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Finding the roots of an equation

  1. Feb 24, 2007 #1
    1. The problem statement, all variables and given/known data

    How do I find the roots of 4x^3+x+5 = 0? It doesn't appear to be in a nice form like many equations in the textbook?
     
  2. jcsd
  3. Feb 24, 2007 #2
    I know one way.. graph it on your graphinc calculator. You see that -1 is a zero of the graph, therefore, (x+1)*something=4x^3+x+5..so divide 4x^3+x+5 by (x+1).. You get 4x^2 + 4x + 5..graph that.. Then you see there are no zeros, meaning that those are imaginary factors.. And I can't simplify 4x^2 + 4x + 5 but
    http://www.hvks.com/Numerical/websolver.htm
    says the imaginary answers are
    X1=(0.5-i1)
    X2=(0.5+i1)
    --
    PS someone remind me how to solve 4x^2 + 4x + 5 :D
     
  4. Feb 24, 2007 #3
    Quadratic formula?????
     
  5. Feb 25, 2007 #4

    Hootenanny

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    Feldoh, although your help is appreciated please do not post complete solutions, never mind incorrect solutions, as it is contrary to our policy.
     
  6. Feb 25, 2007 #5
    Sorry, and on the downside, apparently I can't add :(
     
    Last edited: Feb 25, 2007
  7. Feb 25, 2007 #6

    Hootenanny

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    No problem, don't worry bout it, your a new member so we won't send you to the gallows just yet :tongue: . Seriously though, any help that you can give in the forums is very much appreciated. Welcome to the Forums! :biggrin:
     
  8. Feb 25, 2007 #7
    Thanks for the welcome^^
     
  9. Feb 25, 2007 #8
    Quadratic formula?????

    I hate the quadratic formula.

    To try and simplify 4x^2 + 4x + 5... With quadratic form, I did 4(+/-)[square root of -64]/8
    That simplifies to 4(+/-)/[square root of 8]i/8..

    Now how do we get
    X1=(0.5-i1)
    X2=(0.5+i1) from that..

    I prefer factoring.. by FOILING..4x^2 + 4x + 5.. Or is it impossible to deFoil imaginaries?
     
  10. Feb 25, 2007 #9

    Hootenanny

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    You haven't simplified properly;

    When it actually simplifies to;

    [tex]x = \frac{4\pm 8i}{8}[/tex]
     
  11. Feb 25, 2007 #10
    Oopd.. Forgot to take the square root after de squaring 64 :D

    Is there any way to simplify 4x^2 + 4x + 5 besides quadratic formula?
    I tried completing the square..
    4x^2 + 4x + 5..
    4(x^2 + x)=-5
    (x^2 + x)=-5/4
    (x^2 + x + 0.25)=-1
    (x+0.5)^2=-1
    x+0.5=(+/-)i
    x=-0.5 (+/-)i..

    I got real close.. then got that -0.5 at the end..
     
  12. Feb 25, 2007 #11

    D H

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    None of this talk about quadratics will help the OP solve their problem, as it is a cubic equation: [itex]4x^3+x+5 = 0[/itex]. The mechanics for solving cubics makes the quadratic formula look like baby stuff. Mathworld has a quite complete discussion on the cubic formula: http://mathworld.wolfram.com/CubicFormula.html.
     
  13. Feb 25, 2007 #12

    Hootenanny

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    This answer is actually correct, I missed the minus sign in my previous post (typo sorry :rolleyes: ), you must have made a mistake in your previous method. As an aside completing the square is equivalent to using the quadratic equation (in fact the quadratic equation is derived by completing the square)
     
  14. Feb 25, 2007 #13
    Hmm..?
    http://www.hvks.com/Numerical/websolver.htm
    This also yields +0.5 (+/-)i..
    same as yours of
    https://www.physicsforums.com/latex_images/12/1255823-0.png
    But I still don't know where I went wrong with my
    x=-0.5 (+/-)i.. solution from completing the square..
    ---
    "None of this talk about quadratics will help the OP solve their problem"
    Nope..:D I think the way we did it in class was to assign "p" to the lowest degree, and "q" to the highest degree, find all the factors of each, and the possible zeros are some of "p/q".. Which you can find via graphing..
     
  15. Feb 26, 2007 #14
    Hello..llllllllll
     
  16. Feb 26, 2007 #15
    @pugfug90,

    [tex]\text{The solution of }4x^2+4x+5=0\text{ is }x=\frac{-4\pm 8i}{8}[/tex]​

    so your completing the square solution is absolutely correct. Remember that the quadratic formula is

    [tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]​

    @OP,

    The original equation is a cubic, and while there are clever ways to solve it using plain algebra, the easiest way without a calculator is to try every root in the Rational Root Theorem (http://www.mathwords.com/r/rational_root_theorem.htm). Doing this, you will find, as pugfug90 did, that -1 is a root.

    Once you discover that, you know that [tex](x-(-1))=(x+1)[/tex] is a factor of [tex]4x^3+x+5[/tex] by the Factor Theorem (http://www.purplemath.com/modules/factrthm.htm). Divide through long division or synthetic division to find that

    [tex]4x^3+x+5=(x+1)(4x^2-4x+5)=0[/tex]​

    so the last two roots are the roots of the remaining quadratic and can be solved. In all these problems, the strategy is to try and simplify polynomials with degrees higher than two by finding easy to find roots until you get a product of quadratics which makes finding all the roots possible.
     
  17. Feb 27, 2007 #16
    Does that mean that the online imaginary root calculator is wrong?
     
  18. Feb 27, 2007 #17

    cristo

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    That website gives
    So no, it is not incorrect.
     
  19. Feb 27, 2007 #18
    Yep, the original application of the quadratic formula a few posts up did not seem to use -b but just b.
     
  20. Feb 27, 2007 #19
    How come putting the original 4x^3+x+5 doesn't decompose into -0.5..?
     
  21. Feb 27, 2007 #20

    cristo

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    Because you didn't divide (x+1) into 4x3+x+5 correctly. The cubic factors as 4x3+x+5=(x+1)(4x2-4x+5)
     
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