Abstract algebra functions.

mtayab1994

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?

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?

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?

Mark44

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

So f(x) = ax² + bx + c.

SammyS

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.

Mark44

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.

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.

SammyS

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 ?

mtayab1994

I'm a junior in high school in Morocco, and this is a problem our teacher gave us and told us think about it.

mtayab1994

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?

SammyS

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.

mtayab1994

Yea i solved it . Thank you for your help.

SammyS

What was your result ?