       Solve Limitations

• To: mathgroup at smc.vnet.net
• Subject: [mg62963] Solve Limitations
• From: "Marcelo Mayall" <mmayall at bol.com.br>
• Date: Fri, 9 Dec 2005 05:10:42 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Let's suppose that we are interested in the roots analytic expression of
the following function:
In :=  f = a x + b x^(3/2) + c;
The function Solve could be used:
In :=  sol = Solve[f==0, x];
Defining the values of the constants a, b, c would return the following numeric values:
In :=  froots = Solve[f==0, x]/. {a->1, b->1, c->1} //N
Out =  {{x-> 2.1479}, {x-> -0.57395 + 0.368989 I}, {x-> -0.57395 - 0.368989 I}}
However, f is not null for those values and therefore, these are not the roots of f:
In :=  f/. froots/. {a-> 1, b-> 1, c-> 1} //Chop
Out =  {6.2958, 0, 0}
At first, it seems that the function Solve doesn't take appropriately in
consideration the term in square root.
Some idea to obtain the correct analytic solution of f ??? Or, in fact, this a limitation of the function Solve???

Thanks,
Marcelo Mayall

```

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