How do you solve (2-x)^4+(3-x)^4-(5-x)^4 = 0(2−x)4+(3−x)4−(5−x)4=0 ?
1 Answer
Use a numeric method to find approximations to the zeros:
x_1 ~~ 3.49773x1≈3.49773
x_2 ~~ -10.5327x2≈−10.5327
x_(3,4) ~~ 3.5175+-1.3997ix3,4≈3.5175±1.3997i
Explanation:
Note that:
(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4(a+b)4=a4+4a3b+6a2b2+4ab3+b4
Hence:
(2-x)^4 = 16-32x+24x^2-8x^3+x^4(2−x)4=16−32x+24x2−8x3+x4
(3-x)^4 = 81-108x+54x^2-12x^3+x^4(3−x)4=81−108x+54x2−12x3+x4
(2-x)^4 + (3-x)^4 = 97-140x+78x^2-20x^3+2x^4(2−x)4+(3−x)4=97−140x+78x2−20x3+2x4
(5-x)^4 = 625-500x+150x^2-20x^3+x^4(5−x)4=625−500x+150x2−20x3+x4
(2-x)^4+(3-x)^4-(5-x)^4 = x^4-72x^2+360x-528(2−x)4+(3−x)4−(5−x)4=x4−72x2+360x−528
So the roots of the original equation are the zeros of:
f(x) = x^4-72x^2+360x-528f(x)=x4−72x2+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 ~~ 3.49773x1≈3.49773
x_2 ~~ -10.5327x2≈−10.5327
x_(3,4) ~~ 3.5175+-1.3997ix3,4≈3.5175±1.3997i
