MaplePrimes - answers and comments on Question, solve an equation

Multiplication signs

Markiyan Hirnyk — Wed, 23 Jan 2013 19:04:25 Z

Inserting the omitted multiplication signs, we obtain
> sol := solve(3*a*(1+z)^2 = 2*(a*(1+z)^3+b)^(1/2), z);

> [allvalues(eval(sol, [a = 1, b = 2]))]:
> evalf(%);
[0.1043743377, -0.8917958088 + 0.9522333959 I, -1.876338276,

-0.8917958088 - 0.9522333959 I]
See ?solve , ?solve/details , ?RootOf , ?allvalues , and ?eval for more details.

Two many solutions

Preben Alsholm — Wed, 23 Jan 2013 20:59:02 Z

By using solve with symbolic a and b, you will get 4 "solutions". However, they are not solutions all of them, because solve squares both sides of the equation to get rid of the square root. By doing that it includes the solutions of the equation having opposite sign on one side of the equation.
When a and b are of type numeric, solve is more careful.

restart;
eq:=3*a*(1+z)^2 = 2*(a*(1+z)^3+b)^(1/2);
solve(eq,z);
#Notice the 4th degree polynomial inside RootOf
#Now we do manually as said above:
map(x->x^2,eq);
#We have our 4th degree polynomial (with z instead of _Z):
collect((lhs-rhs)(%),z);
#Now let a and b be concrete from the start:
eq1:=eval(eq,{a=1,b=2});
#solve finds 2 solutions this time (now notice the indices singling out roots of the polynomial):
solve(eq1,z);
convert([%],radical); #If you want to see.
evalf(%);

So one has to be careful.

Symbolic solutions in Maple

Kitonum — Thu, 24 Jan 2013 01:36:29 Z

I think that we should not trust to the symbolic solutions of the equations with the parameters obtained with Maple. Here are two simple examples.

with(RealDomain):

solve(sqrt(x-a)=x, x);

solve(a*x^2-2*x+4=0, x);

Both answers are incorrect. The first equation has two solutions only in the range 0 <=a< 1/4 . For other values of the parameter a there are no solutions or only one solution. This is easily seen by constructing graphs.

The solution of the second equation is correct only for a<>0 and a<=1/4 .

Both equations can be correctly solved in the package Mathematica. Here, for example, the first example:

Your equation 3a(1+z)^2=2(a(1+z)^3+b)^(1/2) with two parameters a and b is much more complex. You can correctly and explicitly solve it in Mathematica by using commands

ToRadicals[

Reduce[3*a*(1 + z)^2 == 2*(a*(1 + z)^3 + b)^(1/2), z, Reals]]

The output is very cumbersome.

solve an equation

golnaz — Wed, 23 Jan 2013 19:54:56 Z

thanks for the answer but i want to solve the equation analytic not numeric

By allvalues command

Markiyan Hirnyk — Wed, 23 Jan 2013 20:05:49 Z

@golnaz Up to ?allvalues , the command allvalues(sol); produces the symbolic solution of the equation under consideration. See its long output in allvalues.mw

analytic not numeric

golnaz — Wed, 23 Jan 2013 20:33:49 Z