# Solve a polynomial involving geometric progression?

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 math.stackexchange.com/questions/392309/… ? – 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
• Did you get the roots as 1, 9/4 and 3/2? – Isomorphism May 18 '13 at 11:20
• 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
• Please help me, I'm stuck. – missiledragon May 18 '13 at 12:23

## 2 Answers

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

• 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

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.