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X^2 - 5x -4(x)^1/2 +13=0

  1. Dec 15, 2016 #1
    • Member warned that an effort must be shown
    1. The problem statement, all variables and given/known data
    x^2 - 5x -4(x)^1/2 +13=0
    how to solve this

    2. Relevant equations


    3. The attempt at a solution
    I couldn't figure out anithing
     
  2. jcsd
  3. Dec 15, 2016 #2
    First get rid of ##\sqrt{x}## by substituting ##y = \sqrt{x}##, Then find a root by guessing. Finally do polynomial long division to get other roots.
     
  4. Dec 15, 2016 #3
    y^4 - 5 y^2 - 4y +13 = 0
    but I cant find root
     
  5. Dec 15, 2016 #4
    Cause there are no real roots. Sorry I just noticed.
    Are you sure you need to solve this in set of real numbers ?
     
  6. Dec 15, 2016 #5
    nope the main question is how many roots it have I was thinking that I could solve it and after that count it
     
  7. Dec 15, 2016 #6
    so how can I count without solving
     
  8. Dec 15, 2016 #7
    Fundamental theorem of algebra.

    Number of roots = degree of polynomial.
     
  9. Dec 15, 2016 #8
    so it is two I guess
     
  10. Dec 15, 2016 #9
    Yes.
     
  11. Dec 15, 2016 #10
    thanks
     
  12. Dec 15, 2016 #11
    Welcome :-p:-p:-p:partytime::partytime::partytime:
     
  13. Dec 15, 2016 #12

    Ray Vickson

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    No: your equation y^4 - 5 y^2 - 4y +13 = 0 is of degree 4, so has 4 roots, not just 2. Then that may (or may not) produce 4 different roots x = y^2. You need to actually check, because sometimes squaring can introduce spurious roots.

    You cannot use the fact that your x-equation has highest power x^2, because it also contains √x and so is not a polynomial equation at all. The root-counting business based on the highest power applies ONLY to polynomials.

    In this case there are, indeed, two roots, but you should not try to conclude this without additional work.
     
    Last edited: Dec 15, 2016
  14. Dec 15, 2016 #13
    I think the total number of roots are two, https://www.wolframalpha.com/input/?i=x^2+-+5x+-4(x)^(1/2)++13=0 .
    Did I miss something ?

    Ok I got it,
    I was not entirely correct but FTA states that Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers (http://www-groups.dcs.st-and.ac.uk/history/HistTopics/Fund_theorem_of_algebra.html).
    So roots are still 2.
     
    Last edited: Dec 15, 2016
  15. Dec 15, 2016 #14

    PeroK

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    If you let ##y = \sqrt{x}## then you get a 4th power polynomial in ##y##, which has 4 complex roots. But, not all these roots may be expressed as ##\sqrt{x}## for some ##x##. For example, take the polynomial:

    ##x^2 -25x +60 \sqrt{x} -36##

    With ##y = \sqrt{x}## this becomes:

    ##y^4 -25y^2 +60y -36##

    This has four real roots: ##y = 1, 2, 3, -6##

    Only ##-6## is not a square root, so the original equation has solutions: ##x = 1, 4, 9##
     
  16. Dec 15, 2016 #15
    I graphed the function and it has no zeroes so I am guessing that any roots it has include imaginary numbers
     
  17. Dec 15, 2016 #16
    So how will tell exact number of roots of this polynomial without finding ?
    It at most have 4 but is there a way to get exact number ?
     
  18. Dec 15, 2016 #17

    PeroK

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    It's still someone else's homework problem!
     
  19. Dec 15, 2016 #18

    epenguin

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    You said at first you had no idea what to do.
    Does this mean the problem has no relation you can see with anything you have been doing in your course? If you tell us what do you have been doing recently that would be one clue. With intractable, ugly and unsatisfactory questions like this, asking the student if he has copied the question/problem out right has often led to simplifications.
     
  20. Dec 16, 2016 #19
    ok I understood that the main thing is that I encountered this problem on quiz in internet and there was no valuable thing I could do so I stated that I couldn't figure out anything but I guess the only solution will be to graph equation .
     
  21. Dec 16, 2016 #20

    PeroK

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    You could assume that the equation has no real solutions (although that doesn't seem to be so easy to verify). Then, you need to know something about complex square roots, assuming you allow ##x## to be complex.
     
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