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Math Help - Roots of a Cubic Polynomial in Arithmetic Sequence

  1. #1
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    Roots of a Cubic Polynomial in Arithmetic Sequence

    Been having some difficulty with the following question:

    Prove that if the roots of:

    x^3-ax^2+bx-c=0 are in arithmetic sequence then 2a^3-9ab+27c=0

    Hence, solve for x in the equation x^3 -12x^2+39x -28=0

    I've tried putting general forms of the roots as:

    (x-a),(x-(a+d)),(x-(a+2d))

    and then substituting them into the initial equation and setting = 0, but it hasn't gotten me anywhere.

    Any help would be greatly appreciated
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Denote the roots x_1 = u-h, x_2 = u , x_3 = u+h, apply the Cardano-Vieta relations and elimitate u and h.
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  3. #3
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    Quote Originally Posted by Wandering View Post
    Been having some difficulty with the following question:

    Prove that if the roots of:

    x^3-ax^2+bx-c=0 are in arithmetic sequence then 2a^3-9ab+27c=0

    Hence, solve for x in the equation x^3 -12x^2+39x -28=0

    I've tried putting general forms of the roots as:

    (x-a),(x-(a+d)),(x-(a+2d))

    and then substituting them into the initial equation and setting = 0, but it hasn't gotten me anywhere.

    Any help would be greatly appreciated
    Alternatively,

    let the roots be E, F, G in arithmetic sequence.

    Then

    (x-E)(x-F)(x-G) expands out to x^3-(E+F+G)x^2+(EF+EG+FG)x-EFG

    which is x^3-ax^2+bx-c

    Since the roots are in arithmetic sequence, then E+F+G=3F

    so a=3F and b=EF+EG+FG and c=EFG

    This allows 2a^3-9ab+27c to simplify to 2(3F)^3-9(3F)(EF+EG+FG)+27c

    giving 54F^3-27F^2(E)-27EFG-27F^2(G)+27EFG

    =54F^3-27F^2(E+G)

    and since E+G=2F, the result follows.

    (sorry, this would be neater with Latex).
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