SOLUTION: to take the 4th root of a -16 isn't possible right? and if it is possible then what would it be?

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Question 420170: to take the 4th root of a -16 isn't possible right? and if it is possible then what would it be?
Found 2 solutions by Theo, richard1234:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
you can take the root of a negative number if the root is odd.

you cannot take the root of a negative number if the root is even.

if you are taking the fourth root and the number is 16, then the root is 2, because 2^4 = 16

if you are taking the fourth root and the number is -16, then the root is not -2, because (-2)^4 = + 16.

can't be done.

not with real numbers.

they invented imaginary numbers to allow you to do things like take the fourth root of -16, but that's another story.

as long as you are dealing with real numbers, the answer is no, you cannot take the fourth root of -16.






Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The fourth roots of -16 are definitely possible, using complex numbers. Let z be a complex number such that z^4 = -16 (we know the magnitude of z must be 2 since 2^4 = 16). We can express z in the form 2e%5E%28i%2Atheta%29 and obtain

z+=+2e%5E%28i%2Atheta%29+=+2%28cos%28theta%29+%2B+i%2Asin%28theta%29%29 (by Euler's formula)

Hence,

z%5E4+=+2%5E4e%5E%284i%2Atheta%29+=+16e%5E%284i%2Atheta%29+=+-16

Therefore, e%5E%284i%2Atheta%29+=+-1. From DeMoivre's theorem, e%5E%284i%2Atheta%29+=+cos%284%2Atheta%29+%2B+i%2Asin%284%2Atheta%29. This number must equal -1, so cos%284%2Atheta%29+=+-1 and sin%284%2Atheta%29+=+0. This implies 4%2Atheta+=+pi+%2B+k%2A2pi --> theta+=+%28pi%2F4%29+%2B+k%2A%28pi%2F2%29, where k is any integer. For now, we can say that theta+=+pi%2F4 and the fourth root of -16 is e%5E%28i%2Api%2F4%29. In a+bi form, this is equal to cos%28pi%2F4%29+%2B+i%2Asin%28pi%2F4%29+=+%28sqrt%282%29%2F2%29+%2B+%28sqrt%282%29%2F2%29i.