If α be a multiple root of the order 3 of the equation x4+bx2+cx+d=0 then α=8d3c

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tswe0uk · Accepted Answer

Step 1Let the roots be α,α,α,−3α so that their sum is 0.Now sum of products of 3 roots taken at a time isα3−3α3−3α3−3α3and this must equal -c and hence α3=c8Now product of roots is −3α4=d so that α4=−d3 and hence α=−d3c8=−8d3cYour question has thus a sign error. You can convince yourself by putting α=1 The desired polynomial is(x−1)3(x+3)=x4−6x2+8x−3so that c=8, d=−3 and −8d3c=1=αI checked your approach and your theorem in question and its proof are correct. Thus α should equal −3c4dLet&#039;s observe that sum of product of roots taken 2 at a time isα2+α2+α2−3α2−3α2−3α2and this should equal b. Thus b=−6α2 and hence−3c4b=−24α3(−24α2)=αSo we haveα=−3c4b=−8d3c

elvis0217t2x · Accepted Answer

Step 1Your answer looks correct to me.We have,(x−m)3(x−n)=x4+bx2+cx+d=0.x4+bx2+cx+d=x4−x3(3m+n)+x2(3m2+3mn)−x(m3+3m2n)+m3n{3m=−n3m(m+n)=bm2(m+3n)=−cm3n=d⟹{−6m2=b8m3=c⟹{m=−3c4b, b≠0n=9c4b, b≠0But, remember that there is a relationship between the coefficients. This implies that the coefficients are not independent. Therefore, to find the root we are looking for, we will use the coefficients that work for us. However, both answers are absolutely correct.{3m=−n3m(m+n)=bm2(m+3n)=−cm3n=d⟹{−3m4=d8m3=c⟹{m=−8d3cn=8dcThis means,3c4b=8d3cc2=329bd

If \alpha be a multiple root of the order 3

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