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Question Number 156921 by cortano last updated on 17/Oct/21

x^3 −4x^2 −3=0 x∈R

x3−4x2−3=0x∈R

Answered by mr W last updated on 17/Oct/21

path 1: ((1/x))^3 +(4/3)((1/x))−(1/3)=0 applying cardano: (1/x)=(((√(((1/6))^2 +((4/9))^3 ))+(1/6)))^(1/3) −(((√(((1/6))^2 +((4/9))^3 ))−(1/6)))^(1/3) ⇒x=(1/( ((((√(337))/(54))+(1/6)))^(1/3) −((((√(337))/(54))−(1/6)))^(1/3) ))≈4.1723 path 2: let x=t+(4/3) t^3 +3×(4/3)t^2 +3×((16)/9)t+(4^3 /3^3 )−4(t^2 +2×(4/3)t+(4^2 /3^2 ))−3=0 t^3 −((16)/3)t−((209)/(27))=0 applying cardano: t=(((√((((209)/(54)))^2 −(((16)/9))^3 ))+((209)/(54))))^(1/3) −(((√((((209)/(54)))^2 −(((16)/9))^3 ))−((209)/(54))))^(1/3) x=(4/3)+((((√(337))/6)+((209)/(54))))^(1/3) −((((√(337))/6)−((209)/(54))))^(1/3) ≈4.1723

path1:(1x)3+43(1x)−13=0applyingcardano:1x=(16)2+(49)3+163−(16)2+(49)3−163⇒x=133754+163−33754−163≈4.1723path2:letx=t+43t3+3×43t2+3×169t+4333−4(t2+2×43t+4232)−3=0t3−163t−20927=0applyingcardano:t=(20954)2−(169)3+209543−(20954)2−(169)3−209543x=43+3376+209543−3376−209543≈4.1723

Commented by cortano last updated on 17/Oct/21

cardano by cortano

cardanobycortano

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