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maths

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the solution set for the following
1.X3+x2-6x-10=0
2.2x2-4x-4=0
3.x4-x3-4x2+x+1=0

  • maths - ,

    1. x^3 + x^2 - 6x - 10 = 0
    let f(x) = x^3 + x^2 - 6x - 10
    f(1) = ... ≠ 0
    f(-1) = ... ≠ 0
    f(2) = 8 + 4 - 12 - 10 ≠ 0
    f(-2) = -8 + 4 + 12 - 10 ≠ 0
    f(5) ≠0, f(-5) ≠ 0
    all rational possiblilities have been exhausted, tough from here on in.
    Wolfram gave me this ....
    http://www.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2+-+6x+-+10+%3D+0
    x = appr 2.6631, and 2 complex roots

    2. I will use completing the square ...
    could use the formula, but that would be harder.
    first divide by 2
    -----> x^2 - 2x = 2
    x^2 - 2x + 1 = 2+1
    (x-1)^2 = 3
    x-1 = ± √3
    x = 1 ± √3

    3. let f(x) = x^4 - x^3 - 4x^2 + x + 1 = 0
    WOW, even nastier than the first one

    Wolfram says this
    http://www.wolframalpha.com/input/?i=x%5E4+-+x%5E3+-+4x%5E2+%2B+x+%2B+1+%3D+0

    What method was suggested to you to solve these?
    Have you learned Newton's Method ?

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