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How to solve the equation algebraically?

Posted 2 years ago

Hello, everyone! Can you please help me to solve the equation algebraically:

(26 + 18 x)/(2 Sqrt[91 + 26 x + 9 x^2]) + (33 + 64 x)/(
2 Sqrt[70 + 33 x + 32 x^2]) + (39 + 74 x)/(
2 Sqrt[49 + 39 x + 37 x^2]) + (40 + 132 x)/(
2 Sqrt[8 + 40 x + 66 x^2]) + (91 + 190 x)/(
2 Sqrt[90 + 91 x + 95 x^2]) == 0.

I have no idea... Wolfram Mathematica cannot solve it too.

POSTED BY: Aleksandr Miller
7 Replies

One can get a polynomial in x using GroebnerBasis.

gbx = First[
   GroebnerBasis[(26 + 18 x)/(2 Sqrt[91 + 26 x + 9 x^2]) + (33 + 
        64 x)/(2 Sqrt[70 + 33 x + 32 x^2]) + (39 + 
        74 x)/(2 Sqrt[49 + 39 x + 37 x^2]) + (40 + 
        132 x)/(2 Sqrt[8 + 40 x + 66 x^2]) + (91 + 
        190 x)/(2 Sqrt[90 + 91 x + 95 x^2]), x]];

It is not small.

Exponent[gbx, x]
Max[Abs[CoefficientList[gbx, x]]]

(* Out[610]= 80

Out[611]= \
1110276819753660237648054091530651383250064170279077722050580451636841\
585205837760192206271706180210400 *)

There are 80 roots. I'll just show the real ones.

N[Select[NSolveValues[gbx == 0, x, WorkingPrecision -> 100], Im[#] == 0 &]]

(* Out[612]= {-3.81511, -1.94173, -1.39884, -1.36244, -0.926541, \
-0.854442, -0.384779, -0.363983, -0.350656, -0.349146, -0.331095, \
-0.325256, -0.322182, -0.310507, -0.292343, -0.282015, -0.275535, \
-0.265719, -0.222805, -0.146951, 0.0258104, 0.455657} *)
POSTED BY: Daniel Lichtblau

As it seems, none of the values except the known unique one yields a zero.

Does anbody know, how to proceed with the GreobnerBasis in case of a set of roots of different polynomials?

POSTED BY: Roland Franzius

Roland is quite right, and I missed the step of weeding out parasite solutions. Here is the corrected approach.

radexpr = (26 + 18 x)/(2 Sqrt[91 + 26 x + 9 x^2]) + (33 + 
     64 x)/(2 Sqrt[70 + 33 x + 32 x^2]) + (39 + 
     74 x)/(2 Sqrt[49 + 39 xradexpr = (26 + 18 x)/(2 Sqrt[91 + 26 x + 9 x^2]) + (33 + 
     64 x)/(2 Sqrt[70 + 33 x + 32 x^2]) + (39 + 
     74 x)/(2 Sqrt[49 + 39 x + 37 x^2]) + (40 + 
     132 x)/(2 Sqrt[8 + 40 x + 66 x^2]) + (91 + 
     190 x)/(2 Sqrt[90 + 91 x + 95 x^2]);
gbx = First[GroebnerBasis[radexpr, x]];
Select[NSolve[gbx == 0, x, W + 37 x^2]) + (40 + 
     132 x)/(2 Sqrt[8 + 40 x + 66 x^2]) + (91 + 
     190 x)/(2 Sqrt[90 + 91 x + 95 x^2]);
gbx = First[GroebnerBasis[radexpr, x]];
Select[NSolve[gbx == 0, x, WorkingPrecision -> 100], Abs[radexpr /. #] < 10^(-20) &]

(* Out[622]= {{x ->
-0.3847788729136109731409292759642111842393961241324969914766953396637\
135729983357283916669163862617368}} *)
POSTED BY: Daniel Lichtblau
Posted 2 years ago

I tried with Maple 2023 ,but 1 hour computation I aborted.

Only numerically:

 eq = (26 + 18 x)/(2 Sqrt[91 + 26 x + 9 x^2]) + (33 + 
       64 x)/(2 Sqrt[70 + 33 x + 32 x^2]) + (39 + 
       74 x)/(2 Sqrt[49 + 39 x + 37 x^2]) + (40 + 
       132 x)/(2 Sqrt[8 + 40 x + 66 x^2]) + (91 + 
       190 x)/(2 Sqrt[90 + 91 x + 95 x^2]);
NSolve[eq, x, Reals]

(*{{x -> -0.384779}}*)
POSTED BY: Updating Name

There is probably no chance to solve algebraically exact for such a sum of rationals with differenent square roots as denominators, because on a common denominator the numerator becomes a sum of square roots of higher order polynomials.

So you have to FindRoot numerically. That is straigthforward-

Plot the expression over a suffiently large interval

Plot[expression, {x,-8,8}]

in order to view the roots. Ther exists a single root near -0.4.

With

sol = FindRoot[ expression, {x, 0}, PrecisionGoal -> 30, WorkingPrecision -> 50  ]`   

you get a number in the form

{x->-0.3...}

Test the solution by

expression /. sol

Don't forget to consult the documentation via Help menu and check.out Solve, Reduce, NSolve, FindRoot and for polynomial systems GroebnerBasis

--

Roland Franzius

POSTED BY: Roland Franzius

Can this be solved approximately? even if I don't get exact solution is there any way to get approximate expression for solution ?

POSTED BY: Debojyoti Mondal
POSTED BY: EDITORIAL BOARD
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