# Solving a complicated equation for approximate analytical solution

I am trying to solve the Following equation for r,

2*a*Q^4 + 5*r^4*(-4*Q^2 - 2*r*(-3*M + r) + 3*c*r^(1 - 3*\[Omega])*(1 + \[Omega])) ==0


Clearly this is unsolvable. But if we substitute a=0 and c=0 we get one of the solutions $$r=\frac{1}{2} \left(\sqrt{9 M^2-8 Q^2}+3 M\right)$$.. Can I obtain approximate analytical solution of the above equation which gives the same value when substitutions a=0 and c=0 are applied. If yes, then how? I have tried AsymptoticSolve, but got no answer. Thanks in advance.

Edit- The cross-post is here

• What are the small parameters which you tried for AsymptoticSolve? Commented May 25 at 8:14
• @Nasser I do not want to put values of a and c. Just want to solve the eq analytically. after getting solution I can put those values to check whether my solution is correct or not. a and c can take different values. so first I need to find solution in explicit form, even if it is approximate. Commented May 25 at 8:57
• @UlrichNeumann a and c Commented May 25 at 8:58
• a,c <<1 small? Commented May 25 at 12:57
• parameter \[Omega]1/3? Commented May 25 at 13:07

If you are looking for an analytical approximation near  r -> 1/2 (3 M + Sqrt[9 M^2 - 8 Q^2]) try

eq = 2*a*Q^4 +5*r^4*(-4*Q^2 - 2*r*(-3*M + r) +3*c*r^(1 - 3*\[Omega])*(1 + \[Omega]))
Asymptotic[eq /. r -> 1/2 (3 M + Sqrt[9 M^2 - 8 Q^2]) + dr, {dr , 0,2}]
soldr=Solve[% , dr]


With additional information given by QP we can get an asymptotic solution

Assuming: a,c small of same order a->eps a,c->eps c eps->0

Solve[eq == 0 /. {a -> 0, c -> 0}, r]
solr0 = %[[-1]] (*  only solution r>0:  r -> 1/2 (3 M + Sqrt[9 M^2 - 8 Q^2]) *)
eqeps = eq /. {a -> eps a, c -> eps c}
soleps = Solve[eqeps == 0, eps][[1]] (* eps[r] *)

Asymptotic[eps /. soleps /. r -> (r /. solr0) + dr, {dr, 0, 1}];
soldr = Solve[% == eps, dr][[1]] /. eps -> 1 // FullSimplify


asymptotic solution ( a,c<<1 )

solr={r -> (r /. solr0) + dr /. soldr}