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SolveBySubstitutingTheSumOfTwoRootsOfLinears(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

Two square roots of linears with one slope, sqrt(L1) and sqrt(L2) with
L1 - L2 = k a constant, rationalised together by their sum
v = sqrt(L1) + sqrt(L2): their difference is k/v, so
sqrt(L1) = (v + k/v)/2, sqrt(L2) = (v - k/v)/2, and
dx = (v^4 - k^2)/(2 m v^3) dv for the slope m.

Remarks

Charlwood's arcsin(sqrt(1 + x) - sqrt(x)) is arcsin(1/v) (v^4 - 1)/(2 v^3),
which parts closes. Asked as written, the remainder parts leaves in x is a nested root
no rule reads, and it was declined after five seconds, and after sixty on a slow
runner. Rubi's 5.3.7 has the arctangent's powers of x beside it.
The identities hold for the principal roots wherever v is not zero: the sum times the
difference is L1 - L2 for any values of the two, so nothing is assumed about
signs. At the question asked or one below it, since it lands on the chain in v.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

SolveBySubstitutingTheSumOfTwoRootsOfLinears(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)