SOLUTION: (z+9)^2/5=-1 Where z is a real number and simplify! Thank You! :)?

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Question 1015214: (z+9)^2/5=-1
Where z is a real number and simplify!
Thank You! :)?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(z+9)^2/5=-1
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z+9 = -1^(5/2)
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z + 9 = i^5
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z + 9 = i
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z = -9+i
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Note: z is not a Real Number
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Cheers,
Stan H.
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Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
http://www.algebra.com/algebra/homework/Exponents-negative-and-fractional/Exponents-negative-and-fractional.faq.question.1015213.html



This problem is the same as the problem I just did for you: See Problem 1015313

John

My calculator said it, I believe it, that settles it