Find all solutions to x^3-9x^2-4=0 to two decimal places? please show step by step.?

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  • 6 months ago

    x^3-9x^2-4=0

    Real solution:

    x = 3 + (29 - 4 sqrt(7))^(1/3) + (29 + 4 sqrt(7))^(1/3)

    Complex solutions:

    x ≈ -0.02443 - 0.66442 i

    x ≈ -0.02443 + 0.66442 i

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  • There's only one real solution

    https://www.wolframalpha.com/input/?i=x%5E3+-+9x%5...

    You can solve it using various methods

    https://en.wikipedia.org/wiki/Cubic_function#Gener...

    ax^3 + bx^2 + cx + d = 0

    a = 1 , b = -9 , c = 0 , d = -4

    p = b^2 - 3ac = 81 - 3 * 1 * 0 = 81

    q = 2b^3 - 9abc + 27a^2 * d = 2 * (-9)^3 - 9 * 1 * (-9) * 0 + 27 * 1^2 * (-4) = 2 * (-729) - 108 = -1458 - 108 = -1566

    C^3 = ((q +/- sqrt(q^2 - 4p^3)) / 2)

    C^3 = ((-1566 +/- sqrt(1566^2 - 4 * 81^3)) / 2)

    C^3 = ((-1566 +/- sqrt(9^2 * 174^2 - 4 * 9^6)) / 2)

    C^3 = ((-1566 +/- sqrt(9^2 * 9 * 58^2 - 4 * 9^6)) / 2)

    C^3 = ((-1566 +/- sqrt(9^3 * (58^2 - 4 * 9^3))) / 2)

    C^3 = ((-1566 +/- sqrt(3^6 * (4 * 29^2 - 4 * 9^3))) / 2)

    C^3 = ((-1566 +/- 3^3 * 2 * sqrt(29^2 - 9^3))) / 2)

    C^3 = ((-1566 +/- 54 * sqrt(841 - 729)) / 2)

    C^3 = -783 +/- 27 * sqrt(112)

    C^3 = -783 +/- 27 * 4 * sqrt(7)

    C^3 = -27 * 29 +/- 27 * 4 * sqrt(7)

    C^3 = 27 * (-29 +/- 4 * sqrt(7))

    C = 3 * (-29 +/- 4 * sqrt(7))^(1/3)

    x = (-1/(3a)) * (C + p/C)

    x = (-1/3) * (3 * (-29 +/- 4 * sqrt(7))^(1/3) + 81 / (3 * (-29 +/- 4 * sqrt(7))^(1/3)))

    x = (-1/3) * (3 * (-29 +/- 4 * sqrt(7))^(1/3) + 27 / (-29 +/- 4 * sqrt(7))^(1/3))

    x = (-1) * ((-29 +/- 4 * sqrt(7))^(1/3) + 9 * (-29 +/- 4 * sqrt(7))^(2/3) / (-29 +/- 4 * sqrt(7)))

    Are you done looking at this hell yet?

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  • 6 months ago

    You're answer would be 9.05

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  • ted s
    Lv 7
    6 months ago

    comment : there is only one real root , not rational , close to 9.05...you can use Newton's methods to approximate the value...start at x = 9

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