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Homework Help: Also indices

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is
    If x= 2+2 2/3 +2 1/3
    solve
    x 3-6x 2+ 6x-2 =0


    2. Relevant equations

    3. The attempt at a solution
    First i tried to substitute the values but its turning out to be really big( and i get quite scared when it turns up like this and so i didn't go any further this way)
    then I tried to make a cube of x-1
    Like this
    x3 -1 -3x 2+ 3x-3x 2+3x-1+x 3=0
    <==>(x-1)3+(x-1)3=x 3
    <==>2(x-1)3=x 3
    Well I can't get any further!!
    Happy if anyone helps!!
     
    Last edited: Dec 16, 2009
  2. jcsd
  3. Dec 16, 2009 #2

    Mark44

    Staff: Mentor

    There's something screwy in your problem description. It's akin to saying "If x = 3, how many things are in a dozen?" There doesn't seem to be any connection between the two statements. Are we supposed to assume that x = 2 + 22/3 + 21/3 is a solution of the second equation?
     
  4. Dec 16, 2009 #3

    Mark44

    Staff: Mentor

    Why did you pick x - 1? I can't think of any good reason to do this.

    Also, I have no idea of what you're trying to do in the work below.
     
  5. Dec 16, 2009 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I presume you mean "show", not "solve". You are trying to show that value of x is a root of the cubic.

    I think you just quit too soon.
    [tex] 2 = \left (\frac {x}{x-1}\right)^3[/tex]
    [tex]\frac x {x-1} = 2^{\frac 1 3}[/tex]

    Solve for x:

    [tex]x = \frac{2^{\frac 1 3}}{2^{\frac 1 3}-1}[/tex]

    Now all you have to do is show this is equal to 2+2 2/3 +2 1/3
     
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