Question 1004388
Without any computation, x = 1 *has* to be the only real solution because the functions x^2 and sqrt(x) are strictly increasing when x >= 0 (note: if x < 0, then the LHS is non-real). Because x = 1 is a solution, it must be the only real solution.


Assuming complex-valued square roots, you can solve by isolating sqrt(x) and squaring:


*[tex \large \sqrt{x} = 2 - x^2]
*[tex \large x = (2 - x^2)^2 = 4 - 4x^2 + x^4]

And so on.