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Question Number 182571 by Khalmohmmad last updated on 11/Dec/22

Answered by aleks041103 last updated on 11/Dec/22

((x^2 .((x^2 .((x^2 ....))^(1/3) ))^(1/3) ))^(1/3) =y⇒y=((x^2 y))^(1/3)   ⇒y^3 =x^2 y⇒y=±x, but obv. y>0⇒y=∣x∣  (√(x^2 +x(√(x^2 +x(√(x^2 +x....))))))=z  ⇒(√(x^2 +xz))=z  ⇒z^2 =x^2 +xz⇒z^2 −xz−x^2 =0  ⇒z_(1/2) =((x±(√(x^2 +4x^2 )))/2)=((1±(√5))/2)x  but z>0⇒z= { ((((1+(√5))/2)x, x>0)),((((1−(√5))/2)x, x<0)) :}  ⇒5= { ((((1+(√5))/2) (x/(∣x∣)),x>0)),((((1−(√5))/2) (x/(∣x∣)),x<0)) :}= { ((((1+(√5))/2),x>0)),((undef x=0)),(((((√5)−1)/2),x<0)) :}  ⇒No solution!

$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} .\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} .\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} ....}}}={y}\Rightarrow{y}=\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} {y}} \\ $$$$\Rightarrow{y}^{\mathrm{3}} ={x}^{\mathrm{2}} {y}\Rightarrow{y}=\pm{x},\:{but}\:{obv}.\:{y}>\mathrm{0}\Rightarrow{y}=\mid{x}\mid \\ $$$$\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}^{\mathrm{2}} +{x}....}}}={z} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} +{xz}}={z} \\ $$$$\Rightarrow{z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{xz}\Rightarrow{z}^{\mathrm{2}} −{xz}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{z}_{\mathrm{1}/\mathrm{2}} =\frac{{x}\pm\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}{x} \\ $$$${but}\:{z}>\mathrm{0}\Rightarrow{z}=\begin{cases}{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x},\:{x}>\mathrm{0}}\\{\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x},\:{x}<\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\mathrm{5}=\begin{cases}{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\frac{{x}}{\mid{x}\mid},{x}>\mathrm{0}}\\{\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\frac{{x}}{\mid{x}\mid},{x}<\mathrm{0}}\end{cases}=\begin{cases}{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},{x}>\mathrm{0}}\\{{undef}\:{x}=\mathrm{0}}\\{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}},{x}<\mathrm{0}}\end{cases} \\ $$$$\Rightarrow{No}\:{solution}! \\ $$

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