
#1
Apr3012, 03:20 PM

P: 584

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? 



#2
Apr3012, 03:44 PM

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#3
Apr3012, 07:01 PM

P: 584





#4
Apr3012, 07:11 PM

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Abstract algebra functions.So f(x) = ax^{2} + bx + c. 



#5
Apr3012, 07:43 PM

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[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. 



#6
Apr3012, 08:05 PM

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#7
May112, 04:14 AM

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#8
May112, 09:39 AM

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[itex]\displaystyle f(a)f(b)=bcac[/itex]A little algebra will give a relationship between the constants D & E and the quantities a+b and c.which becomes[itex]\displaystyle Da^2+Ea+GDb^2EbG=c(ba)[/itex] That will eliminate the constant, G. BTW: What course is this for ? 



#9
May112, 02:29 PM

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#10
May112, 02:41 PM

P: 584

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? 



#11
May112, 11:38 PM

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[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. 



#12
May212, 05:47 AM

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#13
May212, 09:54 AM

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