What are the steps to solve this quartic equation?

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To solve the quartic equation, initial attempts to find rational roots using factors of 9 were unsuccessful. It is suggested that approximating the roots or checking for a typo in the problem may be necessary. The quartic formula is available but is complex and not typically expected for such problems. A calculator provided roots of x=3±√6, which can help in finding all roots and factoring the equation. Further resources, such as Wikipedia and Mathematica, are recommended for additional assistance.
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Homework Statement



Solve the equation http://img593.imageshack.us/img593/2523/msp662119eai22g9fe584bh.gif


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The Attempt at a Solution



I tried to find the factor of the equation by substituting the factor of 9, that is +-1,+-3 and +-9 into the equation but none of them equal 0. I had no ideas how to solve it unless i get one of the roots of the equation... Can anyone give me any hints or tips? Thanks.
 

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I'm afraid you're out of luck then. If this question was from school, maybe they were expecting you to approximate the roots, or there was a typo made somewhere, or they gave you a quadratic factor. If not the only other way is to use the quartic formula, and that's not something you want to do or what they would be expecting of you.

Using a calculator, I got roots x=3\pm \sqrt{6}. This information should be sufficient for you to find all the roots and factorize it into its real quadratic factors or complex linear factors.
 
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You can have a look at this page,http://en.wikipedia.org/wiki/Quartic_equation"
Or using Mathematica,you can get four roots.
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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