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a-b-c-abc-a-2-b-2-c-2-49-ab-bc-ca-




Question Number 217735 by ArshadS last updated on 19/Mar/25
 a + b + c= abc  a^2  + b^2  + c^2  = 49   ab+bc+ca=?   .           .
a+b+c=abca2+b2+c2=49ab+bc+ca=?..
Commented by mr W last updated on 19/Mar/25
there is no unique solution.   maybe you should ask  (ab+bc+ca)_(min) =?  or  (ab+bc+ca)_(max) =?
thereisnouniquesolution.maybeyoushouldask(ab+bc+ca)min=?or(ab+bc+ca)max=?
Answered by mr W last updated on 20/Mar/25
assume a,b,c ∈R  say u=a+b, v=ab  then we have  a+b−abc=−c  ⇒u−cv=−c   ...(i)  (a+b)^2 −2ab=49−c^2   ⇒u^2 −2v=49−c^2    ...(ii)  we can solve for u, v in terms of c.  cu^2 −2u−c(51−c^2 )=0  ⇒u=((1±(√(1+c^2 (51−c^2 ))))/c)  ⇒v=(u/c)+1=2±(√(1+c^2 (51−c^2 )))  1+c^2 (51−c^2 )≥0  c^4 −51c^2 −1≤0  ⇒((51−(√(51^2 +4)))/2)<0≤c^2 ≤((51+(√(51^2 +4)))/2)  ⇒−(√((51+(√(2605)))/2))≤c≤(√((51+(√(2605)))/2))≈7.143    say s=ab+bc+ca, then  s=ab+(a+b)c    =v+cu    =3±2(√(1+c^2 (51−c^2 )))  we see s is not unique.  but c^2 (51−c^2 )≤(((c^2 +51−c^2 )/2))^2 =((51^2 )/4)  we have  s=3+2(√(1+c^2 (51−c^2 )))≤3+2(√(1+((51^2 )/4)))  i.e. s_(max) =3+2(√(1+((51^2 )/4)))=3+(√(2605))  similarly  s=3−2(√(1+c^2 (51−c^2 )))≥3−2(√(1+((51^2 )/4)))  i.e. s_(min) =3−2(√(1+((51^2 )/4)))=3−(√(2605))    summary:  with given conditions we can not  uniquely determine a, b, c and  ab+bc+ca, but we can say that  −(√((51+(√(2605)))/2))≤a, b, c≤(√((51+(√(2605)))/2))≈7.143  3−(√(2605))≤ab+bc+ca≤3+(√(2605))
assumea,b,cRsayu=a+b,v=abthenwehavea+babc=cucv=c(i)(a+b)22ab=49c2u22v=49c2(ii)wecansolveforu,vintermsofc.cu22uc(51c2)=0u=1±1+c2(51c2)cv=uc+1=2±1+c2(51c2)1+c2(51c2)0c451c21051512+42<0c251+512+4251+26052c51+260527.143says=ab+bc+ca,thens=ab+(a+b)c=v+cu=3±21+c2(51c2)weseesisnotunique.butc2(51c2)(c2+51c22)2=5124wehaves=3+21+c2(51c2)3+21+5124i.e.smax=3+21+5124=3+2605similarlys=321+c2(51c2)321+5124i.e.smin=321+5124=32605summary:withgivenconditionswecannotuniquelydeterminea,b,candab+bc+ca,butwecansaythat51+26052a,b,c51+260527.14332605ab+bc+ca3+2605
Commented by ArshadS last updated on 22/Mar/25
Nice sir,thanks a lot!
Nicesir,thanksalot!

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