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Finding root

  1. Oct 19, 2005 #1

    Meh

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    Find all real/imaginary roots to

    x^9=16x
     
  2. jcsd
  3. Oct 19, 2005 #2
    have you tried to solve this yourself?

    try completing the square
     
  4. Oct 19, 2005 #3
    or maybe a diff of squares
     
  5. Oct 19, 2005 #4

    Meh

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    I'm stuck right after i bring the 16x over...
    x^9-16x=0

    Where do I go from here?I have tried this, it was a test question for me today. Didn't get it so just wondering what the answer is.
     
  6. Oct 19, 2005 #5

    Tide

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    HINT 1: Factor! :)

    HINT 2: Think Euler!
     
  7. Oct 19, 2005 #6

    Meh

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    Can someone just give the answer? I don't got a clue on how to factor it : (
     
  8. Oct 19, 2005 #7

    Tom Mattson

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    No, we will not just give the answer.

    Start from your equation: x^9-16x=0

    What's the first thing you should look for when factoring? A common factor.
     
  9. Oct 20, 2005 #8
    x[x^8 -16]=0
    x=0,x^8=16 now solve the latter
     
  10. Oct 20, 2005 #9

    Diane_

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    mathelord - If he does it your way, he'll miss some roots - all of the complex ones, actually.

    Meh - do as Tide suggested. Factoring is the way to go. Let me suggest you go back and review some of the basic factoring patterns - sum of cubes, difference of cubes, things of that nature.
     
  11. Oct 20, 2005 #10
    Meh - if you've dealt with polynomial equations before, you might remember that a polynom of n-th degree has n roots (real or complex or combination of both).
    So P8(x) = x8 - 16 has 8 roots.

    One way of finding them is applying a very useful DeMoivre's Theorem to
    x8 = 16
    and extracting a root of 8th degree.
    If it's not in your course, it's really worth mastering.
    If you do, it will give you a serious sense of satisfaction.

    Otherwise it can be done the way Tide and Diane_ suggested, except that Tide's "Euler hint" may not be needed.
     
    Last edited: Oct 21, 2005
  12. Oct 22, 2005 #11
    Do you know how to solve
    x^2 + 2 = 0 ?
    and
    x^4 + 4 = 0?
    Look at the first one.
    (x+i√2)(x-i√2)=0
    What for the second x^4 + 4=x^4+4x^2+4-4x^2;
    maybe you can continue and find your answer
     
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