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# eliminating irrational (Sqrt) terms in long equation

Posted 9 years ago
 I have the following equation with 15 terms. It has basically 6 variables, namely a, b, c, d, e, f. I wish to remove the square roots from this equation by successively squaring and manipulation. Can Mathematica help in it? R1 Sqrt[a] Sqrt[b] + R2 Sqrt[a] Sqrt[c] + R3 Sqrt[a] Sqrt[d] + R4 Sqrt[a] Sqrt[h] + R5 Sqrt[a] Sqrt[i] + R6 Sqrt[b] Sqrt[c] + R7 Sqrt[b] Sqrt[d] + R8 Sqrt[b] Sqrt[i] + R9 Sqrt[b] Sqrt[h] + R10 Sqrt[c] Sqrt[h] + R11 Sqrt[c] Sqrt[i] + R12 Sqrt[c] Sqrt[d] + R13 Sqrt[d] Sqrt[h] + R14 Sqrt[d] Sqrt[i] + R15 Sqrt[h] Sqrt[i] == 0 Attachments:
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Posted 9 years ago
 Your expression has a different set of variables than a, b, c, d, e, f. (This is not important for your question... just pointing it out.) Here are the parameters in your expression: In:= Cases[ R1 Sqrt[a] Sqrt[b] + R2 Sqrt[a] Sqrt[c] + R3 Sqrt[a] Sqrt[d] + R4 Sqrt[a] Sqrt[h] + R5 Sqrt[a] Sqrt[i] + R6 Sqrt[b] Sqrt[c] + R7 Sqrt[b] Sqrt[d] + R8 Sqrt[b] Sqrt[i] + R9 Sqrt[b] Sqrt[h] + R10 Sqrt[c] Sqrt[h] + R11 Sqrt[c] Sqrt[i] + R12 Sqrt[c] Sqrt[d] + R13 Sqrt[d] Sqrt[h] + R14 Sqrt[d] Sqrt[i] + R15 Sqrt[h] Sqrt[i], Sqrt[z_] -> z, \[Infinity]] // Union Out= {a, b, c, d, h, i} But the overall questions is whether this problem is solvable. For the case of two it is trivial. For the case of 3 variables it is solvable. Here is your left hand side in the case of just 3 variables: In:= R1 Sqrt[a] Sqrt[b] + R2 Sqrt[a] Sqrt[c] + R3 Sqrt[a] Sqrt[d] + R4 Sqrt[a] Sqrt[h] + R5 Sqrt[a] Sqrt[i] + R6 Sqrt[b] Sqrt[c] + R7 Sqrt[b] Sqrt[d] + R8 Sqrt[b] Sqrt[i] + R9 Sqrt[b] Sqrt[h] + R10 Sqrt[c] Sqrt[h] + R11 Sqrt[c] Sqrt[i] + R12 Sqrt[c] Sqrt[d] + R13 Sqrt[d] Sqrt[h] + R14 Sqrt[d] Sqrt[i] + R15 Sqrt[h] Sqrt[i] /. Sqrt[Except[a | b | c]] :> 0 Out= Sqrt[a] Sqrt[b] R1 + Sqrt[a] Sqrt[c] R2 + Sqrt[b] Sqrt[c] R6 So, taking this and working with it in one iteration: In:= Expand[(Sqrt[a] Sqrt[b] R1)^2 == (Sqrt[a] Sqrt[c] R2 + Sqrt[b] Sqrt[c] R6)^2] Out= a b R1^2 == a c R2^2 + 2 Sqrt[a] Sqrt[b] c R2 R6 + b c R6^2 And clearly it requires only one more iteration to remove the Sqrt expressions in this case.Here is your expression's left hand side with 4 variables: In:= R1 Sqrt[a] Sqrt[b] + R2 Sqrt[a] Sqrt[c] + R3 Sqrt[a] Sqrt[d] + R4 Sqrt[a] Sqrt[h] + R5 Sqrt[a] Sqrt[i] + R6 Sqrt[b] Sqrt[c] + R7 Sqrt[b] Sqrt[d] + R8 Sqrt[b] Sqrt[i] + R9 Sqrt[b] Sqrt[h] + R10 Sqrt[c] Sqrt[h] + R11 Sqrt[c] Sqrt[i] + R12 Sqrt[c] Sqrt[d] + R13 Sqrt[d] Sqrt[h] + R14 Sqrt[d] Sqrt[i] + R15 Sqrt[h] Sqrt[i] /. Sqrt[Except[a | b | c | d]] :> 0 Out= Sqrt[a] Sqrt[b] R1 + Sqrt[c] Sqrt[d] R12 + Sqrt[a] Sqrt[c] R2 + Sqrt[a] Sqrt[d] R3 + Sqrt[b] Sqrt[c] R6 + Sqrt[b] Sqrt[d] R7 An algorithmic approach presents itself. 1) pick one of the variables2) collect all terms with the Sqrt of that variable and factor out that Sqrt. 3) Isolate that expression on one side of the equation with all other terms on the other side.4) square both sides (now that variable no longer appears in the overall expression as a Sqrt5) Iterate with the next variable6) once one had gone through all of the variables you will now have an expression with all Sqrts removedAll that you need to do now is code this in a Mathematica function. Note that the collecting of the terms can be done with the Collect function.Note that this approach works for the sort of expression that you have presented: a binomial in only the Sqrts of the parameters. It will not generalize to most other cases.
Posted 9 years ago
 Perfectly replied David Sir. I have found my solution, after following your valuable suggestion.Thanks. Regrats for late reply
Posted 9 years ago
 I am pleased that I was able to help.