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Math Help - solving a quatric equation

  1. #1
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    solving a quatric equation

    I just need to solve a quatric equation like  ax^4 + bx^3 +cx^2+dx+e=0
    I need the 4 roots in terms a,b,c,d,e.
    Can any one please help me solving the quatric equation.

    Thanks
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  2. #2
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    Quote Originally Posted by bjkrishna View Post
    I just need to solve a quatric equation like  ax^4 + bx^3 +cx^2+dx+e=0
    I need the 4 roots in terms a,b,c,d,e.
    Can any one please help me solving the quatric equation.

    Thanks
    Wrong forum.

    Use synthetic division.
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  3. #3
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    all i can say is that its not going to be just solving it man...

    its a pretty long process.

    Its not difficult but it does take time. anyway this is what you need to do:

    First you need to read up on depressed quartic equations. Once you have make your equation into a depressed quartic.

    After youve done that you need to get familiar with Ferrari's solution. It has quite a few outcomes depending on the initial conditions so it might be worth taking a note of that when you show the different methods of solving the polynomial.
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  4. #4
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by bjkrishna View Post
    I just need to solve a quatric equation like  ax^4 + bx^3 +cx^2+dx+e=0
    I need the 4 roots in terms a,b,c,d,e.
    Can any one please help me solving the quatric equation.

    Thanks
    It turns out that solving a quartic boils down to solving a cubic.

    We have  f(x)=ax^4+bx^3+cx^2+dx+e . Let the roots of  f(x) be  \alpha_1, \; \alpha_2, \; \alpha_3, \; \alpha_4 .

    Under the substitution  x=y-\tfrac a4 the quartic becomes  g(y) = y^4+py^2+qy+r , where  p=\tfrac18(-3a^2+8b), \; q=\tfrac18(a^3-4ab+8c), \; r=\tfrac{1}{256}(-3a^4+16a^2b-64ac+256d) .

    The roots of  g(y) are  \alpha_1+\tfrac a4, \; \alpha_2+\tfrac a4, \; \alpha_3+\tfrac a4, \; \alpha_4+\tfrac a4 . Let's call these  \beta_1, \; \beta_2, \; \beta_3, \; \beta_4 respectively.

    Now define  h(x) = x^3-2px^2+(p^2-4r)x+q^2 . Direct verification shows the roots of  h(x) are

     \theta_1 = (\beta_1+\beta_2)(\beta_3+\beta_4)
     \theta_2 = (\beta_1+\beta_3)(\beta_2+\beta_4)
     \theta_3 = (\beta_1+\beta_4)(\beta_2+\beta_3)


    So we have  \theta_1 = (\beta_1+\beta_2)(\beta_3+\beta_4) from above and from  g(x) we get  (\beta_1+\beta_2)+(\beta_3+\beta_4)=0 since the sum of the roots will give the coefficient of the cubic term, and in this case it's zero.

    This gives us  \beta_1+\beta_2 = \sqrt{-\theta_1} and  \beta_3+\beta_4 = -\sqrt{-\theta_1} .

    Similarly
     \beta_1+\beta_3 = \sqrt{-\theta_2} and  \beta_2+\beta_4 = -\sqrt{-\theta_2}
     \beta_1+\beta_4 = \sqrt{-\theta_3} and  \beta_2+\beta_3 = -\sqrt{-\theta_3} .

    Also an easy computation shows  \sqrt{-\theta_1}\sqrt{-\theta_2}\sqrt{-\theta_3} = -q , so the choice of two of the square roots determines the third. Using some algebra we get

     2\beta_1 = \sqrt{-\theta_1}+\sqrt{-\theta_2}+\sqrt{-\theta_3}
     2\beta_2 = \sqrt{-\theta_1}-\sqrt{-\theta_2}-\sqrt{-\theta_3}
     2\beta_3 = -\sqrt{-\theta_1}+\sqrt{-\theta_2}-\sqrt{-\theta_3}
     2\beta_4 = -\sqrt{-\theta_1}-\sqrt{-\theta_2}+\sqrt{-\theta_3}

    From above, knowing the  \beta_i 's lets us know the  \alpha_i 's. Thus we've found the roots of  f(x) .

    And here's the final result after making the substitutions back to  a,b,c,d,e : quartic formula.
    Last edited by chiph588@; April 26th 2010 at 11:04 PM.
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