I'm following up from this question:

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.

I eventually solved the equation (please check @lab bhattacharjee's answer), but I got two answers for $r$, which were 3/2 and 2/3. How exactly do I know which is correct? I've been stuck for ages, so I'd prefer a full explanation rather than hints.

Any help would be appreciated. Thanks.

• Well, we can either write the geometric progression as $a_1, a_2, a_3$, or the other way around: $a_3,a_2,a_1$. The corresponding $r$ then becomes its reciprocal. So, both is correct, and they give the same solution. – Berci May 18 '13 at 13:00
• But is there a way to test whether a solution is correct. I basically have the answer, but I need to eliminate an unwanted solution. – missiledragon May 18 '13 at 13:01
• Plug the results you got for $x$ into the original equation, and check whether that's true or not. – Berci May 18 '13 at 13:02
• Can you please provide a full answer - because I'm not sure what to plug in. It would be greatly appreciated. – missiledragon May 18 '13 at 13:04
• So, $8x^3-38x^2+57x-27=8(x-9/4)(x-3/2)(x-1)$ If you take $(9/4,3/2,1)$ then $r=2/3;$ if you take $(1,3/2,9/4)$ then $r=3/2$ – lab bhattacharjee May 18 '13 at 13:08

Solution :

Since it is a cubic polynomial you can find one of the root in the constant term of the equation : which is here 27 , so factors are 1,3 only. So by putting 1 in equation it gives x -1 as one factor of the equation.

Therefore by dividing the complete polynomial with x -1 you get $8x^2-30x+27$ which after solving gives you two more factors

x = $\frac{3}{2}$ and x =$\frac{9}{4}$ Threfore, 1, $\frac{3}{2} ; \frac{9}{4}$ are the roots of equation.

we can see roots are in G.P with common difference $\frac{3}{2}$