Answer:
y=x+4(1)
Then we have:
(y−1)4+(y+1)4=16
Expanding binomials, odd powers of y cancel, giving
2y4+12y2+2=16
⟹2y4+12y2−14=0
solve this quadratic in y2 for y2 then take square roots on both sides:
y=±−12±122+4⋅2⋅14−−−−−−−−−−−√4−−−−−−−−−−−−−−−−−−−√
substitute into (1):
x=−4±−12±122+4⋅2⋅14−−−−−−−−−−−√4−−−−−−−−−−−−−−−−−−−√
If you evaluate this for all four ± combinations the two real roots turn out to be −5 and −3.
Step-by-step explanation:
