How do you solve #(2-x)^4+(3-x)^4-(5-x)^4 = 0# ?
1 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
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#