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Question-68960

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Question Number 68960 by ajfour last updated on 17/Sep/19
Commented by TawaTawa last updated on 17/Sep/19
Wow, weldone sir, God bless you sir
Wow,weldonesir,Godblessyousir
Commented by ajfour last updated on 17/Sep/19
a=b=c=d=e=−1 λ^2 −λ−1=0 ⇒ λ=((1±(√5))/2) A=−λ^2 −λ−1=−(2+λ) =−(((5±(√5))/2)) Q_1 =q^2 −q−1 , Q_2 =−q^2 −q−1 L_1 =2λq++b(λ+q)+2d =2(((1±(√5))/2))q−((1±(√5))/2)−q−2 =±(√5)q−(((5±(√5))/2)) L_2 =2aλq+c(λ+q)+2e =−2(((1±(√5))/2))q−((1±(√5))/2)−q−2 =−(2±(√5))q−(((5±(√5))/2)) Now AQ_1 ^2 −L_1 L_2 Q_1 +L_1 ^2 Q_2 =0 −(((5±(√5))/2))(q^2 −q−1)^2 −[±(√5)q−(((5±(√5))/2))][−(2±(√5))q−(((5±(√5))/2))] ×(q^2 −q−1) +[±(√5)q−(((5±(√5))/2))]^2 (−q^2 −q−1)=0 ⇒ −2(5±(√5))(q^2 −q−1)^2 +[±2(√5)q−(5±(√5))][2(2±(√5))q+(5±(√5)] ×(q^2 −q−1) −[±2(√5)q−(5±(√5))]^2 (q^2 +q+1)=0 ⇒ let 5±(√5)=n ⇒ −2n(q^2 −q−1)^2 +[−10q+2nq−n] ×[−6q+2nq+n](q^2 −q−1) −[−10q+2nq−n]^2 (q^2 +q+1)=0 ⇒ −2n(q^2 −q−1)^2 + [(2n−10)(2n−6)q^2 −4nq−n^2 ] ×(q^2 −q−1)+ −[(2n−10)^2 q^2 −2n(2n−10)q+n^2 ] ×(q^2 +q+1) = 0 ⇒ −2n(q^4 −2q^3 −q^2 +2q+1)+ +{4(n−5)(n−3)q^4 −4q^3 [(n−5)(n−3)+n] −q^2 [4(n−5)(n−3)+n^2 −4n] +q(n^2 +4n)+n^2 } −[(2n−10)^2 q^2 −2n(2n−10)q+n^2 ] ×(q^2 +q+1)=0 ⇒ q^4 {−2n+4n^2 −32n+60 −4n^2 +40n−100} +q^3 {4n−4n^2 +28n−60 −4n^2 +40n−100 +4n^2 −20n} +q^2 {2n−5n^2 +36n−60 −4n^2 +40n−100 +4n^2 −20n−n^2 } +q{−4n+n^2 +4n 4n^2 −20n−n^2 } (−2n+n^2 −n^2 )=0 Reducing (6n−40)q^4 −(4n^2 −52n+160)q^3 −(6n^2 −58n+160)q^2 +4(n^2 −5n)q−2n=0 Again dividing by 2 (3n−20)q^4 −(2n^2 −26n+80)q^3 −(3n^2 −29n+80)q^2 +(2n^2 −10n)q−n=0 to remind again n=5±(√5) taking n_1 =5+(√5) n^2 (−2q^3 −3q^2 +2q)+ n(3q^4 +26q^3 +29q^2 −10q−1)+ 1(−20q^4 −80q^3 −80q^2 )=0 since n=5±(√5) (n−5)^2 =5 n^2 −10n+20=0 n^2 =10n−20 So, i can write n(−20q^3 −30q^2 +20q) −20(−2q^3 −3q^2 +2q) n(3q^4 +26q^3 +29q^2 −10q−1)+ 1(−20q^4 −80q^3 −80q^2 )=0 ⇒ n(3q^4 +6q^3 −q^2 +10q−1) +(−20q^4 −40q^3 −20q^2 −40q)=0 ⇒ n(3q^4 +6q^3 −q^2 +10q−1) −20(q^4 +2q^3 +q^2 +2q)=0 ________________________ (3n−20)(q^4 +2q^3 +q^2 +2q) −n(4q^2 −4q+1) = 0 _______________________ 3n−20 = −5±3(√5) = N 6n−40=−10±6(√5) −n−20 = −25∓(√5) = −((N+80)/3) 10n−40 = 10±10(√5) = 10(((N+8)/3)) ⇒ q^4 +2q^3 −(((N+80)/(3N)))q^2 +(((10N+80)/(3N)))q−(((N+20)/(3N)))=0
a=b=c=d=e=−1λ2−λ−1=0⇒λ=1±52A=−λ2−λ−1=−(2+λ)=−(5±52)Q1=q2−q−1,Q2=−q2−q−1L1=2λq++b(λ+q)+2d=2(1±52)q−1±52−q−2=±5q−(5±52)L2=2aλq+c(λ+q)+2e=−2(1±52)q−1±52−q−2=−(2±5)q−(5±52)NowAQ12−L1L2Q1+L12Q2=0−(5±52)(q2−q−1)2−[±5q−(5±52)][−(2±5)q−(5±52)]×(q2−q−1)+[±5q−(5±52)]2(−q2−q−1)=0⇒−2(5±5)(q2−q−1)2+[±25q−(5±5)][2(2±5)q+(5±5]×(q2−q−1)−[±25q−(5±5)]2(q2+q+1)=0⇒let5±5=n⇒−2n(q2−q−1)2+[−10q+2nq−n]×[−6q+2nq+n](q2−q−1)−[−10q+2nq−n]2(q2+q+1)=0⇒−2n(q2−q−1)2+[(2n−10)(2n−6)q2−4nq−n2]×(q2−q−1)+−[(2n−10)2q2−2n(2n−10)q+n2]×(q2+q+1)=0⇒−2n(q4−2q3−q2+2q+1)++{4(n−5)(n−3)q4−4q3[(n−5)(n−3)+n]−q2[4(n−5)(n−3)+n2−4n]+q(n2+4n)+n2}−[(2n−10)2q2−2n(2n−10)q+n2]×(q2+q+1)=0⇒q4{−2n+4n2−32n+60−4n2+40n−100}+q3{4n−4n2+28n−60−4n2+40n−100+4n2−20n}+q2{2n−5n2+36n−60−4n2+40n−100+4n2−20n−n2}+q{−4n+n2+4n4n2−20n−n2}(−2n+n2−n2)=0Reducing(6n−40)q4−(4n2−52n+160)q3−(6n2−58n+160)q2+4(n2−5n)q−2n=0Againdividingby2(3n−20)q4−(2n2−26n+80)q3−(3n2−29n+80)q2+(2n2−10n)q−n=0toremindagainn=5±5takingn1=5+5n2(−2q3−3q2+2q)+n(3q4+26q3+29q2−10q−1)+1(−20q4−80q3−80q2)=0sincen=5±5(n−5)2=5n2−10n+20=0n2=10n−20So,icanwriten(−20q3−30q2+20q)−20(−2q3−3q2+2q)n(3q4+26q3+29q2−10q−1)+1(−20q4−80q3−80q2)=0⇒n(3q4+6q3−q2+10q−1)+(−20q4−40q3−20q2−40q)=0⇒n(3q4+6q3−q2+10q−1)−20(q4+2q3+q2+2q)=0________________________(3n−20)(q4+2q3+q2+2q)−n(4q2−4q+1)=0_______________________3n−20=−5±35=N6n−40=−10±65−n−20=−25∓5=−N+80310n−40=10±105=10(N+83)⇒q4+2q3−(N+803N)q2+(10N+803N)q−(N+203N)=0
Commented by TawaTawa last updated on 17/Sep/19
Is this for only a = b = c = d = e = − 1 ?
Isthisforonlya=b=c=d=e=−1?

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