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Math Help - simulatenous equations from quartic root polynomial

  1. #1
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    simulatenous equations from quartic root polynomial

    the question was to find the roots of this equation
    8x^4-2x^3-27x^2+6x+9
    when two of the roots equalled zero.
    so i let the roots be  \alpha, \beta, \gamma, \delta
    therefore \alpha+\beta = 0
    i used the standard sum of roots, product of roots etc technique to work out simultaneous equations and i ended up with this result
     -2\alpha^2\gamma + 8\alpha^2\gamma^2 = 9
    -8\alpha^2+2\gamma-8\gamma^2=-27
    -4\alpha^2\gamma-\alpha^2+4\alpha\gamma^2=-3

    i'm not realli sure if those equations are correct but if they are, i dont realli no how to proceed to figure out gamma and alpha
    if anyone could help me, i would realli realli appreciate it !!! thankyou !!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by THSKluv View Post
    the question was to find the roots of this equation
    8x^4-2x^3-27x^2+6x+9
    when two of the roots equalled zero.
    so i let the roots be  \alpha, \beta, \gamma, \delta
    therefore \alpha+\beta = 0
    i used the standard sum of roots, product of roots etc technique to work out simultaneous equations and i ended up with this result
     -2\alpha^2\gamma + 8\alpha^2\gamma^2 = 9
    -8\alpha^2+2\gamma-8\gamma^2=-27
    -4\alpha^2\gamma-\alpha^2+4\alpha\gamma^2=-3

    i'm not realli sure if those equations are correct but if they are, i dont realli no how to proceed to figure out gamma and alpha
    if anyone could help me, i would realli realli appreciate it !!! thankyou !!
    The rational root theorem suggests you try +/-1/2, +/-1/4, +/-1/8, +/-3/2, +/-3/4, +/-3/8, +/-9, +/-9/2, +/-9/4 and +/-9/8 as possible roots.

    Of these -1/2 and 3/4 are roots.

    CB
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  3. #3
    Junior Member
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    Jun 2009
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    thankyou but i actually havent learnt the rational root theorem and im supposed
    to be solving these equations through the equations for sum of roots, sum of product of roots two at a time, sum of product of roots three at a time and product of roots
    thanks though
    if anyone can help me with my above equations i would realli appreciate it !!
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  4. #4
    MHF Contributor
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    Hi

    You know
    a + b + c + d = 1/4
    ab + ac + ad + bc + bd + cd = -27/8
    abc + abd + acd + bcd = -3/4
    abcd = 9/8
    a + b = 0

    This means
    c + d = 1/4
    ab + cd = -27/8 (since ac + ad + bc + bd = (a+b)(c+d) = 0)
    ab(c + d) = -3/4 (since acd + bcd = (a+b)cd = 0)
    abcd = 9/8
    a + b = 0

    You can find
    ab = -3 (which gives you a and b using a+b=0)
    cd = -3/8 and c+d = 1/4 which gives you c and d
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  5. #5
    Junior Member
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    wow thankyou SO much !!!
    u made the qs so much easier for me
    thanks again !! realli appreciate it =DD
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