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Want to solve a quadratic equation

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Actually, it is told by one of my friend. He asked for the answer of this quadratic equation.

    m^3-m^2+2=0


    2. Relevant equations

    m^3-m^2+2=0

    3. The attempt at a solution

    But, I think we need one more equation to solve it. Is there any way to solve it by only one equation? If yes, please describe it or if not, then tell me the proper reason..
     
  2. jcsd
  3. Feb 9, 2013 #2
    That is not a quadratic equation; it's a cubic equation.
     
  4. Feb 9, 2013 #3
    Okay.. Sorry then. But, is there any possible solution?
     
  5. Feb 9, 2013 #4
    sub in 0, 1, -1 etc and see if you can find a factor
     
  6. Feb 9, 2013 #5
    Well, I don't know how to solve it. Can anyone explain me??
     
  7. Feb 9, 2013 #6

    SteamKing

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    There's also a formula (rather involved).
     
  8. Feb 9, 2013 #7

    Ray Vickson

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    Phospho has already given you some valuable suggestions. Have you tried them?
     
  9. Feb 9, 2013 #8
    I said that I didn't understand it. I know how to solve quadratic equation, but don't know how to solve cubic equation. That is what I am saying, can anyone explain it.
     
  10. Feb 9, 2013 #9
    While the formula for finding the roots of a cubic function would do the trick, I'd suggest you do it using the Factor Theorem/The Remainder Theorem, and use any method of division to reduce the polynomial.
     
  11. Feb 9, 2013 #10

    Dick

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    If you can find one root, say r, of the equation then m-r is a factor of the polynomial. You should be able to then write it as (m-r)*(quadratic)=0. Then solve the quadratic. But you have to find the one root r, first. There is one. Guess it.
     
  12. Feb 9, 2013 #11

    Ray Vickson

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    That is not what he said and not what he meant. Why can't you just substitute m = 0, m = 1 or m = -1 into the equation to see if the equation is satisfied?

    Nobody is asking you to solve a cubic equation---although people have pointed out to you that there are formulas to do it (just Google 'cubic equation'). What people are asking you to do is to test some m-values to see if they work.
     
  13. Feb 9, 2013 #12

    Mark44

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    The easiest way is to look for a potential linear of the form x - a, where a is a specific number. If you can find a linear factor, the other factor of your cubic will be a quadratic, and you can use the quadratic formula to find its roots.

    Here's a link to a web page on polynomial long division - http://www.purplemath.com/modules/polydiv3.htm

    For your problem, if there are any linear factors with rational coefficients, they must be of the form (x - 1), (x + 1), (x - 2), or (x + 2).
     
  14. Feb 9, 2013 #13

    Mark44

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    This is not a calculus problem, so I'm moving it to the precalc section.
     
  15. Feb 10, 2013 #14
    I searched for the cubic equation solver online. I found results like this

    m1= 2.73205080756888
    m2= -0.73205080756888
    m3= 1

    I saw the method but didn't get it. I have to study it from the basic. btw, thnx for the help..
     
  16. Feb 10, 2013 #15

    Ray Vickson

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    Have you checked to see if these work? All three are WRONG!

    On several occasions you have been asked to try some simple values such as m = -1, m = 1, etc., to see if they solve the equation. You have refused to do this, claiming you do not know how to solve cubic equations. As I said already, just checking if a given value "works" is not solving---it is just "checking".

    What people were trying to do is to *teach*you how to solve cubic equations, but you have not been cooperative.
     
  17. Feb 10, 2013 #16
    OKay... I tried to put m= -1, 1 and 0..

    It does not satisfy the equation.
     
  18. Feb 10, 2013 #17

    SammyS

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    Try m = -1 again.

    Show the steps, please.
     
  19. Feb 10, 2013 #18

    Ray Vickson

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    That is not true. Try again. Perhaps you are looking at a different equation? Are you sure your equation is ##m^3 - m^2 + 2 = 0?##
     
    Last edited: Feb 10, 2013
  20. Feb 10, 2013 #19

    Dick

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    Oh come on, m=(-1) does.
     
  21. Feb 10, 2013 #20

    rcgldr

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    Don't try to use all three numbers as the three roots to the equation, just try the -1 as one of the roots.
     
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