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The second itemfirst items of More Information in the documentation of Solve says :

Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].    
  • 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 or an inequality in the variables vars, inequality as well as an expression testtests like e.g. Positive[x]Positive or NonNegative[x]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:

  1. 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}
    
  2. using Part (shorthand [[]]) with e.g. x > 0 or with the 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} 
    
  • 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} 
    

WeThe 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.

The second item of More Information in the documentation of Solve says :

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

Every expri can be an equation or an inequality in the variables vars, as well as an expression test like e.g. Positive[x] or NonNegative[x] etc., thus we can do simply 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:

  1. 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}
    
  2. using Part (shorthand [[]]) with e.g. x > 0 or with the 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} 
    

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 e.g.

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.

2 extended discussion
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The second item of More Information in the documentation of Solve says :

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

Every expri can be equations an equation or inequalitiesan inequality in the variables of vars, as well as an expression test like e.g. Positive[x] or NonNegative[x] etc., thus we can do simply 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:

Solve[ x^2 - 1 == 0 && x > 0, x ]
{{x -> 1}}
  1. 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}
    
  2. using Part (shorthand [[]]) with e.g. x > 0 or with the 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} 
    

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 e.g. Reduce[x^2 - 1 == 0 && x > 0, x].

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

The second item of More Information in the documentation of Solve says :

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

expri can be equations or inequalities in the variables of vars, thus we can do simply :

Solve[ x^2 - 1 == 0 && x > 0, x ]
{{x -> 1}}

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 e.g. Reduce[x^2 - 1 == 0 && x > 0, x].

The second item of More Information in the documentation of Solve says :

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

Every expri can be an equation or an inequality in the variables vars, as well as an expression test like e.g. Positive[x] or NonNegative[x] etc., thus we can do simply 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:

  1. 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}
    
  2. using Part (shorthand [[]]) with e.g. x > 0 or with the 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} 
    

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 e.g.

Reduce[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x][[All, 2]]
1
source | link

The second item of More Information in the documentation of Solve says :

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

expri can be equations or inequalities in the variables of vars, thus we can do simply :

Solve[ x^2 - 1 == 0 && x > 0, x ]
{{x -> 1}}

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 e.g. Reduce[x^2 - 1 == 0 && x > 0, x].