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- Feb 14, 2012

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Find all values of $m$ for which the equation $16x^4-mx^3+(2m+17)x^2-mx+16=0$ has four distinct roots forming a geometric progression.

Attempt:

We're told that the four distinct roots of the equations above forming a geometric progression. So, I let the roots of the equation above as $a, ar, ar^2$, and $ar^3$ where $a$ and $r$ are the first term and common ratio of the geometric progression respectively.

These are the three equations that I determined by using Vieta's formulas and and I can't see where these equations could lead me to the values of m...

$\displaystyle a^2r^3=1$

$\displaystyle a(1+r)(1+r^2)=\frac{m}{16}$

$\displaystyle\left(\frac{(1+r)}{r}\right)^2(r^2-r+1)=\frac{2m-15}{16}$

The farthest that I could go in the effort of finding the values of m by blending all these equations together is the following:

$\displaystyle \left(\frac{(1+r)}{r}\right)^2\left[\frac{(r^2-r+1)(r^2+r+1)}{r^2}+1\right]=\frac{m^2}{16}$

and as you can see, this doesn't seem right because I couldn't eliminate either the variables $m$ or $r$ to solve for either of them and thus, the values of m remain unknown.

Any hints, please?

Thanks in advance.