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

  1. Dec 15, 2016 #1
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    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.
     
  22. Dec 17, 2016 #21

    epenguin

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    In that case probably somebody else has garbled the question.

    It is very unlikely that anybody had any practical need, e.g. an engineering problem, to solve this equation.
    It is even more unlikely that any pure mathematician needed to solve it for some fundamental question.
    And it is pretty unlikely that it would be set as an exercise for students to illustrate some method or principle, such as, say, the solution of quartic equations. Such exercises are usually given with 'nice' solutions.

    So the most likely thing is that some dim student has tried to get his homework solved for him and is even too dim and lazy to get the question right. Now maybe you have experienced what it is like to be a homework helper. We often get careless mistranscribed misunderstood etc. questions and often are able to work out what the real question must have been, but in this case I can't be bothered to try.

    (Unless the trick point:devil: is that graphing the equation it looks like it has a double root, but in fact this is false, it really has a minimum where the function is a very small positive number (when you do the graph on a scale that shows up the other minimum and maximum).

    I guess It can have been slightly useful to you if you hadn't realised the only hope was to make the variable √x (which we have called y) and then you have quartic equation in that, which can be solved algebraically. What we would do and have done is play with it a bit to see if there are any special features that enables us to spot a factorisation, whole number solution etc. There arent any. So then it is always possible to solve a quartic algebraically. But there is little point in doing so in the general case, and no one ever does it for practical purposes. (OK I did once meet someone who had done it for an aeronautical engineering problem ,but I think he did it for fun.) Anyone would do it numerically with a calculator or or graph on a calculator as you say is usually revealing but slightly deceptive in this case, see above.

    Useful for you to know about is you can always find the total number of real roots, and also if wanted the number between any two values of the unknown by the method of Sturm. Simpler but usually less informative is Descartes's rule which here tells you you can have at most two positive real and at most two negative real roots. Not much but not nothing. Then if anybody really wants to to do it I think you can do this example requiring knowledge only of elementary differentiation and solving quadratics, but it involves nasty numbers and does not look great fun, but I might have missed something. You may find it slightly advantageous to use as variable not y = √x but √x = 1/y and then your quartic is
    1 - 5y2 - 4y3+ 13y4

    This has a convenient maximum at y = 0. You can reduce things to quadratics by eliminating between the polynomial and its derivative. But it is still a bore.
     
    Last edited: Dec 18, 2016
  23. Dec 18, 2016 #22

    epenguin

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    So I think a brute force way of doing it is to solve the quadratic we can get for the minima of the quartic P(y), substitute these solutions into P(y) and find out what it is at these minima - according to computation they should both be > 0, corresponding to no real roots of P.
    However I wanted to avoid that brute force thing, and never actually solve an equation, but still get the nature of the roots from elementary arguments. I am almost sure I have now seen and that with the elimination process I sketched, one can eliminate y from the equations for the minima and get a quadratic in the values of P at these extrema; then one ought to be able to see and that these must be real positive without solving. But is still a slog and I don't have time to complete this before Christmas and make no promises - anyone is welcome to beat me to it.

    Enjoy and happy Christmas. :oldsmile:
     
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