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Math Help - complex roots confusion

  1. #16
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    Re: complex roots confusion

    Quote Originally Posted by anthonye View Post
    So going back to the equation x^4 - 3x^3 - x^2 - x + 3 = 0.
    How can I tell from just the equation how many roots of each type it will have?
    Look at this solution.
    The polynomial has degree four and has all real coefficients. That tells us that if there are any complex roots (i.e. the imaginary parts non-zero) they must occur in conjugate pairs. Thus there could be zero, two, or four complex roots. In fact there are two. There are also two irrational roots. See the web link above.
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  2. #17
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    Re: complex roots confusion

    Ok thank you again plato it's a question on a web site and it say that it has four complex roots
    thats what was confusing me.
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  3. #18
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    Re: complex roots confusion

    Any sites were I can go and test myself on this matter?
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  4. #19
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    Re: complex roots confusion

    Quote Originally Posted by anthonye View Post
    Any sites were I can go and test myself on this matter?
    Are you sure that you copied the equation correctly?
    You can use WolframAlpha to solve such equations.
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  5. #20
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    Re: complex roots confusion

    Yeah your right that x^2 should be possitive
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  6. #21
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    Re: complex roots confusion

    Hi;
    Question state the number of complex roots,the possible number of real roots
    of x^3 + 4x^2 + 5x -1 = 0

    complex roots = 3 why(complex numbers include real numbers)

    real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)

    are my statements correct?
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  7. #22
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    Re: complex roots confusion

    Quote Originally Posted by anthonye View Post
    Ok getting there now.

    This is what I have:

    For the equation x^4 - 3x^3 - x^2 - x + 3 = 0

    Find the number of complex roots,The possible number of real roots
    and the possible number of rational roots.


    complex roots = 4 (The reason for my answer is complex numbers are almost any numbers)
    Yes, that is true.

    real roots = 0,2,4 ( the reason for my answers are 0 the answer may all be complex) and this is where i'm confused

    real numbers are on the number line -3,-2,-1,0,1,2,3ect
    Well, not just integers, of course, numbers like " \pi", " -\sqrt{3}", etc. are real numbers. And you can say a little bit more. By "DeCarte's rule of signs", there can be either 0 or 2 positive real roots (because the sign changes at " -3x^3" and at "+3"- two sign changes. DesCarte's rule of signs says the number of positive real roots is at most the number of sign changes but may be less by an even number. Now, replacing "x" by "-x" swaps positive and negative roots and (-x)^4- 3(-x)^3+ (-x)^2- (-x)+ 3= x^4+ 3x^3+ x^2+ x+ 3 which also has 0 or 2 sign changes.

    But if a complex number is say 3(imaginary part=0) is 3 complex or
    real?
    Both. The set of all real numbers is a subset of the set of all complex numbers.

    Thanks.
    For the question about rational roots, you can use the "rational roots theorem": if the rational number \frac{m}{n}, with m and n integers, of the polynomial equation a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0, all coefficients being integers, then n must evenly divide the "leading coefficient", a_n and m must evenly divide the "constant term", a_0.

    Here, x^4 - 3x^3 - x^2 - x + 3 = 0 so the "leading coefficient" is 1 which is evenly divisible only by 1 and -1. The "constant term" is 3 which is evenly divisible only by 1, -1, 3, and -3. That is, the only possible rational roots are the integers 1, -1, 3, and -3. You can put each into the polynomial to see which, if any, actually satisfy the equation.
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  8. #23
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    Re: complex roots confusion

    Did you miss my last post.
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  9. #24
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    Re: complex roots confusion

    Quote Originally Posted by anthonye View Post
    Question state the number of complex roots,the possible number of real roots
    of x^3 + 4x^2 + 5x -1 = 0
    complex roots = 3 why(complex numbers include real numbers)
    real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)
    are my statements correct?
    I think that you have misunderstood the whole point of this discussion.
    Here is a direct quote: " the cubic polynomial x^3 + x has two complex roots"
    Now that polynomial has three roots, one real and two complex.
    As an accident of history, the language of roots of a polynomial we use the words complex root to mean a complex number with a nonzero imaginary part. Thus x^3 + 4x^2 + 5x -1 = 0, has at least one real root and at most two complex roots.

    NOTE: above we are talking about complex roots, not complex numbers as such.
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  10. #25
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    Re: complex roots confusion

    Yes exactly plato two complex and one real thats what I got because complex
    roots always come in pairs so that web page is wrong.
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  11. #26
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    Re: complex roots confusion

    Quote Originally Posted by anthonye View Post
    Yes exactly plato two complex and one real thats what I got because complex
    roots always come in pairs so that web page is wrong.
    What web page are you talking about?
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  12. #27
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    Re: complex roots confusion

    Quote Originally Posted by anthonye View Post
    Hi;
    Question state the number of complex roots,the possible number of real roots
    of x^3 + 4x^2 + 5x -1 = 0

    complex roots = 3 why(complex numbers include real numbers).

    real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)

    are my statements correct?
    Yes, because non-real solutions to an equation with real coefficients must come in complex conjugate pairs, there must be either 2 or no non-real solutions. That leaves either 1 or 3 real roots. And we can say if there only one real root it must be positive. If all three roots are real, one is positive and the other two negative.
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  13. #28
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    Re: complex roots confusion

    Hi;
    Just need my work checking(I'm sure it's all correct now)


    For each equation state the number of complex roots,the possible number or real roots
    and the possible rational roots.

    1) -x^4 = 0 (complex roots 0 or2 or 4)
    (real roots 2 or 4)
    (rational roots 0)



    2) 2x^5 - 4x^4 - 4x^2 + 5 = 0 (complex roots 0 or 2 or 4)
    (real roots 1 or 3 or 5)
    (rational roots + or - 5/2 or 1/2 or 1 or 5)


    3)4x + 8 = 0 (complex roots 0)
    (real roots 1)
    (rational roots + or - 1/4 or 1/2 or 1 or 2 or or 8)


    4)-2x^6 - x^2 + x - 7 = 0 (complex roots 0 or 2 or 4 or 6)
    (real roots 2 or 4 or 6)
    (rational roots + or - 1 or 1/2 or 7 or 7/2)


    5)x^10 + x^8 - x^4 + 3x^2 - x + 1 = 0 (complex roots 0 or 2 or 4 or 6 or 8 or 10)
    (real roots 0or 2 or 4 or 6 or 8 or 10)
    (rational roots + or - 1)
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  14. #29
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    Re: complex roots confusion

    Please someone Please check my work above

    Thank you.
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  15. #30
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    Re: complex roots confusion

    We don't see your answers in order to check them. Add your responses.
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