Current Events > Brutal math question. What are the values of x such that [x^3]=[x^2] ?

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FLOUR
05/31/22 6:34:12 PM
#1:


Note that [ ] represents the floor function. So for instance, [1.7]=1 and [e]=2 .

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DrizztLink
05/31/22 6:35:04 PM
#2:


Many.

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TerrifyingRei
05/31/22 6:35:12 PM
#3:


[Confused screaming]

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Gobstoppers12
05/31/22 6:35:57 PM
#4:


I don't do math lol

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pogo_rabid
05/31/22 6:36:39 PM
#5:


Nice try. Not doing your homework for you, buster.

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ForsakenHermit
05/31/22 6:36:41 PM
#6:


1

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WaterLink
05/31/22 6:37:49 PM
#7:


does x have to be an integer? because if not I'm pretty sure there's infinite amounts.

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Garlands_Soul
05/31/22 6:38:58 PM
#8:


0 also

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MattcheteGuy
05/31/22 6:39:08 PM
#9:


the answer to every single math question is either 69, 420, about tree fiddy or some combination of any of those.

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FLOUR
05/31/22 6:39:54 PM
#10:


WaterLink posted...
does x have to be an integer? because if not I'm pretty sure there's infinite amounts.


Definitely not. You can express the solution as one or more intervals.

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#11
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WaterLink
05/31/22 6:49:54 PM
#12:


FLOUR posted...
Definitely not. You can express the solution as one or more intervals.

then I guess the negative cubed root of 2 through the cubed root of 2, non inclusive.

or maybe 0 though the cubed root of two since negatives remain negative when cubed. idk it's been a while but it's somewhere in that ballpark

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AngelsNAirwav3s
05/31/22 7:00:18 PM
#13:


[0, 1.25992]
[1.41421, 1.44225]

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Lunar_Savage
05/31/22 7:03:44 PM
#15:


1, 0, and the absolute value of |-1| all fit.

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thronedfire2
05/31/22 7:04:12 PM
#16:


42

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EmbraceOfDeath
05/31/22 7:12:24 PM
#17:


Would it be the interval (x^(1/3) - x^(1/2)), where x is an integer?

Never mind, way off.

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8964752
05/31/22 7:16:15 PM
#18:


0 x < 2^(1/3)
2^(1/2) x < 3^(1/3)

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warlock7735
05/31/22 7:30:26 PM
#19:


[0, cubert(2)), [sqrt(2), cubert(3))

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