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

[evagelos]

Homework Statement

The equation is:$$\frac{(a-x)^4+(x-b)^4}{(a+b-2x)^2}=\frac{a^4+b^4}{(a+b)^2}$$

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.

 

[Susanne217]

Homework Statement

The equation is:$$\frac{(a-x)^4+(x-b)^4}{(a+b-2x)^2}=\frac{a^4+b^4}{(a+b)^2}$$

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 $$x = \frac{2ab}{a+b}$$ or $$\frac{a^2+b^2}{a+b}$$ or x = 0

just remember that $$(a-x)^4 = ((a-x)^2)^2$$

 

[evagelos]

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

I get that x is either $$x = \frac{2ab}{a+b}$$ or $$\frac{a^2+b^2}{a+b}$$ 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??

 

[Susanne217]

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

 

[evagelos]

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.

 

[Susanne217]

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.