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How to solve for an unknown in two different denominators?

In this equation: $F = \frac{GMm_3}{r_2} - \frac{Gmm_3}{d-r_2}.$

Everything is given except for $r_2$. How do I solve for $r_2$? Is it possible?

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    If you put $f=0$ in Arturo Magidin's answer, the quadratic term disappears, so not in that case. And if the denominators are really $r_2^2$ and $(d-r_2)^2$ you will just have the ratio of squares with no linear term.2011-07-25

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First, rewrite the right hand side as a single fraction by adding the fractions; factoring out $Gm_3$ simplifies matters a bit: $\begin{align*} f &= \frac{GMm_3}{r_2} - \frac{Gmm_3}{d-r_2}\\ &= Gm_3\left(\frac{M}{r_2} - \frac{m}{d-r_2}\right)\\ &= Gm_3\left(\frac{M(d-r_2) - mr_2}{r_2(d-r_2)}\right). \end{align*}$ Then clear denominators by cross multiplying, and collect appropriate powers of $r_2$; you'll end up with a quadratic equation in $r_2$ that can be solved using the quadratic formula or other methods: $\begin{align*} f & = Gm_3\left(\frac{Md - (M+m)r_2}{r_2(d-r_2)}\right)\\ fr_2(d-r_2) &= Gm_3\left(Md - (M+m)r_2\right)\\ fdr_2 - f(r_2)^2 &= Gm_3Md - Gm_3(M+m)r_2\\ 0&= GMm_3d - \Bigl(Gm_3(M+m)+fd\Bigr)r_2 + f(r_2)^2. \end{align*}$

Added. If $f=0$, as you now write, then the equation becomes $0 = Gm_3\left(\frac {Md - (M+m)r_2}{r_2(d-r_2)}\right).$ If $G\neq 0$ and $m_3\neq 0$, then this holds if and only if the numerator is $0$, if and only if $0 = Md - (M+m)r_2,$ which is easy to solve.

If, as Ross suggests, the denominators should be squared and $f=0$, then you would instead get $0 = M(d-r_2)^2 - m{r_2}^2$ which yields a quadratic equation in $r_2$ again, namely $(M-m){r_2}^2 - 2Mdr_2 + Md^2 = 0.$

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    true, I am hypnotized by the appearance of Newton's law as opposed to an arbitrary equation. These conditions will then be satisified.2011-07-25
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Try multiplying each side by both denominators, i.e. both $r_2$ and $d-r_2$, so that there are no fractions left. Combine like terms - in other words, group the terms with $r_2^2$ in them, the terms with $r_2$ in them, and the terms with no $r_2$'s in them. Then, treating $r_2$ as a variable we are solving for, we can apply the quadratic formula.

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You can, but you will probably use the quadratic formula.

$f = \frac{GMm_3}{r_2} - \frac{Gmm_3}{d-r_2}.$

$f \cdot (r_2)(d-r_2) = GMm_3 \cdot (d-r_2) - Gmm_3 \cdot r_2$

$ -f \cdot r_2^2 + (fd + GMm_3 + Gmm_3) \cdot r_2 - GMm_3 d = 0$

And this is quadratic in $r_2$, so solve using the quadratic formula. Aha - just as I am ready to post, this tells me that Zev has posted a similar answer. Well, this is his, except Texed up.

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A brief plea for symmetry!

The problem probably came from a situation in which there are two distances, $r_1$ and $r_2$, whose sum is $d$. It could make life easier to use $r_1$ instead of $d-r_2$. We then need to specify additionally that $r_1+r_2=d$.

So instead of a single equation, we have a system of two equations. But there is much more symmetry. Calculations often are more pleasant, and the calculation is more likely to reveal structural information.