Solve a polynomial involving geometric progression?

Question

I have had trouble with this question:

"Solve the equation $8x^3 - 38x^2 + 57x -27 = 0$" if the roots are in geometric progression.

Any help would be appreciated.

So what did you learn from http://math.stackexchange.com/questions/392309/solve-the-roots-of-a-cubic-polynomial ? — Isomorphism, May 18 '13 at 11:05
I got to the point where I divided 1+r+r^2, but when I try to form a new quadratic equation - I end up with an imaginary number. I'm only asking the question to see if I'm missing anything important. — missiledragon, May 18 '13 at 11:06
Yes, I got them - but I got 3/2 as well as 2/3 in my quadratic equation. How do I choose? — missiledragon, May 18 '13 at 12:12

lab bhattacharjee · Accepted Answer · 2013-05-18T11:12:52.900

1

Let the roots be $a, a\cdot r, a\cdot r^2$

So using Vieta's formula $a+ a\cdot r+ a\cdot r^2=\frac{38}8\implies a(1+r+r^2)=\frac{19}4$

and $a( a\cdot r)+ a\cdot r(a\cdot r^2)+a\cdot r^2(a)=\frac{57}8\implies a^2r(1+r^2+r)=\frac{57}8$

Divide to get $ar=\frac{\frac{57}8}{\frac{19}4}=\frac32$ as $a(1+r+r^2)\ne0$

Put $a=\frac3{2r}$ in the first equation

edited May 18 '13 at 11:12

answered May 18 '13 at 11:07

lab bhattacharjee

Thanks, I know you already answered a similar question - but for some reason I ended with weird numbers. Thanks for clearing it up! – missiledragon May 18 '13 at 11:09
Now, what's wrong with this method, that caused the down-vote? – lab bhattacharjee May 18 '13 at 11:14
I didn't downvote, I'll upvote now. – missiledragon May 18 '13 at 11:18
The question/request is to the down-voter to disclose the mistake that eluded me – lab bhattacharjee May 18 '13 at 11:20
I got that r=3/2,2/3 in a quadratic equation. How do I tell which is correct? – missiledragon May 18 '13 at 11:46
@missiledragon, if $r=\frac32,a=1, ar^2=\left(\frac32\right)^2=\frac94$ and what if $r=\frac23?$ – lab bhattacharjee May 18 '13 at 11:48
let us continue this discussion in chat – lab bhattacharjee May 18 '13 at 11:49
If r = 2/3, then a=9/4 – missiledragon May 18 '13 at 11:50
I still don't understand it. I got different answers. Is 2/3 or 3/2 correct, and why?
Please help :(
– missiledragon May 18 '13 at 12:25

score 1 · Answer 2 · answered May 18 '13 at 11:07

1

Since the roots are in geometric progression, you have $x_2^2=x_1x_3$. From Theorem of Viet $x_1x_2x_3=27$. Hence $x_2^3=27$ etc.

answered May 18 '13 at 11:07

Boris Novikov

2 Answers2