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Question Number 68422 by ajfour last updated on 10/Sep/19
Answered by ajfour last updated on 10/Sep/19
C=10m3p2+b(m3+6m2p+3mp2)+c(3m2+6mp+p2)+d(6m+4p)+10eD=10m2p3+b(3m2p+6mp2+p3)+c(m2+6mp+3p2)+d(4m+6p)+10eC−D=0gives−10m2p2(m−p)+b[(m3−p3)+6mp(m−p)−3mp(m−p)]+c[3(m2−p2)−(m2−p2)]+d[6(m−p)−4(m−p)]+0=0Ifmwouldbepthenz=m=potherwisedividingbym−pgives10m2p2+b(m2+4mp+p2)+c[2(m+p)]+2d=0.....(I)(ifm≠−peven)eq.(i)×(m+p)gives10m2p2(m+p)+b[m3+4m2p+mp2+m2p+4mp2+p3]+c[2m2+4mp+2p2]+d(2m+2p)=0......(i)NowC+D=0gives−10m2p2(m+p)+b[m3+p3+6mp(m+p)+3mp(m+p)]+c(4m2+4p2+12mp)+d(10m+10p)+20e=0....(II)(II)−(i)gives−b[4mp(m+p)]+c[2m2+8mp+2p2]+d[8(m+p)]+20e=0⇒2bmp(m+p)+c[(m+p)2+2mp]+4d(m+p)+10e=0....(III)Nowletstransformeq.(I)&(III)letm+p=s,andmp=gEq.(I)−10g2+b(s2+2g)+2cs+2d=0Eq.(III)−2bgs+c(s2+2g)+4ds+10e=0⇒g=−cs2+4ds+10e2(bs+c)substitutingforgineq.(I)−10(cs2+4ds+10e)2−4b(cs2+4ds+10e)(bs+c)+4(bs2+2cs+2d)(bs+c)2=0⇒5(cs2+4ds+10e)2−2b(bs+c)(cs2+4ds+10e)+2(bs+c)2(bs2+2cs+2d)=0Thisisadegree4eq.ins.Soitseemsishallobtainsandg=−cs2+4ds+10e2(bs+c).m,p=s2±s24−gNowwethenhaveAt5+Bt4+Et+F=0ort5+(BA)t4+(EA)t+FA=0butifm=pclearlyz=mhenceallcoefficientsgetzeroandweneedtochangetoz=mt+pt−1;thiscaseishalltakeupinanotherpost..
Commented by TawaTawa last updated on 10/Sep/19
wow,Godblessyousirandincreaseyourknowledge.Iwilllearnthiswithanexample.
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