Imaginary Solutions to Radical Equations

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SUMMARY

The equation Sqrt(x) + 1 = 0 has no solution in the realm of real numbers, as the square root function is defined to yield only positive real numbers. Conversely, the equation x^2 + 1 = 0 lacks solutions in real numbers unless the imaginary unit i is introduced. Defining a unit for Sqrt(x) + 1 = 0 is deemed unnecessary due to its impracticality. In complex numbers, however, the square root function allows for multi-valued solutions, where -1 can be a square root of 1, leading to x = 1 as the only solution for Sqrt(x) = -1.

PREREQUISITES
  • Understanding of complex numbers and the imaginary unit i
  • Familiarity with the properties of square root functions
  • Knowledge of multi-valued functions in mathematics
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of complex numbers and their applications
  • Learn about multi-valued functions and their implications in mathematics
  • Explore the definitions and properties of square root functions in both real and complex domains
  • Investigate the historical context and practical applications of imaginary numbers
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Mathematicians, students studying advanced algebra, educators teaching complex numbers, and anyone interested in the theoretical aspects of radical equations.

Repetit
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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?
 
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\sqrt{i^{4}} +1 = 0
 
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.
 
Hmm, I see, that makes sense. Thanks to both of you!
 

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