}, \frac{1}{m + k}$ to be in GP.

Now if $a, b, c$ are in GP, $\frac{b}{a} = \frac{c}{b} \implies b^2 = ac$. This means we could have $\frac{1}{m - k} \frac{1}{m + k} = \frac{1}{m^2}$, or that $m^2 - k^2 = m^2$ where $k = 0$, and hence the $3$ roots would all be the same, which is invalid.

We also could have: $$\frac{1}{m(m -k)} = \frac{1}{(m + k)^2} \implies m^2 - mk = m^2 + 2mk + k^2 \implies -3mk = k^2 \implies k = -3m.$$

as $k \ne 0$. Note that the last case $\frac{1}{m(m + k)} = \frac{1}{(m - k)^2}$ is equivalent to the case before with $k \mapsto -k$, and $k$ could be anything except $0$.

Hence our roots are $4m, m, -2m$ and $2b^2 + 3ac = 0$ as previously mentioned.