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Solve this simple equation

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data

    ##z^4-6z^2+1=0##

    2. Relevant equations



    3. The attempt at a solution

    ##z^{2}=\frac{6\pm \sqrt{36-4}}{2}=3\pm 2\sqrt{2}##. Now this has to be wrong already yet I don't know why..

    Wolfram gives me some solutions ##1+\sqrt{2}##, ##1-\sqrt{2}##, ##-1+\sqrt{2}##, ##-1-\sqrt{2}##...

    Now how can I find them without wolfram alpha?
     
  2. jcsd
  3. Apr 10, 2014 #2

    jbunniii

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    No, it's fine. Note that you have solved for ##z^2##, not ##z##. If the goal is to solve for ##z##, you need to do some more work. Start with one of your solutions, say ##z^2 = 3 + 2 \sqrt{2}##. There are two values of ##z## that satisfy this equation, namely ##z = \pm \sqrt{3 + 2 \sqrt{2}}##. Note that this expression can be simplified. Indeed, ##(1 + \sqrt{2})^2 = 3 + 2 \sqrt{2}##, so ##\sqrt{3 + 2 \sqrt{2}} = 1 + \sqrt{2}## is one solution. See if you can find the other solutions. There should be four since the original polynomial ##z^4 -6z^2 + 1## has degree 4.
     
  4. Apr 10, 2014 #3
    See, I couldn't see that. Now It is obvious. Thanks!
     
  5. Apr 10, 2014 #4

    LCKurtz

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    Your original question asked how to get the roots without Wolfram. Once you know that you have $$
    z=\pm\sqrt{3+2\sqrt 2}$$it is easy enough to verify that ##z=1+2\sqrt 2## works, but it's another thing to find that form in the first place. You might find it instructive to set$$
    (a+b\sqrt 2)^2 = 3+2\sqrt 2$$square both sides, and solve for ##a## and ##b##.
     
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