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I am a Mathematica beginner. I want to know the X. X satisfied with below equation. enter image description here

$T,M,λso,tt,R$ are positive numbers.And $-1<X<1$. So I wrote code;

enter image description here

But Mathematica8 will not response. What should I do next? Is it possible to know the answer analytically? (Of course, I can do numerical, but I want to know the answer analytically.)

Solve[2*T^2 - 8*\[Lambda]so^2 - (2*((T^2*(M^2 + tt^2))/2 - (8*M*R*T*X*\[Lambda]so)/Sqrt[1 - X^2] - 32*R^2*\[Lambda]so^2))/Sqrt[(T^2*(M^2 + tt^2)*(1 + X))/2 + 8*M*R*T*Sqrt[1 - X^2]*\[Lambda]so + 4*R^2*(tt^2 + 8*\[Lambda]so^2 - 8*X*\[Lambda]so^2)] == 0,X]
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1 Answer 1

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My approach is convert the equation to polynomial system, it seems that the result is root of sextic equation.

expr = 2 T^2 - 8 λso^2 - (2 (1/2 T^2 (M^2 + tt^2) - (8 M R T X λso)/Sqrt[1 - X^2] - 32 R^2 λso^2))/Sqrt[1/2 T^2 (M^2 + tt^2) (1 + X) + 8 M R T Sqrt[1 - X^2] λso +  4 R^2 (tt^2 + 8 λso^2 - 8 X λso^2)];
expr /. {Sqrt[1-X^2]->Y, 1/Sqrt[1-X^2]->1/Y} //Together // Numerator // Collect[#,Sqrt[_]]&
#1^2 == #2^2 & @@ %
Solve[{%, X^2+Y^2==1}, X, {Y}] // AbsoluteTiming

enter image description here

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  • $\begingroup$ Could you post an English version of the output? $\endgroup$ Commented Sep 6, 2017 at 3:43
  • $\begingroup$ @J.M. No problem. $\endgroup$
    – chyanog
    Commented Sep 6, 2017 at 3:48
  • $\begingroup$ Thank you very much. But I could not understand what we have done all. $\endgroup$
    – Sakurai.JJ
    Commented Sep 6, 2017 at 5:07

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