Question

How do you solve `(2-x)^4+(3-x)^4-(5-x)^4 = 0` ?

Answer 1

Use a numeric method to find approximations to the zeros:

`x_1 ~~ 3.49773`

`x_2 ~~ -10.5327`

`x_(3,4) ~~ 3.5175+-1.3997i`

Explanation:

Note that:

`(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4`

Hence:

`(2-x)^4 = 16-32x+24x^2-8x^3+x^4`

`(3-x)^4 = 81-108x+54x^2-12x^3+x^4`

`(2-x)^4 + (3-x)^4 = 97-140x+78x^2-20x^3+2x^4`

`(5-x)^4 = 625-500x+150x^2-20x^3+x^4`

`(2-x)^4+(3-x)^4-(5-x)^4 = x^4-72x^2+360x-528`

So the roots of the original equation are the zeros of:

`f(x) = x^4-72x^2+360x-528`

By the rational root theorem, any rational zeros of this quartic are factors of `528`. That gives us rather a lot of factors to check, and none works. It is somewhat quicker to find approximations to the zeros using a method like Durand-Kerner.

Here's an example C++ program for this quartic...

Using this program I found approximations:

`x_1 ~~ 3.49773`

`x_2 ~~ -10.5327`

`x_(3,4) ~~ 3.5175+-1.3997i`