Question

(x+3)^4 + (x+5)^4= 16


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Answer 1

Step-by-step explanation:

Substitute:

y=x+4(1)(1)y=x+4

Then we have:

(y−1)4+(y+1)4=16(y−1)4+(y+1)4=16

Expanding binomials, odd powers of yy cancel, giving

2y4+12y2+2=162y4+12y2+2=16

⟹2y4+12y2−14=0⟹2y4+12y2−14=0

solve this quadratic in y2y2 for y2y2 then take square roots on both sides:

y=±−12±122+4⋅2⋅14−−−−−−−−−−−√4−−−−−−−−−−−−−−−−−−−√y=±−12±122+4⋅2⋅144

substitute into (1)(1):

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