# Abstract algebra functions.

by mtayab1994
Tags: abstract, algebra, functions
 P: 584 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?
Mentor
P: 21,400
 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?
P: 584
 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?

Mentor
P: 21,400
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.
Emeritus
HW Helper
PF Gold
P: 7,819
 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.
Mentor
P: 21,400
 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.
P: 584
 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.
Emeritus
HW Helper
PF Gold
P: 7,819
 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 ?
P: 584
 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.
 P: 584 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?
Emeritus
HW Helper
PF Gold
P: 7,819
 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.
P: 584
 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.
Emeritus