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Math Help - Higher degree equations

  1. #1
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    Higher degree equations

    (2x) / (1+x^2)^2 = 2/5

    I completely forgot how to solve equations with x^4
    can someone tell me how to do this?
    I take the denominator to the other side and factored it and got an equation with x to the 4th degree.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfyz View Post
    (2x) / (1+x^2)^2 = 2/5

    I completely forgot how to solve equations with x^4
    can someone tell me how to do this?
    I take the denominator to the other side and factored it and got an equation with x to the 4th degree.
    what was the equation you got? show us and we'll take it from there, there is no one way to attack a quartic, we must do it by case

    EDIT: ok, this is not a "nice" quadratic, it does not have integer roots. so if you have to solve this by hand i see no other way to do it at the moment other than to use the quartic formula. which is a mess. you can use Cardano's or Galois' method (your professor is cruel)
    Last edited by Jhevon; September 22nd 2007 at 10:13 PM.
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    what was the equation you got? show us and we'll take it from there, there is no one way to attack a quartic, we must do it by case

    EDIT: ok, this is not a "nice" quadratic, it does not have integer roots. so if you have to solve this by hand i see no other way to do it at the moment other than to use the quartic formula. which is a mess. you can use Cardano's or Galois' method (your professor is cruel)
    I have to solve for x for the equation above.
    I did it on the calculator but i thought there was a way to do it by hand.
    I guess not! (I don't think my teacher wants me to use the quartic..) lol, I think calculator is fine.. I hope
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfyz View Post
    I have to solve for x for the equation above.
    I did it on the calculator but i thought there was a way to do it by hand.
    I guess not! (I don't think my teacher wants me to use the quartic..) lol, I think calculator is fine.. I hope
    ok, good. no worries then. because doing it by hand, as you can see, is a pain. sometimes we can do it pretty easily by hand, but not in this case
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  5. #5
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    Quote Originally Posted by xfyz View Post
    (2x) / (1+x^2)^2 = 2/5

    I completely forgot how to solve equations with x^4
    can someone tell me how to do this?
    I take the denominator to the other side and factored it and got an equation with x to the 4th degree.
    Let x=\tan y then we have,
    \frac{2\tan y}{(1+\tan^2 y)^2} = \frac{2}{5}
    Thus,
    \frac{2\tan y}{\sec^4 y} = \frac{2}{5}
    Thus,
    2\tan y\cos ^4 y = \frac{2}{5}
    Which means,
     2 \sin y \cos y \cos^2 y = \frac{2}{5}
    Identities,
    \sin 2y \cos^2 y = \frac{2}{5}
    More,
    (1 - 2\cos^2 y)(\cos^2 y) = \frac{2}{5}
    Let z=\cos^2 y to get,
    (1-2z)z = \frac{2}{5}.
    The above is quadradic which is solvable.
    Then you need to convert \cos into \tan .
    And then take arctangent to get final answer.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    Let x=\tan y then we have,
    \frac{2\tan y}{(1+\tan^2 y)^2} = \frac{2}{5}
    Thus,
    \frac{2\tan y}{\sec^4 y} = \frac{2}{5}
    Thus,
    2\tan y\cos ^4 y = \frac{2}{5}
    Which means,
     2 \sin y \cos y \cos^2 y = \frac{2}{5}
    Identities,
    \sin 2y \cos^2 y = \frac{2}{5}
    More,
    (1 - 2\cos^2 y)(\cos^2 y) = \frac{2}{5}
    Let z=\cos^2 y to get,
    (1-2z)z = \frac{2}{5}.
    The above is quadradic which is solvable.
    Then you need to convert \cos into \tan .
    And then take arctangent to get final answer.
    Clever! I always forget to look for something like this.

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    ...you can use Cardano's [method]...
    For the record, Cardano's method solves cubic equations, not quartic.

    -Dan
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  8. #8
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    And it is not called the Galois method. That links derived the method for 4th degree equations using Galois Theory.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    i was speaking loosely after all. i wasn't sure what the method's were called. i just named them by the name i saw on the webpage
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