What are the real roots to the given power equation: 3(x-4)^4/3+16=64

asked Feb 21, 2019 179k views

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3(x-4)^4/3+16 = 64
3(x-4)^4/3 = 48
(x-4)^4/3 = 16
(x-4)^4 = 4096
(x-4)^2 = 64
x^2-8x+16 = 64
x^2-8x-48 = 0
(x+4)(x-12) = 0
x = -4 or x = 12

Dbobrowski answered Feb 26, 2019

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