I used Solve[Sqrt[x] + Sqrt[1 - x^2] == Sqrt[2 - 3 x - 4 x^2], x, Reals] // ToRadicals, then I got one real solution.

I have just online. Thank you four your answer.

@minthao_2011 I expect you've understood my answer so I'm glad I could help a bit. In case you find something unclear try to expose the problem more clearly in your question. I find there might be some misleading issues, however I think recent updates of Solve are quite reasonable and in general it works better than formerly.

Perhaps about my English, therefore my question is not clearly. My question is "How many solutions when I solve the equation Sqrt[x] + Sqrt[1 - x^2] == Sqrt[2 - 3 x - 4 x^2] in the real domain?

Thank you very much for your answers and your helps.

@minthao_2011 In general for transcendental equations it might be difficult to decide how many solutions there are, however in our case the equation appears to have two solutions regardless if we search for them in reals or in complexes.

I am a teacher. When I teach for my students, in this case, when we solve by hand, we always conclude the given solution has only real solution (solve in real domain). Am I wrong?

@minthao_2011 I don't think so, certain simple polynomial equations have also complex solutions (if coefficients are real there are conjugate pairs of solutions), thus you should specify what kind of equations you consider and then you have to check carefully all solutions. But as far as an equation is equivalent to a polynomial one the problem simplifies cosiderably.

In my country, when define solution (root) of an equation, first, it belongs to the domain of definition of that equation. In my equation, the solution x -> -1 - Sqrt[2] is omit.

Hi, we can discuss the problem.

@minthao_2011 What is the domain of definition of the equation given?

Why did you exclude this solution -1 - Sqrt[2] ?

I used to find the domain of definition of the equation (because I solve in Real domain) Reduce[{x >= 0 && 1 - x^2 >= 0 && 2 - 3 x - 4 x^2 >= 0}, x]

In a school when children don't know about complex numbers, then it is used this kind of reasoning, however this -1 - Sqrt[2] is also a correct solution

But when you supplement an equation with another conditions then they may rule out this solution -1 - Sqrt[2]

However the supplemented system is not equivalent to the original one

Of course, If I solve that equation in complex domain, I receive solution -1 - Sqrt[2].

I don't agree, -1 - Sqrt[2] is also a real number

This ambiguity comes from the fact that the original equation and the supplemented system are not equivalent

I agree with you -1 - Sqrt[2] is a real number, but it is not satisfying the conditions of Reduce[{x >= 0 && 1 - x^2 >= 0 && 2 - 3 x - 4 x^2 >= 0}, x].

These conditions are added to the original equation

they are not a consequence of the equation Sqrt[x] + Sqrt[1 - x^2] == Sqrt[2 - 3 x - 4 x^2]

Can you explain for me, how is a solution of an equation? Perhaps, my views and you are different.

I demonstrated in my answer with FillSimplify that -1 - Sqrt[2] is also a solution to Sqrt[x] + Sqrt[1 - x^2] == Sqrt[2 - 3 x - 4 x^2]

However it is a negative number thus Sqrt[x] and Sqrt[1 - x^2] are complex

It is another story, in school children don't know about numbers Sqrt[x] for negative x

thus for the sake of simplicity you have to postulate that x is positive

Yes. You understand me.

Nevertheless the complete set of solutions includes also negative numbers.

Thus, my view and your view are different.

OK.

@Artes. Can I disscus my problem in Mapleprimes?

What is it Mapleprimes?

mapleprimes.com

Of course you can

If you agree, I will post my question at there.

I don't know if you understand the problem correctly

Remamber that Mathematica works on complex numbers by default

I understand your answer correctly.

But did you agree with my arguments?

Yes.

@Ar

@Artes Will you help me in next my questions?

If you have another questions I will help when I have time, but recently I was quite a busy

Therefore I couldn't answer too many questions

Thank you. Now I must go best. Good bye.

Bye