Finding real roots for an eigth-degree polynomial

Question

The context for this problem is identical to my first question. Unfortunately my previous result for $L$ was inexact since it was a relatively simple approximation. This time I have managed to produce a very convincing approximation for $L$. Alike last time, solving for $W$ provides the $n$th solution if you find the $n$th root of a specific [octive] polynomial:

$96x^8-192Lx^7+(768π^2 A^2+128L^2 ) x^6+(-64L^3-640A^2 π^2 L) x^5+(2655π^4 A^4-640L^2 π^2 A^2+32L^4 ) x^4+(384A^2 L^3 π^2-578A^4 Lπ^4 ) x^3+(4476π^6 A^6-2760L^2 π^4 A^4+128L^4 π^2 A^2 ) x^2+(456A^6 Lπ^6-278A^4 L^3 π^4 )x+(3136π^8 A^8-3700L^2 π^6 A^6+1089L^4 π^4 A^4 )=0$

Since my investigation is based on the assumption that there is only one value for $W$, given the values of $A$ and $L$, ergo there should only be two real roots (one should be the positive value for $W$ and the other should be the negative value for $W$.)

My question is, can you find the positive real root for the eight order polynomial above?

Answer 1

Let's define a polynomial p of one variable x depending on two parameters A and L:

p[x_, A_, L_] := 
    96 x^8 − 192 L x^7 + (768 π^2 A^2 + 128 L^2) x^6 
    + (−64 L^3 − 640 A^2 π^2 L) x^5 + (2655 π^4 A^4 − 640 L^2 π^2 A^2 + 32 L^4) x^4 
    + (384 A^2 L^3 π^2 − 578 A^4 L π^4) x^3 
    + (4476 π^6 A^6 − 2760 L^2 π^4 A^4 + 128 L^4 π^2 A^2) x^2 
    + (456 A^6 L π^6 − 278 A^4 L^3 π^4) x 
    + (3136 π^8 A^8 − 3700 L^2 π^6 A^6 + 1089 L^4  π^4 A^4)

An appropriate tool for finding positive solutions is Solve (one can use Reduce also) with the MaxExtraConditions -> All option, moreover we need the assumption x > 0 and aditionally domain specification Reals, otherwise A and L might be complex in general, making the task too involved.
So we have:

TraditionalForm[ sol = Solve[ p[x, A, L] == 0 && x > 0, x, Reals, 
                              MaxExtraConditions -> All]]

There are three positive real solutions under quite involved conditions ( the second arguments of ConditionalExpression), however the first solution is in a zero measure subset of variable space A , L thus in fact generically there are only two positive solutions.
We can verify that the conditions are not satisfied e.g. for A == 1 && L ==1, e.g.

Simplify[ sol, A == 1 && L == 1]

 {{x -> Undefined}, {x -> Undefined}, {x -> Undefined}}