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I am trying to get a solution of equation with many parameters. The general solution does not seem to reduce to a specific one.

The specific solution with all parameters known is

Block[{t = 1, λ = 1, m = 1/4, β = 2}, Solve[2 m^2 + 4 m t + 3 t^2 + λ^2 - 4 t (m + t) x + (t^2 - λ^2) (2 x^2 - 1) == β^2/2, x]]

This gives $x=5/8$.

This is the general solution

sol = Solve[2 m^2 + 4 m t + 3 t^2 + λ^2 - 4 t (m + t) x + (t^2 - λ^2) (2 x^2 - 1) == β^2/2, x] // FullSimplify

Inserting the parameters gives Indeterminate:

sol /. {t -> 1, λ -> 1, m -> 1/4, β -> 2} // N

I tried even to find the Limit when $t\rightarrow\lambda$, but it still gives Indeterminate.

Block[{m = 1/4, β = 2}, Limit[sol[[1, 1, 2]], t -> λ]]
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  • $\begingroup$ 1. Please put a bit more effort in typesetting your question properly, and using a more informative title. 2. Block[{m = 1/4, \[Beta] = 2, \[Lambda] = 1}, Limit[sol[[1, 1, 2]], t -> \[Lambda]]] works. 3. Block[{m = 1/4, \[Beta] = 2}, Limit[sol[[1, 1, 2]], t -> \[Lambda], Assumptions -> \[Lambda] > 0]] also works, but the Assumptions should not be necessary, and e.g. Assumptions -> \[Lambda] < 0 leads to incorrect result, which looks like a bug. $\endgroup$ Commented Nov 4 at 8:59
  • $\begingroup$ Thanks@xzczd, I edit the text. $\endgroup$ Commented Nov 4 at 9:15
  • $\begingroup$ You are asking why the solution a quadratic in x with a parametrized leading coefficient acts funny when you substitute a value that kills the leading coefficient? I'd be concerned if it acted any other way. $\endgroup$ Commented Nov 4 at 18:01

2 Answers 2

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For the parameter values problem,Solve could not get the full-dimensional component,we have to use Reduce.

sol = Reduce[ 2 m^2 + 4 m t + 3 t^2 + λ^2 - 4 t (m + t) x + (t^2 - λ^2) (2 x^2 - 1) == β^2/2, x] /. {t -> 1, λ -> 1, m -> 1/4, β -> 2}

x == 5/8

  • Compare with
Solve[2 m^2 + 4 m t + 3 t^2 + λ^2 - 4 t (m + t) x + (t^2 - λ^2) (2 x^2 - 1) == β^2/2, x, MaxExtraConditions -> All] /. {t -> 1, λ -> 1, m -> 1/4, β -> 2}
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I do not see any problems with the solution provided by Solve, limit gives correct result too.

Limit[SolveValues[ 2 m^2 + 4 m t + 3 t^2 + λ^2 - 4 t (m + t) x + (t^2 - λ^2) (2 x^2 - 1) == β^2/2, x], Flatten[ ToRules /@ {t == 1, λ == 1, m == 1/4, β == 2}]] 
{5/8, Indeterminate}
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  • $\begingroup$ Limit is indeed having trouble with symbolic λ, see 3rd item of my comment above. $\endgroup$ Commented Nov 4 at 9:29
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    $\begingroup$ @xzczd I think that would be a different question. OP seems to ask why particular solution can not be directly substituted to the equation. But it can if the limit is used properly. $\endgroup$ Commented Nov 4 at 9:50

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