Pre Calc

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Determine K and solve the equation x^3-kx^2+3x+54=0, if one of its zeros is triple of another.

  • Pre Calc -

    let the roots be a, 3a and b
    (x-a)(x-3a)(x-b) = x^3 - kx^2 + 3x + 54

    looking at the last term -3a^2b = 54
    a^2b = -18 , the only square factors of -18 are 1 and 9
    case 1: a=1, b = -18
    (x-1)(x-3)(x+18) would be the expression
    this give us x^3 + 14x^2 - 69x + 54
    which would not match up the x terms

    case 2: a=3, b=-2
    (x-3)(x-9)(x+2) would be the expression for
    x^3 - 10x^2 + 3x + 54

    this matches if k = 10

    and the roots are 3,9 and -2

    (there should be an easier way to do this)

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