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I have two equations:

$3192 =\frac{a\left(r^6-1\right)}{r-1}$

$3704 = \frac{a\left(r^7-1\right)}{r-1}$

I need to solve for $r$ and $a$. How would I do so? Substitution leads to a really messy equation.

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One thing that should really be emphasized from Andre's answer: there isn't a unique solution here, but instead several values of r that work. If this isn't the problem you were given but instead the equations you derived for your problem, it's possible that you missed some simplifying step or assumption along the way that narrows down the possible values and conceivably would lead to a much easier equation. –  Steven Stadnicki May 29 '12 at 15:20
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2 Answers 2

up vote 4 down vote accepted

By manipulation we get $3704(r^6-1)=3192(r^7-1)$, except that $r=1$ should presumably not be viewed as a solution of the original equation.

So in principle we can divide both sides by $r-1$ and reach an equation of degree $6$.

But let's not do this, just rewrite as $3192r^7 -3704r^6+512=0$, or more simply $399r^7-463r^6 +64=0$. Emboldened by the simple nature of $64$, we may look for a rational root other than $1$. A bit of fooling around yields the root $r=-2/3$. Perhaps that root is useful to you, though probably not, since it is negative. It is easy to find the corresponding $a$.

Now (in principle) divide the polynomial $399r^7-463r^6 +64$ by $r-1$, then by $r+2/3$. We reach a degree $5$ polynomial that does not appear to have a special enough structure. I did not find another rational root. So we end up with a quintic. There is another real root, which can be approximated by a numerical procedure, but which I do not have a closed form for.

Remark: I did a rough calculation. The positive root is fairly close to $1$, a bit under. Dividing by $r-1$ is useful so that a numerical method will not get wrongly sucked into $1$.

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The format of the question makes me think that this has to do with geometric series, so I'm not sure why you thought negative roots would be unhelpful here. Also it's worth noting that the roots you gave are the only real roots. –  Eric Stucky May 29 '12 at 7:03
    
I thought of the problem as arising in an applied setting, where $r$ would likely be positive. –  André Nicolas May 29 '12 at 7:07
    
about $0.98916$ –  Henry May 29 '12 at 7:59
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The 'trick' here is that $\frac{r^6-1}{r-1}$ and $\frac{r^7-1}{r-1}$ are are both partial sums of geometric series; $\frac{r^7-1}{r-1}=r^6+r^5+r^4+r^3+r^2+r+1$ and similarly for the other term, so by taking the difference you get $ar^6=512$; plugging this into either equation gives a polynomial you can solve to find the relevant values of $r$. From there, you can follow along with Andre's answer.

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