Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 149238 by ajfour last updated on 04/Aug/21

z^3 −z=c z^4 −z^2 =cz let z=x+q x^4 +4qx^3 +6q^2 x^2 +4q^3 x+q^4 −x^2 −2qx−q^2 −cx−cq=0 ⇒ x^4 +4qx^3 +(6q^2 −1)x^2 + (4q^3 −2q−c)x+q^4 −q^2 −cq=0 let q^2 =1/6 ⇒ x^4 +4qx^3 −(((4q)/3)+c)x−(5/(36))−cq=0 ≡ (x^2 +mx+k)(x^2 +sx+h)=0 ⇒ m+s=4q h+k+ms=0 mh+ks=−(((4q)/3)+c) kh=−(5/(36))−cq h, k= −((ms)/2)±(√(((m^2 s^2 )/4)+(5/(36))+cq)) m(−((ms)/2)+D)+s(−((ms)/2)−D) +((4q)/3)+c=0 (m−s)D−((ms)/2)(4q)+((4q)/3)+c=0 now let′s assume ms=p & since m+s=4q (m−s)^2 =(8/3)−4p ⇒ ((8/3)−4p)((p^2 /4)+(5/(36))+cq) =(2pq−((4q)/3)−c)^2 p^3 +((5/9)+4cq−(8/9)−4cq)p +(((4q)/3)+c)^2 −(8/3)((5/(36))+cq)=0 ⇒ p^3 −(p/3)+((14)/(27))=0 D_0 =((7/(27)))^2 −((1/9))^3 ....

$${z}^{\mathrm{3}} −{z}={c} \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} ={cz} \\ $$$${let}\:\:{z}={x}+{q} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} +\mathrm{6}{q}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{3}} {x}+{q}^{\mathrm{4}} \\ $$$$−{x}^{\mathrm{2}} −\mathrm{2}{qx}−{q}^{\mathrm{2}} −{cx}−{cq}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} +\left(\mathrm{6}{q}^{\mathrm{2}} −\mathrm{1}\right){x}^{\mathrm{2}} + \\ $$$$\:\:\:\:\left(\mathrm{4}{q}^{\mathrm{3}} −\mathrm{2}{q}−{c}\right){x}+{q}^{\mathrm{4}} −{q}^{\mathrm{2}} −{cq}=\mathrm{0} \\ $$$${let}\:\:{q}^{\mathrm{2}} =\mathrm{1}/\mathrm{6}\:\:\Rightarrow \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} −\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right){x}−\frac{\mathrm{5}}{\mathrm{36}}−{cq}=\mathrm{0} \\ $$$$\equiv\:\left({x}^{\mathrm{2}} +{mx}+{k}\right)\left({x}^{\mathrm{2}} +{sx}+{h}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{m}+{s}=\mathrm{4}{q} \\ $$$$\:\:\:\:\:\:{h}+{k}+{ms}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{mh}+{ks}=−\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right) \\ $$$$\:\:\:\:\:\:{kh}=−\frac{\mathrm{5}}{\mathrm{36}}−{cq} \\ $$$${h},\:{k}=\:−\frac{{ms}}{\mathrm{2}}\pm\sqrt{\frac{{m}^{\mathrm{2}} {s}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+{cq}} \\ $$$${m}\left(−\frac{{ms}}{\mathrm{2}}+{D}\right)+{s}\left(−\frac{{ms}}{\mathrm{2}}−{D}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}=\mathrm{0} \\ $$$$\left({m}−{s}\right){D}−\frac{{ms}}{\mathrm{2}}\left(\mathrm{4}{q}\right)+\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}=\mathrm{0} \\ $$$${now}\:{let}'{s}\:{assume}\:\:{ms}={p} \\ $$$$\&\:{since}\:\:{m}+{s}=\mathrm{4}{q} \\ $$$$\left({m}−{s}\right)^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{4}{p} \\ $$$$\Rightarrow\:\:\left(\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{4}{p}\right)\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+{cq}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{2}{pq}−\frac{\mathrm{4}{q}}{\mathrm{3}}−{c}\right)^{\mathrm{2}} \\ $$$$\:\:{p}^{\mathrm{3}} +\left(\frac{\mathrm{5}}{\mathrm{9}}+\mathrm{4}{cq}−\frac{\mathrm{8}}{\mathrm{9}}−\mathrm{4}{cq}\right){p} \\ $$$$\:\:\:\:\:\:\:+\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right)^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{3}}\left(\frac{\mathrm{5}}{\mathrm{36}}+{cq}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{3}} −\frac{{p}}{\mathrm{3}}+\frac{\mathrm{14}}{\mathrm{27}}=\mathrm{0} \\ $$$${D}_{\mathrm{0}} =\left(\frac{\mathrm{7}}{\mathrm{27}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{3}} \\ $$$$.... \\ $$

Commented by Tawa11 last updated on 04/Aug/21

Longest time sir.

$$\mathrm{Longest}\:\mathrm{time}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com