equation of 4th degree

[kanali]

I was trying to solve the equation:

\(\displaystyle x^4+(a-x)^4=b^4\) by expanding the terms but that leads to a complicated 4th degree eqaution.

Do you think i am in the right track ??

 

[tkhunny]

What track are you on? It's not any more complicated than any other quartic equation.

 

[kanali]

What track are you on? It's not any more complicated than any other quartic equation.

If we expand we have:

\(\displaystyle x^4+a^4+4a^3x+6a^2x^2+4ax^3+x^4=b^4\).That is where i get stuck.I find it difficult to solve this equation

 

[tkhunny]

Couple of things...

1) Why are you trying to solve this? Is it a direct problem? Is it a result of something else you are doing? Is it a challenge problem from somehwere?

2) Hint: Solutions are trivial for x = 0 and x = a/2

3) Do we know ANYTHING about 'a' and 'b'? b < a < 0 is a problem.

4) Just Real solutions? Do we believe that there ARE some?

 

[kanali]

Couple of things...

1) Why are you trying to solve this? Is it a direct problem? Is it a result of something else you are doing? Is it a challenge problem from somewhere?

2) Hint: Solutions are trivial for x = 0 and x = a/2

3) Do we know ANYTHING about 'a' and 'b'? b < a < 0 is a problem.

4) Just Real solutions? Do we believe that there ARE some?

1) These are problems i come across in an old book of mathematics

2)I think here ,we are not considering trivial solutions.As for the cases where a=0 or b=0 or a and b are equal to zero.

we are considering the case where \(\displaystyle ab\neq 0\)


The problem here is : can we find a method that will solve these kind of problems>

Certainly we looking for all kinds of solutions whether real or complex

 

[Denis]

If we expand we have:
\(\displaystyle x^4+a^4+4a^3x+6a^2x^2+4ax^3+x^4=b^4\)

Your equation: 2x^4 + 4ax^3 + 6a^2x^2 + 4a^3x + a^4 - b^4 = 0

Standard quartic: Ax^4 + Bx^3 + Cx^2 + Dx + E = 0

A = 2, B = 4a, C = 6a^2, D = 4a^3, E = a^4 - b^4

Your question:
> The problem here is : can we find a method that will solve these kind of problems ?

Answer: none needed; just solve the quartic. If you forgot how, use Google.

 

[kanali]

If we expand we have:
\(\displaystyle x^4+a^4+4a^3x+6a^2x^2+4ax^3+x^4=b^4\)

Your equation: 2x^4 + 4ax^3 + 6a^2x^2 + 4a^3x + a^4 - b^4 = 0

Standard quartic: Ax^4 + Bx^3 + Cx^2 + Dx + E = 0

A = 2, B = 4a, C = 6a^2, D = 4a^3, E = a^4 - b^4

Your question:
> The problem here is : can we find a method that will solve these kind of problems ?

Answer: none needed; just solve the quartic. If you forgot how, use Google.

.

Thank you.

By the way there is a substitution technique as the google suggests

 

[tkhunny]

It is a quartic equation. There is a general solution.