S.O.S. Mathematics CyberBoard

[Bitupon]

Question: Solve 27(x^4)-195(x^3)+494(x^2)-520x+192=0, the roots being
in G.P

My Solution:-Since the equation is of degree 4 therefore it will have
four roots
Let the roots be a/(r^3),a/r,ar,a(r^3)

Then,
{a/(r^3)}+{a/r}+ar+a(r^3)=195/27=65/9


{(a^2)/(r^4)}+(a^2)+(a^2)(r^4)+(a^2)+{(a^2)/(r^2)}+(a^2)(r^2)=494/27

{(a^3)/(r^3)}+(a^3)(r^3)+(a^3)r+{(a^3)/r}=520/27

a^4=192/27=64/9

How to simplify from here onwards to arrive at the solutions?

[bugzpodder]

First of all, you shouldn't assume that the common ratio is r^2. The roots could after all be complex numbers, however, I largely suspect that this isnt the case in this situation.
Secondly, we can see its non-zero, since all four roots are non-zero, to justify division of r.

In either case, What you do is solve for the variable a from the last equation and plug it into one of the other equations. Verify all equations.
_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:

where

[Soroban]

Hello, Bitupon!

[bugzpodder]

the last two equation factors



but i am betting that the original poster skipped a phrase: all roots must be real
_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:

where

[jinydu]

I still don't see what's wrong with letting the common ratio be . What do non-real roots have to do with it?

The way I see it, Bitupon has used a bit of ingenuity to narrow down the possible values of a to only four possible values. He should be able to complete the problem by testing all four values in the other three equations until he arrives at a with a pair of values (a, r) that is consistent with the original quartic.