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Question Number 193617 by SaRahAli last updated on 17/Jun/23

Answered by witcher3 last updated on 17/Jun/23

x(x−3)=x^2 −3x (x−1)(x−2)=x^2 −3x+2 t=x^2 −3x ⇔t(t+2)+1=0⇒(t+1)^2 =0⇒t=−1 x^2 −3x+1=0⇒x∈{((3−(√5))/2),((3+(√5))/2)}

x(x−3)=x2−3x(x−1)(x−2)=x2−3x+2t=x2−3x⇔t(t+2)+1=0⇒(t+1)2=0⇒t=−1x2−3x+1=0⇒x∈{3−52,3+52}

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