Menu Close

in-question-34992-this-had-to-be-solved-t-4-8t-3-2t-2-8t-1-0-here-another-concept-worked-I-tried-some-t-1-a-b-c-d-t-2-a-b-c-d-t-3-a-b-c-d-t-3-a-b-c-d-t-t-1-t-t-2-t-t-3-t-t-4-0-leads

Tinku Tara 0 Comments
Pin




Question Number 36219 by MJS last updated on 30/May/18
in question 34992 this had to be solved: t^4 +8t^3 +2t^2 −8t+1=0 here another concept worked (I tried some) t_1 =a+b+c+d t_2 =a+b−c−d t_3 =a−b+c−d t_3 =a−b−c+d (t−t_1 )(t−t_2 )(t−t_3 )(t−t_4 )=0 leads to t^4 −4at^3 +2(3a^2 −b^2 −c^2 −d^2 )t^2 + +4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)t+ +a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 ) 1. −4a=8 2. 2(3a^2 −b^2 −c^2 −d^2 )=2 3. 4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)=−8 4. a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 )=1 1. a=−2 2. b^2 +c^2 +d^2 =11 3. b^2 +c^2 +d^2 +bcd=5 4. ... 3. −2. bcd=−6 now I tried by chance if there are easy real solutions b^2 =p, c^2 =q, d^2 =r, 0≤p≤q≤r p+q+r=11 this leads to p=2, q=3, p=6 ∣b∣=(√2), ∣c∣=(√3), ∣d∣=(√6) with one or three out of a, b, c negative because of bcd=−6 because of the nature of t_1 , t_2 , t_3 , t_4 these lead to the same solutions I took b=−(√2), c=−(√3), d=−(√6) t_1 =−2−(√2)−(√3)−(√6) t_2 =−2−(√2)+(√3)+(√6) t_3 =−2+(√2)−(√3)+(√6) t_4 =−2+(√2)+(√3)−(√6)
inquestion34992thishadtobesolved:t4+8t3+2t2−8t+1=0hereanotherconceptworked(Itriedsome)t1=a+b+c+dt2=a+b−c−dt3=a−b+c−dt3=a−b−c+d(t−t1)(t−t2)(t−t3)(t−t4)=0leadstot4−4at3+2(3a2−b2−c2−d2)t2++4(a(−a2+b2+c2+d2)−2bcd)t++a4+b4+c4+d4−2(a2(b2+c2+d2)+b2(c2+d2)+c2d2)1.−4a=82.2(3a2−b2−c2−d2)=23.4(a(−a2+b2+c2+d2)−2bcd)=−84.a4+b4+c4+d4−2(a2(b2+c2+d2)+b2(c2+d2)+c2d2)=11.a=−22.b2+c2+d2=113.b2+c2+d2+bcd=54.…3.−2.bcd=−6nowItriedbychanceifthereareeasyrealsolutionsb2=p,c2=q,d2=r,0⩽p⩽q⩽rp+q+r=11thisleadstop=2,q=3,p=6∣b∣=2,∣c∣=3,∣d∣=6withoneorthreeoutofa,b,cnegativebecauseofbcd=−6becauseofthenatureoft1,t2,t3,t4theseleadtothesamesolutionsItookb=−2,c=−3,d=−6t1=−2−2−3−6t2=−2−2+3+6t3=−2+2−3+6t4=−2+2+3−6
Commented by Rasheed.Sindhi last updated on 30/May/18
Creative work , I think! �� �� ✌
Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18
t^4 +8t^3 +2t^2 −8t+1=0 t^2 +8t+2−(8/t)+(1/t^2 )=0 t^2 +(1/t^2 )+8(t−(1/t))+2=0 (t−(1/t))^2 +2+8(t−(1/t))+2=0 k=t−(1/t) k^2 +8k+4=0 k=((−8±(√(64−16)) )/2) =((−8±4(√3) )/2) =−4±2(√3) k=−4+2(√3) and −4−2(√3) now t−(1/t)=k t^2 −1=tk t^2 −tk−1=0 t=((k±(√(k^2 +4)) )/2) =((k+(√(k^2 +4)) )/2) and ((k−(√(k^2 +4)) )/2) now put value of k=−4+2(√3) t=((−4+2(√3) ±(√(16−16(√3) +12+4)))/2) t=((−4+2(√3) ±(√(32−16(√3) )))/2) t=−2+(√3) ±(√(8−4(√3) )) t=−2+(√3) +(√(8−4(√3) )) and−2+(√3) −(√(8−4(√3) )) now put k=−4−2(√3) you will get two value of t so total 2+2=4 roots of t found
t4+8t3+2t2−8t+1=0t2+8t+2−8t+1t2=0t2+1t2+8(t−1t)+2=0(t−1t)2+2+8(t−1t)+2=0k=t−1tk2+8k+4=0k=−8±64−162=−8±432=−4±23k=−4+23and−4−23nowt−1t=kt2−1=tkt2−tk−1=0t=k±k2+42=k+k2+42andk−k2+42nowputvalueofk=−4+23t=−4+23±16−163+12+42t=−4+23±32−1632t=−2+3±8−43t=−2+3+8−43and−2+3−8−43nowputk=−4−23youwillgettwovalueoftsototal2+2=4rootsoftfound

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *