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 Solve[{274.6 = a*Sqrt[x + h] + c, 370 = a*Sqrt[x + h + 1000] + c, 
        159.3 = a*Sqrt[z + h] + c, 161.5 = a*Sqrt[z + h + 1] + c, 
        74.2 = a*Sqrt[1/12 + h] + c}, {a, x, h, c, z}]

Set::setraw: Cannot assign to raw object 274.6. Set::setraw: Cannot assign to raw object 370. Set::setraw: Cannot assign to raw object 159.3. General::stop: "Further output of Set::setraw will be suppressed during this calculation." Solve::naqs: c+a Sqrt[h+x]&&c+a Sqrt[1000+h+x]&&c+a Sqrt[h+z]&&c+a Sqrt[1+h+z]&&c+a Sqrt[1/12+h] is not a quantified system of equations and inequalities. >>

Output Solve[{c + a Sqrt[h + x], c + a Sqrt[1000 + h + x], c + a Sqrt[h + z], c + a Sqrt[1 + h + z], c + a Sqrt[1/12 + h]}, {a, x, h, c, z}]

Any help would be much appreciated.

I have tried using the double ==, however that just gives me a blank output of

{}
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closed as off-topic by Sjoerd C. de Vries, Mr.Wizard Sep 16 '13 at 8:47

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3  
And ... why do you think that there is a solution? BTW, you should use == –  belisarius Sep 16 '13 at 5:00
    
blank means that this function doesnot have a solution –  tintin Sep 16 '13 at 6:51
1  
There are no solutions. There are two "parasite" solutions, that is, they solve the system one would get by making new variables from the radicals e.g. newvarxh for Sqrt[x+h], and adding equations such as "newvarxh^2==x+h. You can find these by using the Solve option VerifySolutions->False. –  Daniel Lichtblau Sep 16 '13 at 13:41
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2 Answers 2

At times it is not possible to obtain an exact solution, consider for instance Solve[x Tan[x]==5,x]. In such cases one can obtain an approximate result with functions like FindRoot.

FindRoot[{274.6 == a*Sqrt[x + h] + c, 370 == a*Sqrt[x + h + 1000] + c,
159.3 == a*Sqrt[z + h] + c, 161.5 == a*Sqrt[z + h + 1] + c, 
74.2 == a*Sqrt[1/12 + h] + 
c}, {{a, .1}, {x, .1}, {h, .1}, {c, .1}, {z, .1}}, 
MaxIterations -> 400]

Functions like FindRoot are based on numerical methods which converge after a number of iterations. That's why I used MaxIterations -> 400. Note that the result would be different with MaxIterations->50. Imaginary parts of the solution can be deleted with Chop. Perhaps this is an unrefined solution but it should get you somewhere:-).

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When you remove your errors and replace you approximate numbers with exact numbers, then Reduce seem to indicate clearly that no solution exists

Reduce[{
  2746/10 == a*Sqrt[x + h] + c,
  370 == a*Sqrt[x + h + 1000] + c,
  1593/10 == a*Sqrt[z + h] + c,
  1615/10 == a*Sqrt[z + h + 1] + c,
  742/10 == a*Sqrt[1/12 + h] + c}, {a, x, h, c, z}]

(* False *)
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I thought that Reduce will also seek for an exact result, which in this particular problem might not exist. –  Gregory Rut Sep 16 '13 at 8:40
    
@GregoryRut But Reduce tells you if it cannot solve for a solution of e.g. an transcendental equation like Reduce[x Tan[x] == 5, x]. I confident that it does return False only, when there is no solution. –  halirutan Sep 16 '13 at 9:08
    
Ah, right. Thanks. –  Gregory Rut Sep 16 '13 at 9:17
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