Solving Challenging 4th, 3rd, 2nd, and 1st Degree Equations

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Homework Help Overview

The discussion revolves around solving a complex equation involving polynomial expressions of various degrees. The equation presented involves terms raised to the fourth power and is set equal to a fraction involving constants and variables.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss expanding the left-hand side of the equation and express difficulty in solving the resulting polynomial of varying degrees. There are questions regarding the assumptions made about the solutions for x, including potential multiple solutions and the nature of those solutions.

Discussion Status

Participants are actively engaging with the problem, asking for clarification on the reasoning behind proposed solutions and expressing confusion about the steps involved. Some guidance has been offered regarding the expansion of terms, but there is no clear consensus on the solutions or methods to proceed.

Contextual Notes

There are indications of confusion regarding the number of solutions and the nature of the polynomial equation, as well as a mention of personal circumstances affecting participation in the discussion.

evagelos
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Homework Statement


The equation is:[tex]\frac{(a-x)^4+(x-b)^4}{(a+b-2x)^2}=\frac{a^4+b^4}{(a+b)^2}[/tex]


Homework Equations





The Attempt at a Solution



I expanded the terms on the L.H.S and ended up with an equation of 4th,3rd,2nd ,1st degree equation very difficult to solve.
 
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evagelos said:

Homework Statement


The equation is:[tex]\frac{(a-x)^4+(x-b)^4}{(a+b-2x)^2}=\frac{a^4+b^4}{(a+b)^2}[/tex]


Homework Equations





The Attempt at a Solution



I expanded the terms on the L.H.S and ended up with an equation of 4th,3rd,2nd ,1st degree equation very difficult to solve.

Am I correct to assume that you are suppose to find a solution with respect to x?

I get that x is either [tex]x = \frac{2ab}{a+b}[/tex] or [tex]\frac{a^2+b^2}{a+b}[/tex] or x = 0

just remember that [tex](a-x)^4 = ((a-x)^2)^2[/tex]
 
Last edited:
Susanne217 said:
Am I correct to assume that you are suppose to find a solution with respect to x?

I get that x is either [tex]x = \frac{2ab}{a+b}[/tex] or [tex]\frac{a^2+b^2}{a+b}[/tex] or x = 0


Yes you right ,but how did you get those solutions?

But is it not there a forth solution ,or one of them is double??
 
evagelos said:
Yes you right ,but how did you get those solutions?

But is it not there a forth solution ,or one of them is double??

1) Use the fact that (x-a)^4 = ((x-a)^2)^2 and then solve with respect to x
 
Susanne217 said:
1) Use the fact that (x-a)^4 = ((x-a)^2)^2 and then solve with respect to x

I am sorry can you elaborate a little more ,i cannot follow.
 
evagelos said:
I am sorry can you elaborate a little more ,i cannot follow.

Sorry I didn't report back to yesterday had an upset stomac :(

You expand the lefthand side as I showed you and then solve it respect to x.
 

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