When dealing with imaginary and complex numbers, are there 2 solutions?
I.e.: x^2=4
x=2,-2
For: x^2=-4
x=2i or -2i ?
(Is there a 2i and -2i, like for real numbers?)
try it
(2i)^2 = 4 i^2 = 4(-1) = -4
(-2i)^2 = 4 i^2 = 4(-1) =-4 sure enough
In other words, yes
Thank you!
When dealing with imaginary and complex numbers, the equation x^2 = 4 indeed has two solutions in the real number system: x = 2 and x = -2. However, when working with complex numbers, there is an additional possibility involving the imaginary unit "i."
The imaginary unit "i" is defined as the square root of -1, which means that i^2 = -1. With this in mind, when solving the equation x^2 = -4, we can rewrite it as x^2 + 4 = 0.
To find the solutions, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.
In this case, a = 1, b = 0, and c = 4. Substituting these values into the quadratic formula, we get:
x = (-0 ± √(0^2 - 4 * 1 * 4)) / 2 * 1
= (± √(-16)) / 2
= (± 4i) / 2
= ± 2i
Therefore, when solving x^2 = -4 in the complex number set, we get two solutions: x = 2i and x = -2i.
To summarize, for equations involving complex numbers, there can indeed be two solutions. In this case, the solutions are x = 2i and x = -2i.