Result for f(a+b+c) = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

In summary: Homework Statement f is a quadratic function from the second degree and f(a)=bc;f(b)=ac;f(c)=ab Homework EquationsCalculate : f(a+b+c) The Attempt at a SolutionIn summary, the problem asks for the value of f(a+b+c) for a quadratic function f with coefficients of the form ax^2+bx+c, given that f(a) = bc, f(b) = ac, and f(c) = ab. It is not possible to simply add the values of f(a), f(b), and f(c) to find f(a+b+c). Instead, we must use the fact that a quadratic function can be written in the form ax^2+bx+c
  • #1
mtayab1994
584
0

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?
 
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  • #2
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?

No, you can't do that. You haven't used the fact that f is a quadratic function. You know what that means, right?
 
  • #3
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?

What do you mean a quadratic function to the second degree is written in the form ax^2+bx+c that's correct right?
 
  • #4
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?
Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared".

So f(x) = ax2 + bx + c.
 
  • #5
Mark44 said:
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:
[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
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]​
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:
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.
 
  • #7
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.

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.
 
  • #8
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.
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 ?
 
  • #9
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]
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 ?

I'm a junior in high school in Morocco, and this is a problem our teacher gave us and told us think about it.
 
  • #10
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
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?
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.
 
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  • #12
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.

Yea i solved it . Thank you for your help.
 
  • #13
mtayab1994 said:
Yea i solved it . Thank you for your help.
What was your result ?
 

FAQ: Result for f(a+b+c) = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

What is abstract algebra?

Abstract algebra is a branch of mathematics that studies algebraic structures, such as groups, rings, and fields, and their properties. It is a generalization of classical algebra, which deals with numbers and their operations.

What are algebraic structures?

Algebraic structures are sets of elements and operations defined on those elements. Examples include groups, rings, and fields. These structures have specific properties and rules that govern how the elements can be combined.

What is a function in abstract algebra?

In abstract algebra, a function is a rule that assigns one element from a set to another element from a different set. It is often represented by an equation or expression, and can be used to describe relationships between elements in an algebraic structure.

What is the difference between a group, a ring, and a field?

A group is an algebraic structure that consists of a set of elements and a single operation that satisfies specific properties, such as closure and associativity. A ring is a structure with two operations, usually addition and multiplication, that also satisfy specific properties. A field is a ring with additional properties that make it possible to perform division.

What are some real-world applications of abstract algebra?

Abstract algebra has many applications in computer science, cryptography, and physics. For example, public key encryption algorithms use concepts from abstract algebra, and the structure of atoms can be described using group theory. It also has applications in coding theory, where algebraic structures are used to efficiently transmit and store data.

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