Imaginary Solutions to Radical Equations

[Repetit]

Does the equation Sqrt(x) + 1 = 0 have a solution? I would say that it doesnt. But the equation x^2 + 1 = 0 doesn't have a solution either, unless you define the imaginary unit i as the solution to the equation.

So why don't one define some unit which is the solution to the equation Sqrt(x) + 1 = 0? Is it simply because it is of no practical use?

 

[Sourabh N]

\sqrt{i^{4}} +1 = 0

 

[HallsofIvy]

Of course, i⁴= 1!

Repetit, if you are referring to the square root function as defined for real numbers, then \sqrt{x} is specifically DEFINED as "the positive real number whose square is x". By that definition, it is impossible that \sqrt{x}= -1.

If, however, you are referring to the square root function as defined for complex numbers, where 'multi-valued' functions are allowed, then -1 is one of the two square roots of 1. The only solution to the equation \sqrt{x}= -1 is x= 1.

 

[Repetit]

Hmm, I see, that makes sense. Thanks to both of you!