The discussion revolves around a quadratic function defined by specific values for f(a), f(b), and f(c), with the goal of calculating f(a+b+c). Participants explore the implications of the quadratic nature of the function and the relationships between the variables.
There is ongoing exploration of the function's properties, with some participants providing guidance on how to approach the problem. Multiple interpretations of the quadratic function and its coefficients are being considered, but no consensus has been reached regarding the method to solve for f(a+b+c).
Participants note the ambiguity in the definitions of the variables and coefficients, as well as the constraints imposed by the problem's context. One participant mentions being a junior in high school in Morocco, indicating the educational setting for this discussion.
mtayab1994 said:Homework Statement
f is a quadratic function from the second degree and [tex]f(a)=bc;f(b)=ac;f(c)=ab[/tex]
Homework Equations
Calculate : [tex]f(a+b+c)[/tex]
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 said:No, you can't do that. You haven't used the fact that f is a quadratic function. You know what that means, right?
Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared".mtayab1994 said: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 said:Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared".
So f(x) = ax2 + bx + c.
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.SammyS said: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]
SammyS said: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.
SammyS said: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.
Letting [itex]\displaystyle f(x)=Dx^2+Ex+G[/itex], you might look at [itex]\displaystyle f(a)-f(b)[/itex] for instance.mtayab1994 said: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 said: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]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+G-Db^2-Eb-G=c(b-a)[/itex]
That will eliminate the constant, G.
BTW: What course is this for ?
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.mtayab1994 said: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 said: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.
What was your result ?mtayab1994 said:Yea i solved it . Thank you for your help.