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Prove that x^4+x^2=1

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whats x ?
variable :P
i mean is it cube root of unity ? or imaginary number i or something else ? if not, then u just said its variable, so x^4+x^2 can't equal 1.

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Other answers:

or do u need to solve for x ???
I don't know that is what I am trying to figure out... this was the question in my paper and I left it blank
just prove it to 1
by means of anything
good luck with that :)
i think question is incomplete...we can solve for x, but can't prove that =1...
x= 1 :P
nopes, x not =1 :P
then x= -1 :P
nopes, x not =-1 :P
x^4 +x^2=1 => => x^4 +x^2 -1 = 0 => (x +0.786)*(x -0.786)*(x^2 +1.618) = 0 Real solutions: Root 1: -0.786 Root 2: 0.786 Complex roots: Root 3: 0+1.272 * i Root 4: 0-1.272 * i
are u solving that for x ,saura ?
there you go ^^
u copied that ^
yeah should i give the source ? :P
it is equal to 1
take a screenshot of your question
|dw:1353238271333:dw|Looks like no solution
1 Attachment
koli, any luck with screenshot ?
Let x^2=y
thats a good substitution to solve for x
u get quadratic in y
|dw:1353238793125:dw| So, only two real roots
so for any of the 4 values of x we found above, u can prove that x^4+x^2= 1
how it is equal to 1??
solving for x in x^4+x^2=1 means ginding vales of x for which x^4+x^2 equals 1 so, for the 4 values we found, x^4+x^2 will equal 1.
what ?

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