Question Number 186241 by a.lgnaoui last updated on 02/Feb/23
$${x}+\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}? \\ $$
Commented by MJS_new last updated on 02/Feb/23
$$\mathrm{are}\:\mathrm{you}\:\mathrm{unable}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{quadratic}? \\ $$
Commented by a.lgnaoui last updated on 02/Feb/23
$${The}\:{value}\:\left(\mathrm{5}{x}\right)^{\mathrm{5}} =?,\:{without}\:{calcul}\:{of}\:{x}. \\ $$
Commented by a.lgnaoui last updated on 02/Feb/23
$${I}\:{want}\:{by}\:{this}\:{simple}\:{equation}\:{to} \\ $$$${evaluate}\:\left(\mathrm{5}{x}\right)^{\mathrm{5}} \:{without}\:{calcul}\:{of}\:{x}? \\ $$$$ \\ $$
Answered by Frix last updated on 02/Feb/23
$${x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\mathrm{i} \\ $$
Answered by MJS_new last updated on 02/Feb/23
$${x}+\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{2}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{know} \\ $$$$\left({x}^{\mathrm{2}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right){x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right){x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)={x}^{\mathrm{5}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{5}} =\mathrm{1}\:\Rightarrow\:\left(\mathrm{5}{x}\right)^{\mathrm{5}} =\mathrm{3125} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 02/Feb/23
$${good} \\ $$