## Abstract algebra functions.

1. The problem statement, all variables and given/known data

f is a quadratic function from the second degree and $$f(a)=bc;f(b)=ac;f(c)=ab$$

2. Relevant equations

Calculate : $$f(a+b+c)$$

3. The attempt at a solution

Can we say that $$f(a+b+c)=f(a)+f(b)+f(c)$$ and the go on from there plugging in the values of each one are do i have to do something else?

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 Quote by mtayab1994 1. The problem statement, all variables and given/known data f is a quadratic function from the second degree and $$f(a)=bc;f(b)=ac;f(c)=ab$$ 2. Relevant equations Calculate : $$f(a+b+c)$$ 3. The attempt at a solution Can we say that $$f(a+b+c)=f(a)+f(b)+f(c)$$ and the go on from there plugging in the values of each one are do i have to do something else?
No, you can't do that. You haven't used the fact that f is a quadratic function. You know what that means, right?

 Quote by Mark44 No, you can't do that. You haven't used the fact that f is a quadratic function. You know what that means, right?
What do you mean a quadratic function to the second degree is written in the form ax^2+bx+c that's correct right?

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## Abstract algebra functions.

 Quote by mtayab1994 What do you mean a quadratic function to the second degree is written in the form ax^2+bx+c that's correct right?
Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared".

So f(x) = ax2 + bx + c.

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 Quote by Mark44 Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared". So f(x) = ax2 + bx + c.
Of course, for your function, you will want to different variable names for your coefficients, perhaps:
$\displaystyle f(x)=Dx^2+Ex+G$
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.

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 Quote by SammyS Of course, for your function, you will want to different variable names for your coefficients, perhaps:$\displaystyle f(x)=Dx^2+Ex+G$
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.
 Quote by SammyS 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.

 Quote by SammyS Of course, for your function, you will want to different variable names for your coefficients, perhaps:$\displaystyle f(x)=Dx^2+Ex+G$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.
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.

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 Quote by mtayab1994 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.
Letting $\displaystyle f(x)=Dx^2+Ex+G$, you might look at $\displaystyle f(a)-f(b)$ for instance.
$\displaystyle f(a)-f(b)=bc-ac$
which becomes
$\displaystyle Da^2+Ea+G-Db^2-Eb-G=c(b-a)$
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 ?

 Quote by SammyS Letting $\displaystyle f(x)=Dx^2+Ex+G$, you might look at $\displaystyle f(a)-f(b)$ for instance.$\displaystyle f(a)-f(b)=bc-ac$which becomes$\displaystyle Da^2+Ea+G-Db^2-Eb-G=c(b-a)$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 ?
I'm a junior in high school in Morocco, and this is a problem our teacher gave us and told us think about it.

 Ok so this is what i got: $$f(a+b+c)=Da^{2}+Ea+G+Db^{2}+Db+G+Dc^{2}+Dc+G$$ then you factor it out and you get: $$D(a^{2}+b^{2}+c^{2})+E(a+b+c)+3G=c(a+b)+ab$$ Now can we use substitution for a+b+c or what from here on?

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 Quote by mtayab1994 Ok so this is what i got: $$f(a+b+c)=Da^{2}+Ea+G+Db^{2}+Db+G+Dc^{2}+Dc+G$$ then you factor it out and you get: $$D(a^{2}+b^{2}+c^{2})+E(a+b+c)+3G=c(a+b)+ab$$ Now can we use substitution for a+b+c or what from here on?
No. You're still treating this as if $\displaystyle f(a+b+c)=f(a)+f(b)+f(c)\,,$ which is definitely not the case.

$\displaystyle f(a+b+c)=D(a+b+c)^2+E(a+b+c)+G$

and of course,
$\displaystyle (a+b+c)^2=a^2+2ab+b^2+2bc+c^2+2ac\ .$

Etc.

 Quote by SammyS No. You're still treating this as if $\displaystyle f(a+b+c)=f(a)+f(b)+f(c)\,,$ which is definitely not the case. $\displaystyle f(a+b+c)=D(a+b+c)^2+E(a+b+c)+G$ and of course, $\displaystyle (a+b+c)^2=a^2+2ab+b^2+2bc+c^2+2ac\ .$ Etc.
Yea i solved it . Thank you for your help.

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 Quote by mtayab1994 Yea i solved it . Thank you for your help.