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?
[Math] How to solve for an unknown in two different denominators
algebra-precalculusphysics
Best Answer
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 $$(M-m){r_2}^2 - 2Mdr_2 + Md^2 = 0.$$