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Easy Question

  1. Feb 24, 2005 #1
    find x:

    [tex]1+8x^{-1}+15x^{-2}[/tex]

    I've got so far:

    [tex]\frac{x^2+8x+15}{x^2}[/tex]
     
    Last edited: Feb 24, 2005
  2. jcsd
  3. Feb 24, 2005 #2
    find x for
    [tex]1+8x^{-1}+15x^{-2}[/tex]

    and so far you got
    [tex]\frac{x^2+8x+15}{x^2}[/tex]

    x can be anything, all real number, or even complex number.... apple, orange, banana.... just put what ever you want
     
  4. Feb 24, 2005 #3
    sorry :) LOL ....ya a banana.....ok just simplify
     
  5. Feb 24, 2005 #4
    If they're asking for all real numbers of x, then x can be anything but zero. Otherwise, like vincentchan said, x can be anything, even complex.
     
  6. Feb 24, 2005 #5
    sorry :) LOL ....ya a banana.....ok just simplify.
     
  7. Feb 24, 2005 #6
    (x+3)(x+5)=x^2+8x+15
     
  8. Feb 24, 2005 #7
    vincent I know that but there is a x^2 underneath!

    The answer in the book is

    [tex](1+\frac{5}{x})(1+\frac{3}{x})[/tex]
     
  9. Feb 24, 2005 #8
    can you simplify from my solution to your text book's one?
    (x+3)(x+5)/x^2=(1+3/x)(1+5/x)
     
  10. Feb 24, 2005 #9
    ;) got it thanks mr.
     
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