Lennard-Jones Potential - distance of closest approach

[xn_]

Here's my solution to the problem of finding the distance of closest approach for two atoms approaching each other in a Lennard-Jones potential, starting with velocities (+/-) v_0:

They start with energy T₀ = (1/2)m(2v₀²) which is conserved. Thus,
E = T + V = (1/2)m(2v²) + A/r¹² - B/r⁶ = T₀. At the distance of closest approach, v=0 and we have:
T₀ = A/r_t¹² - B/r_t⁶. If we let x = r_t⁶ and rearrange:
T₀x² + Bx - A = 0
which has the solutions
x = (1/2T₀)(-B +/- sqrt(B² + 4AT_0)

My question is - what, physically, do each of these roots correspond to? They're both real, but of opposite sign - is the negative one a kind of 'virtual' turning point for the atoms passing through each other (tunnelling?) and coming out the other side?

 

[kuruman]

The solutions are not all real. You have defined $x=r_t^6$. Note that the solution $x=\frac{1}2{T_0}(-B-\sqrt{B^2+4AT_0})$ is negative assuming of course that A and B are positive. That cannot be for an even power of a number.