Pluto Mass: M Diameter: D | Charon: Mass: (1/8)M Diamater: (1/2)D
In order to find where there would be no net gravitational force we need to find where the force of gravity would be equal to each other
f(g) = Gm1(M)/r^2
f(p) = f(c) where p = Pluto and c = Charon
G(m1)(M)/r^2 = G(m1)(M/8)/(d-r)^2 --> start canceling variables
1/r^2 = (1/8)/(d-r)^2
1/r^2 = 1/(8)(d^2 - 2dr +r^2)
r^2 = 8d^2 - 16dr + 8r^2
7r^2 - 16dr + 8d^2 = 0
Solve for r using quadratic formula
r = 16d +- sqrt((16d)^2 - 4(7)(8d^2))/(14)
r = 16d +- sqrt(256d^2 - 224d^2) / 14
r = 16d +- sqrt(32d^2) / 14
r has to be a fraction of d because if it was greater than d, it would be on the other side of Charon
so
r = 16d - 5.6568d / 14 = 10.3432d/14 = ~0.7388d