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Find-all-pair-x-y-of-real-numbers-that-are-the-solutions-to-the-system-x-4-2x-3-y-1-4-3-y-4-2y-3-x-1-4-3-

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Question Number 119849 by benjo_mathlover last updated on 27/Oct/20
Find all pair(x,y) of real numbers that are the solutions to the system { ((x^4 +2x^3 −y=−(1/4)+(√3))),((y^4 +2y^3 −x=−(1/4)−(√3))) :}
Findallpair(x,y)ofrealnumbersthatarethesolutionstothesystem{x4+2x3−y=−14+3y4+2y3−x=−14−3
Answered by 1549442205PVT last updated on 27/Oct/20
Adding up two the equations we get x^4 +2x^3 −y+(1/4)+y^4 +2y^3 −x+(1/4)=0 ⇔(x^2 +x−(1/2))^2 +(y^2 +y−(1/2))^2 =0 ⇔ { ((x^2 +x−(1/2)=0)),((y^2 +y−(1/2)=0)) :} Δ=1+4.0.5=3,so x= { ((x=((−1±(√3))/2))),((y=((−1±(√3))/2))) :} Thus the solutions of given system are (x,y)∈{(((−1−(√3))/2),((−1−(√3))/2)),(((−1−(√3))/2),((−1+(√3))/2)) ,(((−1+(√3))/2),((−1−(√3))/2)),(((−1+(√3))/2),((−1+(√3))/2))}
Addinguptwotheequationswegetx4+2x3−y+14+y4+2y3−x+14=0⇔(x2+x−12)2+(y2+y−12)2=0⇔{x2+x−12=0y2+y−12=0Δ=1+4.0.5=3,sox={x=−1±32y=−1±32Thusthesolutionsofgivensystemare(x,y)∈{(−1−32,−1−32),(−1−32,−1+32),(−1+32,−1−32),(−1+32,−1+32)}

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