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y-x-3-x-c-find-the-roots-0-c-2-3-3-




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

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