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SolveByTheReciprocalBesideARootOfAQuadratic​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

A power of the variable below the bar beside a root of a quadratic, by
x = 1/t: 1/(x^n sqrt(q0 + q1 x + q2 x^2)) is
-sgn(t) t^(n - 1)/sqrt(q0 t^2 + q1 t + q2), a polynomial over the root of the
quadratic with its coefficients reversed, which the rules for those answer.

Remarks

1/(x sqrt(1 - (a + b x)^2)) was answered and 1/(x^2 sqrt(...)) was not:
the repeated factor is what nothing reads, and it is what a round of parts against
acos(a + b x)/x^4 leaves. SolveByReciprocalSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) is the
same substitution for a different shape -- a palindromic quartic under the root, where
the reciprocal maps the quartic to itself -- and reads nothing here.
 The radicand is assembled rather than substituted into: Q(1/t) is
R(t)/t^2 with R the reversed quadratic, so the root is
R^(p/2) |t|^(-p), and writing the quotient inside the root instead leaves a
nesting nothing downstream reduces. The modulus is a sgn(t) in front for an odd
p, constant between its zeros; with t = 1/x it is sgn(x), which is
what the answers of this family carry anyway.
https://github.com/asc-community/AngouriMath/issues/718

























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