Find all pairs (a,b) of real numbers such that
a + b = 10
a^4 + b^4 = 2800
a + b = 10 → b = 10 - a (1)
a^4 + b^4 = 2800 (2)
Sub (1) into (2) and we have that
a^4 + ( 10 - a)^4 = 2800
a^4 + a^4 - 40a^3 + 600a^2 -4000a + 10000 = 2800
2a^4 - 40a^3 + 600a^2 - 4000a - 7200 = 0
a^4 - 20a^3 + 300a^2 - 2000a - 3600 = 0
The (real) solutions to this are
a ≈ 11.451 b ≈ -1.451
or
a ≈ -1.451 b ≈ 11.451