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Thread: A quintic equation!

  1. #1
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    A quintic equation!

    Solve this quintic equation: $\displaystyle x^5+ax^3 + \frac{1}{5}a^2x + b = 0, \ \ a,b \in \mathbb{R}.$

    Note: You don't have to find the roots. Just show us how to find them!
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  2. #2
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    Let me try it !


    $\displaystyle \cos(5x) = \cos(2x+3x) = \cos(2x)\cos(3x) - \sin(2x)\sin(3x) $

    $\displaystyle = ( 2\cos^2(x) - 1)[4\cos^3(x) - 3\cos(x)] - 2\sin(x)\cos(x) \sin(x) [ 3 - 4(1-\cos^2(x))] $

    $\displaystyle = = ( 2\cos^2(x) - 1)[4\cos^3(x) - 3\cos(x)] - 2\cos(x)(1- \cos^2(x)) [ 4\cos^2(x) -1 ] $

    $\displaystyle = \cos^5(x) [ 8 + 2(4) ] + \cos^3(x)[ -4 -6 -2(4) -2 ] + \cos(x) [ 3 + 2 ] $

    $\displaystyle \cos(5x) = 16\cos^5(x) - 20\cos^3(x) + 5\cos(x) $

    The next step is substituting $\displaystyle \cos(x) = ku $ , $\displaystyle k $ is a constant

    $\displaystyle
    16(ku)^5 - 20(ku)^3 + 5(ku) - \cos(5x) = 0
    $

    $\displaystyle u^5 - \frac{5}{4k^2}u^3 + \frac{5}{16k^4} u - \frac{1}{16k^5}\cos(5x) = 0 $

    Let $\displaystyle \frac{5}{4k^2} = -a \implies k= \sqrt{\frac{-5}{4a} }$

    then let's see what the coefficient of $\displaystyle u $ is

    $\displaystyle \frac{5}{ 16 ( \frac{5}{4a} )^2} = \frac{a^2}{5} $

    Which equals to that in the equation $\displaystyle u^5 + au^3 + \frac{a^2}{5}u + b = 0 $

    and finally compare the remaining coefficient : $\displaystyle x^0 $

    we will see that $\displaystyle - \frac{1}{16k^5}\cos(5x) = b $ so that we can obtain the values of $\displaystyle x $ , the root are $\displaystyle \frac{1}{k}\cos(x_1) , \frac{1}{k}\cos(x_2) , \frac{1}{k}\cos(x_3) ,\frac{1}{k}\cos(x_4) ,\frac{1}{k}\cos(x_5) $
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  3. #3
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    Here is another method :

    We have $\displaystyle a^5 + b^5 = (a+b)^5 - 5ab(a+b)^3 + 5(ab)^2(a+b) $ or

    $\displaystyle (a+b)^5 - 5ab(a+b)^3 + 5(ab)^2(a+b) - (a^5 + b^5) = 0$

    Sub. $\displaystyle a+b = x $

    $\displaystyle x^5 - 5(ab)x^3 + 5(ab)^2 x - (a^5 +b^5) = 0$

    Again , let $\displaystyle ab = \frac{-A}{5}$

    we will obtain $\displaystyle x^5 + Ax^3 + \frac{A^2}{5}x - (a^5+b^5) = 0
    $

    and let $\displaystyle B = -(a^5+b^5) $

    $\displaystyle (ab)^5 = - \frac{A^5}{3125} $

    Let $\displaystyle \alpha = a^5 , \beta = b^5 $

    we will find

    $\displaystyle \alpha + \beta = -B $ and
    $\displaystyle \alpha \beta = - \frac{A^5}{3125}$

    we know $\displaystyle \alpha , \beta $ are the roots of the quadratic equation $\displaystyle 3125z^2 + 3125B z - A^5 = 0 $

    After finding them , the roots of the quintic equation can also be found .

    For example , let $\displaystyle A = -5 , B= 2 $

    $\displaystyle \alpha = -1 , \beta = -1 $

    $\displaystyle a = - 1 , b = -1 \implies x = -2 $

    which satisfies $\displaystyle (-2)^5 - 5(-2)^3 + 5(-2) + 2 = 0$
    Last edited by simplependulum; Aug 29th 2009 at 12:52 AM.
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  4. #4
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    probably an easier way is to just substitute $\displaystyle x=z-\frac{a}{5z}.$

    this problom was solved over 300 years ago by some mathematician!
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  5. #5
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    this problom was solved over 300 years ago by some mathematician!
    yeah.
    Quintic Equation -- from Wolfram MathWorld
    Last edited by mr fantastic; Sep 18th 2009 at 08:40 AM. Reason: Restored original reply
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