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Question-193617




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)}
$$\mathrm{x}\left(\mathrm{x}−\mathrm{3}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{3x} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2} \\ $$$$\mathrm{t}=\mathrm{x}^{\mathrm{2}} −\mathrm{3x} \\ $$$$\Leftrightarrow\mathrm{t}\left(\mathrm{t}+\mathrm{2}\right)+\mathrm{1}=\mathrm{0}\Rightarrow\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{t}=−\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{x}\in\left\{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$

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