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Question Number 220193 by behi834171 last updated on 07/May/25
x^8 =21x+13 ; x∈R x=?
$$\:\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{8}} =\mathrm{21}\boldsymbol{{x}}+\mathrm{13}\:\:\:\:\:\:\:\:\:\:;\:\:\:\:\boldsymbol{{x}}\in{R} \\ $$$$\:\:\:\:\boldsymbol{{x}}=? \\ $$
Answered by SdC355 last updated on 07/May/25
f(z)=z^8 −21z−13 f(−1)f(0)<0....? f(−1)=1+21−13=22−13=9 f(0)=−13 ∴ f(−1)f(0)<0 and f(0)f(1)<0.....? f(0)=13 f(1)=1−21−13=−33 ∴ f(0)f(1)<0 z_(n+1) =z_n −((f(z_n ))/(f^((1)) (z_n ))) z^8 −21z−13=0 Solution z_(n+1) =z_n −((z_n ^8 −21z_n −13)/(8z_n ^7 −21)) and Let z_1 =1 lim_(n→∞) z_(n+1) =−0.6180339887498948 First Solution and Let z_1 ′=2 lim_(n→∞) z_(n+1) =1.6180339887498948 Second Solution
$$\:{f}\left({z}\right)={z}^{\mathrm{8}} −\mathrm{21}{z}−\mathrm{13} \\ $$$${f}\left(−\mathrm{1}\right){f}\left(\mathrm{0}\right)<\mathrm{0}….? \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{1}+\mathrm{21}−\mathrm{13}=\mathrm{22}−\mathrm{13}=\mathrm{9} \\ $$$${f}\left(\mathrm{0}\right)=−\mathrm{13} \\ $$$$\therefore\:{f}\left(−\mathrm{1}\right){f}\left(\mathrm{0}\right)<\mathrm{0} \\ $$$$\mathrm{and}\: \\ $$$${f}\left(\mathrm{0}\right){f}\left(\mathrm{1}\right)<\mathrm{0}…..? \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{13} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{21}−\mathrm{13}=−\mathrm{33} \\ $$$$\therefore\:{f}\left(\mathrm{0}\right){f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${z}_{{n}+\mathrm{1}} ={z}_{{n}} −\frac{{f}\left({z}_{{n}} \right)}{{f}^{\left(\mathrm{1}\right)} \left({z}_{{n}} \right)} \\ $$$${z}^{\mathrm{8}} −\mathrm{21}{z}−\mathrm{13}=\mathrm{0}\:\:\mathrm{Solution} \\ $$$${z}_{{n}+\mathrm{1}} ={z}_{{n}} −\frac{{z}_{{n}} ^{\mathrm{8}} −\mathrm{21}{z}_{{n}} −\mathrm{13}}{\mathrm{8}{z}_{{n}} ^{\mathrm{7}} −\mathrm{21}} \\ $$$$\mathrm{and}\:\mathrm{Let}\:{z}_{\mathrm{1}} =\mathrm{1}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{z}_{{n}+\mathrm{1}} =−\mathrm{0}.\mathrm{6180339887498948}\:\:\mathrm{First}\:\mathrm{Solution} \\ $$$$\mathrm{and}\:\mathrm{Let}\:{z}_{\mathrm{1}} '=\mathrm{2} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{z}_{{n}+\mathrm{1}} =\mathrm{1}.\mathrm{6180339887498948}\:\mathrm{Second}\:\mathrm{Solution} \\ $$
Commented by behi834171 last updated on 07/May/25
thank you so much sir. any roots?
$${thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$$${any}\:{roots}? \\ $$$$ \\ $$
Answered by Frix last updated on 07/May/25
ϕ^2 =ϕ+1 ϕ^4 =ϕ^2 +2ϕ+1=3ϕ+2 ϕ^8 =9ϕ^2 +12ϕ+4=21ϕ+13 ⇒ x_1 =ϕ=(1/2)+((√5)/2) Similar x_2 =−(1/ϕ)=(1/2)−((√5)/2) x^8 −21x−13=0 (x^2 −x−1)(x^6 +x^5 +2x^4 +3x^3 +5x^2 +8x+13)=0 F_1 =0, F_2 =1, F_n =F_(n−2) +F_(n−1) ∀n≥3 x^n −F_n x−F_(n+1) =0 ⇒ Solutions are always x=(1/2)±((√5)/2)
$$\varphi^{\mathrm{2}} =\varphi+\mathrm{1} \\ $$$$\varphi^{\mathrm{4}} =\varphi^{\mathrm{2}} +\mathrm{2}\varphi+\mathrm{1}=\mathrm{3}\varphi+\mathrm{2} \\ $$$$\varphi^{\mathrm{8}} =\mathrm{9}\varphi^{\mathrm{2}} +\mathrm{12}\varphi+\mathrm{4}=\mathrm{21}\varphi+\mathrm{13} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =\varphi=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{Similar}\:{x}_{\mathrm{2}} =−\frac{\mathrm{1}}{\varphi}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{8}} −\mathrm{21}{x}−\mathrm{13}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\left({x}^{\mathrm{6}} +{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{13}\right)=\mathrm{0} \\ $$$$ \\ $$$${F}_{\mathrm{1}} =\mathrm{0},\:{F}_{\mathrm{2}} =\mathrm{1},\:{F}_{{n}} ={F}_{{n}−\mathrm{2}} +{F}_{{n}−\mathrm{1}} \forall{n}\geqslant\mathrm{3} \\ $$$${x}^{{n}} −{F}_{{n}} {x}−{F}_{{n}+\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{Solutions}\:\mathrm{are}\:\mathrm{always}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 08/May/25
x^8 =21x+13 ; x=? let x^2 =ax+b x^4 =a^2 x^2 +2abx+b^2 =a^2 (ax+b)+2abx+b^2 =a^3 x+2abx+ab+b^2 =(a^3 +2ab)x+ab+b^2 x^8 =(a^3 +2ab)^2 x^2 +2(a^3 +2ab)(ab+b^2 )x+(ab+b^2 )^2 =(a^3 +2ab)^2 (ax+b)+2(a^3 +2ab)(ab+b^2 )x+(ab+b^2 )^2 ={a(a^3 +2ab)^2 +2(a^3 +2ab)(ab+b^2 )}x+b(a^3 +2ab)^2 +(ab+b^2 )^2 { ((a(a^3 +2ab)^2 +2(a^3 +2ab)(ab+b^2 )=21)),((b(a^3 +2ab)^2 +(ab+b^2 )^2 =13)) :} a,b=? a=1,b=1 satisfy the system Hence, x^2 =x+1⇒x^2 −x−1=0 x=((1±(√(1+4)))/2) =((1±(√5))/2)
$${x}^{\mathrm{8}} =\mathrm{21}{x}+\mathrm{13}\:;\:{x}=? \\ $$$$\: \\ $$$${let}\:{x}^{\mathrm{2}} ={ax}+{b} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{4}} ={a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{abx}+{b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} \left({ax}+{b}\right)+\mathrm{2}{abx}+{b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{3}} {x}+\mathrm{2}{abx}+{ab}+{b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right){x}+{ab}+{b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{8}} =\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)\left({ab}+{b}^{\mathrm{2}} \right){x}+\left({ab}+{b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)^{\mathrm{2}} \left({ax}+{b}\right)+\mathrm{2}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)\left({ab}+{b}^{\mathrm{2}} \right){x}+\left({ab}+{b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\left\{{a}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)^{\mathrm{2}} +\mathrm{2}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)\left({ab}+{b}^{\mathrm{2}} \right)\right\}{x}+{b}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)^{\mathrm{2}} +\left({ab}+{b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\begin{cases}{{a}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)^{\mathrm{2}} +\mathrm{2}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)\left({ab}+{b}^{\mathrm{2}} \right)=\mathrm{21}}\\{{b}\left({a}^{\mathrm{3}} +\mathrm{2}{ab}\right)^{\mathrm{2}} +\left({ab}+{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{13}}\end{cases}\:{a},{b}=? \\ $$$${a}=\mathrm{1},{b}=\mathrm{1}\:{satisfy}\:{the}\:{system} \\ $$$${Hence}, \\ $$$${x}^{\mathrm{2}} ={x}+\mathrm{1}\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}\:=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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