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Math Help - Zeros

  1. #1
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    Zeros

    Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there?

    How do I figure this out?
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  2. #2
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    There can be as many zeros as the number of the degree

    so there are 3 or less zeros

    the descartes rule of signs states that if you count the sign changes of the function, the number of sign changes (or minus two of that number) is the number of positive roots (or roots at 0)

    so x cubed - 2x has 1 sign change, so there must be 1 positive zero (if there are 3 sign changes, however, there could be 3 OR 1 positive zero)

    take f(-x) and count the sign changes to find out the number of negative zeros

    it becomes -x cubed + 2x and there is 1 sign change again, so there must be one negative zero

    because there must be 3 zeros total, and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero

    EDIT oh you said rational and irrational (not complex) woops i wasn't paying attention might not have answered your question

    IN THAT CASE:

    we have x cubed - 2x
    factor out an x
    x (x squared - 2)
    x=0 is a rational root
    x squared=2
    absolute value of x = sq. root of 2
    x = positive or negative sq. root of 2
    so there is a root at 0, sq. root of 2, and negative sq. root of 2

    so 2 irrational and 1 rational roots

    wait a minute where is the complex root? can anyone answer my question now? i've never had this happen before...
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  3. #3
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    Hello,
    Quote Originally Posted by juldancer View Post
    Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there?

    How do I figure this out?
    Factorise your cubic.

    x^3-2x=x(x^2-2)

    use the difference of 2 squares : x^2-2=(x-\sqrt{2})(x+\sqrt{2})

    Hence x^3-2x=x(x-\sqrt{2})(x+\sqrt{2})

    ----------------------------------------
    In order to know how many there are, you can use Descartes' rule of signs : http://www.purplemath.com/modules/drofsign.htm
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  4. #4
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    Quote Originally Posted by juldancer View Post
    Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there?

    How do I figure this out?
    x^3-2x=x(x-\sqrt{2})(z+\sqrt{2})

    So now you tell us how many rational and irrational roots there are.

    RonL
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  5. #5
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    Can anyone answer my error? I factored correctly and I've used DesCartes rule of Signs perfectly billions of times before...maybe I've just forgotten how to do it correctly but I can't find where I made my error
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    Hi mikedwd,

    and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero
    This is not possible. If the coefficients of a polynomial are real, then if there is a complex zero, there is another complex zero, namely its conjugate.

    The mistake is, in my opinion, in the fact that you count 0 as +0 and not -0. but I'm sorry, I don't know this rule enough to be able to spot more precisely the error...
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  7. #7
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    Quote Originally Posted by mikedwd View Post
    Can anyone answer my error? I factored correctly and I've used DesCartes rule of Signs perfectly billions of times before...maybe I've just forgotten how to do it correctly but I can't find where I made my error
    There is one change of sign and one positive root. No problem.

    (a zero root added to a polynomial does not alter the number of times that the signs change, and so is undetectable by the rule of signs)

    RonL
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  8. #8
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    hmm I'm sure I was told that a zero root is included with positive roots in the rule of signs, but I suppose that was wrong

    anyway yes I just realized (im a bit slow today apparently) that you cannot have 1 complex zero...
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