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Question-66476




Question Number 66476 by aliesam last updated on 15/Aug/19
Commented by kaivan.ahmadi last updated on 15/Aug/19
t=4x+1⇒  (((t−1)^2 )/(((√t)−1)^2 ))=t+2⇒((((√t)−1)^2 ((√t)+1)^2 )/(((√t)−1)^2 ))=t+2⇒  ((√t)+1)^2 =t+2⇒t+1+2(√t)=t+2⇒2(√t)=1⇒(√t)=(1/2)⇒  t=(1/4)⇒  x=((t−1)/4)=(((1/4)−1)/4)=((−3)/(16))
$${t}=\mathrm{4}{x}+\mathrm{1}\Rightarrow \\ $$$$\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{t}}−\mathrm{1}\right)^{\mathrm{2}} }={t}+\mathrm{2}\Rightarrow\frac{\left(\sqrt{{t}}−\mathrm{1}\right)^{\mathrm{2}} \left(\sqrt{{t}}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{t}}−\mathrm{1}\right)^{\mathrm{2}} }={t}+\mathrm{2}\Rightarrow \\ $$$$\left(\sqrt{{t}}+\mathrm{1}\right)^{\mathrm{2}} ={t}+\mathrm{2}\Rightarrow{t}+\mathrm{1}+\mathrm{2}\sqrt{{t}}={t}+\mathrm{2}\Rightarrow\mathrm{2}\sqrt{{t}}=\mathrm{1}\Rightarrow\sqrt{{t}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow \\ $$$${x}=\frac{{t}−\mathrm{1}}{\mathrm{4}}=\frac{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}}{\mathrm{4}}=\frac{−\mathrm{3}}{\mathrm{16}} \\ $$$$ \\ $$

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