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Abstract algebra functions.

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data

    f is a quadratic function from the second degree and [tex]f(a)=bc;f(b)=ac;f(c)=ab[/tex]

    2. Relevant equations

    Calculate : [tex]f(a+b+c)[/tex]


    3. The attempt at a solution

    Can we say that [tex]f(a+b+c)=f(a)+f(b)+f(c)[/tex] and the go on from there plugging in the values of each one are do i have to do something else?
     
    Last edited by a moderator: Apr 30, 2012
  2. jcsd
  3. Apr 30, 2012 #2

    Mark44

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    No, you can't do that. You haven't used the fact that f is a quadratic function. You know what that means, right?
     
  4. Apr 30, 2012 #3
    What do you mean a quadratic function to the second degree is written in the form ax^2+bx+c that's correct right?
     
  5. Apr 30, 2012 #4

    Mark44

    Staff: Mentor

    Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared".

    So f(x) = ax2 + bx + c.
     
  6. Apr 30, 2012 #5

    SammyS

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    Of course, for your function, you will want to different variable names for your coefficients, perhaps:
    [itex]\displaystyle f(x)=Dx^2+Ex+G[/itex]​

    Are you sure you aren't working with a more limited form of a quadratic? ... It's possible that is the case due to the relationships that were stated.
     
  7. Apr 30, 2012 #6

    Mark44

    Staff: Mentor

    That ambiguity occurred to me, too, but I'm leaning toward the view that a, b, and c in the problem description are the same as the coefficients of the terms in the quadratic. I could be wrong, though.
     
  8. May 1, 2012 #7
    Yes i could work with other variables too say alpha β or gamma. And the a,b,and c have nothing to do with the form of the quadratic equation itself.
     
  9. May 1, 2012 #8

    SammyS

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    Letting [itex]\displaystyle f(x)=Dx^2+Ex+G[/itex], you might look at [itex]\displaystyle f(a)-f(b)[/itex] for instance.
    [itex]\displaystyle f(a)-f(b)=bc-ac[/itex]
    which becomes​
    [itex]\displaystyle Da^2+Ea+G-Db^2-Eb-G=c(b-a)[/itex]​
    A little algebra will give a relationship between the constants D & E and the quantities a+b and c.

    That will eliminate the constant, G.

    BTW: What course is this for ?
     
  10. May 1, 2012 #9
    I'm a junior in high school in Morocco, and this is a problem our teacher gave us and told us think about it.
     
  11. May 1, 2012 #10
    Ok so this is what i got:

    [tex]f(a+b+c)=Da^{2}+Ea+G+Db^{2}+Db+G+Dc^{2}+Dc+G[/tex] then you factor it out and you get:

    [tex]D(a^{2}+b^{2}+c^{2})+E(a+b+c)+3G=c(a+b)+ab[/tex]

    Now can we use substitution for a+b+c or what from here on?
     
  12. May 1, 2012 #11

    SammyS

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    No. You're still treating this as if [itex]\displaystyle f(a+b+c)=f(a)+f(b)+f(c)\,,[/itex] which is definitely not the case.

    [itex]\displaystyle f(a+b+c)=D(a+b+c)^2+E(a+b+c)+G[/itex]

    and of course,
    [itex]\displaystyle (a+b+c)^2=a^2+2ab+b^2+2bc+c^2+2ac\ .[/itex]

    Etc.
     
    Last edited: May 1, 2012
  13. May 2, 2012 #12
    Yea i solved it . Thank you for your help.
     
  14. May 2, 2012 #13

    SammyS

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    What was your result ?
     
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