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

[mtayab1994]

Homework Statement

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

Homework Equations

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

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?

 

[Mark44]

Homework Statement

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

Homework Equations

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

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?

 

[mtayab1994]

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?

 

[Mark44]

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.

 

[SammyS]

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.

 

[Mark44]

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.

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.

 

[mtayab1994]

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.

 

[SammyS]

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 ?

 

[mtayab1994]

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.

 

[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?

 

[SammyS]

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.

 

[mtayab1994]

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.

 

[SammyS]

Yea i solved it . Thank you for your help.

What was your result ?