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    Plz try this:
    Find integer roots of this equation
    x^2000 + 2000^1999 = x^1999 + 2000^2000.
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    x=2000
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    It's easy to quote the root, but plz explain your work.
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    (Original post by BCHL85)
    It's easy to quote the root, but plz explain your work.

    Exaplian the work?! :confused: Surely if you look at it logically you see that x=2000
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    (Original post by saiyamana)
    Exaplian the work?! :confused: Surely if you look at it logically you see that x=2000
    Yeah I know ... but the question is find the roots ..it means find all the roots of the equation. He didn't prove that it's the only root of the equation. That's what Im looking for.
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    (Original post by BCHL85)
    Plz try this:
    Find integer roots of this equation
    x^2000 + 2000^1999 = x^1999 + 2000^2000.
    2000 is obvious from the equation symmetry. however I'm not sure about the other 1999 roots and whether any of them are integers.
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    (Original post by mik1w)
    2000 is obvious from the equation symmetry. however I'm not sure about the other 1999 roots and whether any of them are integers.
    Right, 2000 is obviously, but if it's just like that ... I wouldn't put it here. Solving the equation is not only show a root of that equation.
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    x^2000 - x^1999 = 2000^2000 - 2000^1999
    (x^1999)(x-1) = (2000^1999).(1999)
    its clear that x=2000 is the only possible solution. (for your peace of mind, take x<2000. then you have, for example, (1500^1999)(1499)=(2000^1999).(1 999), which is impossible since both terms on the R.H.S are greater than those on the L.H.S. Similarly, if you take x>2000, then you'll get two terms on the L.H.S which are both definitely greater than both terms on the R.H.S.)
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    There has to be another real root as the polynomial's degree is even and 2000 isn't a repeated root. The other root is negative - and it is the only other real root.
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    This is the way I did it:
    x^2000 - x^1999 = 2000^2000 - 2000^1999
    (x^1999)(x-1) = (2000^1999).(1999)
    Then as (x,x-1)=1 and the 1999/LHS and 1999 is a prime, therefore x or x-1 must be divisible by 1999.
    If 1999/x then the LHS will be divisible by 1999^1999 but the RHS is not, so 1999/x-1
    So x-1>=1999 and x>=2000, so the LHS>=2000^1999 . 1999
    So the only solution is x=2000
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    No-one' saying the other solution will be an integer.

    A deg 2000 polynomial with have 2000 roots, counting multiplicities and the complex ones will come in complex conjugates.

    So a degree 2000 poly cannot have one real root with no repetitions
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    (Original post by RichE)
    No-one' saying the other solution will be an integer.

    A deg 2000 polynomial with have 2000 roots, counting multiplicities and the complex ones will come in complex conjugates.

    So a degree 2000 poly cannot have one real root with no repetitions
    This is true. I'm unclear on how to find the other value.
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    f(x) = x^2000 - x^1999.
    f'(x) = 2000x^1999-1999x^1998 = x^1998(2000x-1999)
    ->f'(x) = 0 -> x = 0 and x = 1999/2000.
    f(0) = 0, f(1999/2000) = -1.84
    So this equation has 2 real roots, one is greater than 1999/2000 ( x= 2000) and one is smaller than 0, which I haven't found
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    Oh no! I was wrong. It has only 1 real root because f(x) = 2000^2000 - 2000^1999 = 2000^1999.1999 > 0.
 
 
 
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