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