Solve an equation in $\mathbb{R}^+$

Question

How can I solve this equation (both numerically and literally) only in the positive reals $\mathbb{R}^+$?

Solve[x == (v0 - (A CD t v0^2 ρ)/(4m)) Cos[θ] t, t]

And for example, is there a way to have an output like this :

52.0756

and not like this :

{{t -> -52.3918}, {t -> 52.0756}}

?

Answer 1

The first items of More Information in the documentation of Solve says :

• The system expr in Solve[expr,vars] can be any logical combination of:

  lhs == rhs               equations 
  lhs != rhs               inequations 
  lhs > rhs or lhs >= rhs  inequalities 
  expr ∈ dom               domain specifications 
  ForAll[x,cond,expr]      universal quantifiers 
  Exists[x,cond,expr]      existential quantifiers 
  

• Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].

Every expri can be an equation, inequality as well as an expression tests like e.g. Positive or Negative etc., thus we can do simply e.g. Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x], but to get only the list of solutions (without Rules ) there are at least two ways:

• using ReplaceAll (shorthand /.) (mentioned by Markus Roellig) with the condition x > 0 :

  x/.Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x]
  

  {1, 2, 3}
  

• using Part (shorthand [[]]) with e.g. x > 0 or with an expression test like Positive, NonNegative etc.:

  Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && Positive[x], x][[All, 1, 2]]
  

   {1, 2, 3} 
  

The above ways can be mixed, e.g. : x /. Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x][[3]]. We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use it like e.g.

Reduce[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x][[All, 2]]

Answer 2

You could use ReplaceAll (i.e. /.) and Select

Select[x /. Solve[x^2 - 1 == 0, x], Positive]

gives

{1}

It is a list (List) not a single number. You might not know how many positive solutions exist:

Select[x /. NSolve[(x - 1) (x + 3) (x - 3) == 0, x], Positive]

{1., 3.}

Edit:

Using Part or its short-hand notation [[]] you can select parts from the list:

Part[{1}, 1]

1

{1}[[1]]

1

Part[{1.,3.},2]

3.