Question Number 140604 by ajfour last updated on 10/May/21
$${y}={x}^{\mathrm{3}} −{x}−{c}\:\:\:;\:{find}\:{the}\:{roots}. \\ $$$$\:\:\mathrm{0}\leqslant{c}\leqslant\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Answered by ajfour last updated on 11/May/21
Commented by ajfour last updated on 11/May/21
$${AE}\:=\:\mathrm{1}\:={EH} \\ $$$${EP}\:=\:\boldsymbol{{x}}\:=\:{tan}\theta \\ $$$$\:{EQ}\:=\:\boldsymbol{{x}}^{\mathrm{2}} =\:{tan}^{\mathrm{2}} \theta \\ $$$$\:{HQ}=\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\boldsymbol{{BH}}=\:\boldsymbol{{c}}\:\:\:\left(\:{not}\:=\:\mathrm{2}{c}\:\:{as}\:{in}\:{image}\right) \\ $$$$\:\angle{BHQ}=\theta \\ $$$$\:\Rightarrow\:\frac{{BH}}{{HQ}}=\:{tan}\theta\:=\:\frac{{c}}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:{x} \\ $$$$\Rightarrow\:\:\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1}\right)=\boldsymbol{{c}} \\ $$$${lets}\:{find}\:{one}\:{x}\:{in}\:{terms}\:{of}\:\boldsymbol{{c}}. \\ $$$$\boldsymbol{{A}}{lso}\:\Rightarrow\:\:{x}^{\mathrm{2}} −\mathrm{1}=\frac{{c}}{{x}} \\ $$$${Let}\:{circumradius}\:\boldsymbol{\rho}\: \\ $$$${OE}\:=\:\frac{\boldsymbol{{c}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\rho^{\mathrm{2}} =\:\frac{{c}^{\mathrm{2}} }{\mathrm{4}}+{x}^{\mathrm{4}} \\ $$$${also}\:\:\:\:\boldsymbol{\rho}\:=\:\frac{{c}}{\mathrm{2}}+{x} \\ $$$$\Rightarrow\:\:\left(\rho−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{4}} +\frac{{c}^{\mathrm{2}} }{\mathrm{4}}=\:\rho^{\mathrm{2}} \\ $$$${let}\:\:\rho−\frac{{c}}{\mathrm{2}}\:=\:{s}\:;\:\:\frac{{c}}{\mathrm{2}}=\:{k} \\ $$$$\Rightarrow\:\:{s}^{\mathrm{4}} +{k}^{\mathrm{2}} ={s}^{\mathrm{2}} +\mathrm{2}{ks}+{k}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{s}^{\mathrm{3}} −{s}−\mathrm{2}{k}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{s}^{\mathrm{3}} −{s}−{c}\:=\:\mathrm{0} \\ $$$$……… \\ $$