Kenneth S. answered • 08/06/17
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The left side is the fourth root of [(x-2)(x+2)(x2+4)].
This radicand must be positive and therefore the domain is
(-infinity,-2] or [2,infinity).
The left side can be 0 or more, never negative since it represents a principal fourth root.
Therefore the solution set is (-infinity,-2] ∪ 2.
No x value > 2 works, because the left side will be the fourth root of x4 - 16, which will always be < x.
This is quite a nice problem!
Mako O.
In short how did you calculate the domain.
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08/06/17
Andrew M.
x2 + 4 will always be positive regardless.
For the 4th root on left side to be positive either
(x-2) and (x+2) are both negative ... giving domain (-infinity, -2]
or
(x-2)(x+2) are both positive... giving domain [2, infinity)
The -2 and 2 are included since the left can also be equal to zero
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08/06/17
Kenneth S.
Andrew has answered your further questions very well. (In a radicand that must be non-negative, there could be two factors that are each negative, thus making the radicand positive over certain intervals, as cited.)
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08/06/17
Mako O.
08/06/17