x-1-x-5-and-x-y-4-find-x-2-1-y-2-

Question Number 191451 by mathlove last updated on 24/Apr/23

$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{5}\:\:{and}\:{x}\centerdot{y}=\mathrm{4}\:{find} \\ $$$${x}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} =? \\ $$

Answered by mr W last updated on 24/Apr/23

x^2 −5x+1=0 x=((5±(√(21)))/2) x^2 =5x−1 ((1/y))^2 =((x/4))^2 =(x^2 /(16)) x^2 +((1/y))^2 =((17x^2 )/(16))=((17)/(16))(5x−1) =((17)/(16))(5×((5±(√(21)))/2)−1)=((391±85(√(21)))/(32))

$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{5}\pm\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{5}{x}−\mathrm{1} \\ $$$$\left(\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} =\left(\frac{{x}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{\mathrm{16}} \\ $$$${x}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} =\frac{\mathrm{17}{x}^{\mathrm{2}} }{\mathrm{16}}=\frac{\mathrm{17}}{\mathrm{16}}\left(\mathrm{5}{x}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{17}}{\mathrm{16}}\left(\mathrm{5}×\frac{\mathrm{5}\pm\sqrt{\mathrm{21}}}{\mathrm{2}}−\mathrm{1}\right)=\frac{\mathrm{391}\pm\mathrm{85}\sqrt{\mathrm{21}}}{\mathrm{32}} \\ $$

Commented by mehdee42 last updated on 24/Apr/23

$${sir}\:{W} \\ $$$${please}\:{guide}\:{me}\:{how}\:{can}\:{i}\:{submit}\:\:{a}\:{question}\:{in}\:{the}\:{group}? \\ $$

Commented by Tinku Tara last updated on 24/Apr/23

$$\mathrm{Just}\:\mathrm{press}\:+\:\mathrm{button}\:\mathrm{at}\:\mathrm{top}. \\ $$

Commented by mr W last updated on 24/Apr/23

$${to}\:{compose}\:{a}\:{new}\:{question}\:{just}\:{press} \\ $$$$“+''\:{button}\:{at}\:{the}\:{top}; \\ $$$${to}\:{post}\:{an}\:{image}\:{as}\:{a}\:{question}\:{just} \\ $$$${press}\:“{image}''\:{button}\:{at}\:{the}\:{top}. \\ $$

Commented by mr W last updated on 24/Apr/23