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

1 Answer
Aug 5, 2016

Answer:

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...

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Using this program I found approximations:

#x_1 ~~ 3.49773#

#x_2 ~~ -10.5327#

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