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Solve-for-a-a-2-1-a-a-1-a-2-




Question Number 6590 by Temp last updated on 04/Jul/16
Solve for a  ((a^2 +1)/a)=(a/(1−a^2 ))
$$\mathrm{Solve}\:\mathrm{for}\:{a} \\ $$$$\frac{{a}^{\mathrm{2}} +\mathrm{1}}{{a}}=\frac{{a}}{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$
Commented by nburiburu last updated on 04/Jul/16
1−a^4 =a^2   0=a^4 +a^2 −1  t=a^2   0=t^2 +t−1  t_1 =((−1+(√(1+4)))/2)=(((√5)−1)/2)  t_2 =((−1−(√(1+4)))/2)=((−(√5)−1)/2)    a_1 =(√t_1 )=(√(((√5)−1)/2))  a_2 =−(√t_1 )=−(√(((√5)−1)/2))  a_3 =(√t_2 )=(√(((√5)+1)/2))i  a_4 =−(√t_2 )=−(√(((√5)+1)/2))i
$$\mathrm{1}−{a}^{\mathrm{4}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{0}={a}^{\mathrm{4}} +{a}^{\mathrm{2}} −\mathrm{1} \\ $$$${t}={a}^{\mathrm{2}} \\ $$$$\mathrm{0}={t}^{\mathrm{2}} +{t}−\mathrm{1} \\ $$$${t}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$${t}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{−\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${a}_{\mathrm{1}} =\sqrt{{t}_{\mathrm{1}} }=\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}} \\ $$$${a}_{\mathrm{2}} =−\sqrt{{t}_{\mathrm{1}} }=−\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}} \\ $$$${a}_{\mathrm{3}} =\sqrt{{t}_{\mathrm{2}} }=\sqrt{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}{i} \\ $$$${a}_{\mathrm{4}} =−\sqrt{{t}_{\mathrm{2}} }=−\sqrt{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}{i} \\ $$$$ \\ $$

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