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Question Number 116824 by bemath last updated on 07/Oct/20

Given a>b>0 , a&b real number such that a^2 −ab+b^2 =7 and a−ab+b=−1. find the value of a^2 −b^2

Givena>b>0,a&brealnumbersuchthata2−ab+b2=7anda−ab+b=−1.findthevalueofa2−b2

Answered by 1549442205PVT last updated on 08/Oct/20

From the hypothesis we have { ((a^2 −ab+b^2 =7(1))),(( a−ab+b=−1(2).)) :} a−ab+b=−1⇒b=((a+1)/(a−1))(3).Replace into (1)we get a^2 +(((a+1)/(a−1)))^2 −((a^2 +a)/(a−1))=7 ⇔a^2 +((a^2 +2a+1)/(a^2 −2a+1))−((a^2 +a)/(a−1))=7 ⇒a^4 −2a^3 +a^2 +a^2 +2a+1−(a^3 −a) =7a^2 −14a+7⇔a^4 −3a^3 −5a^2 +17a−6=0 ⇔(a−3)(a−2)(a^2 +2a−1)=0 ⇔a∈{2,3,−1+(√2)}.Replace into (1) we get b∈{3,2,−1−(√2)}but since a>b>0,we get b=2 Thus,a^2 −b^2 =5

Fromthehypothesiswehave{a2−ab+b2=7(1)a−ab+b=−1(2).a−ab+b=−1⇒b=a+1a−1(3).Replaceinto(1)wegeta2+(a+1a−1)2−a2+aa−1=7⇔a2+a2+2a+1a2−2a+1−a2+aa−1=7⇒a4−2a3+a2+a2+2a+1−(a3−a)=7a2−14a+7⇔a4−3a3−5a2+17a−6=0⇔(a−3)(a−2)(a2+2a−1)=0⇔a∈{2,3,−1+2}.Replaceinto(1)wegetb∈{3,2,−1−2}butsincea>b>0,wegetb=2Thus,a2−b2=5

Commented by bemath last updated on 07/Oct/20

sir given condition a >b>0 why a^2 −b^2 negative sir?

sirgivenconditiona>b>0whya2−b2negativesir?

Commented by 1549442205PVT last updated on 07/Oct/20

Thank Sir.I mistaked and corrected

ThankSir.Imistakedandcorrected

Answered by MJS_new last updated on 07/Oct/20

a^2 +b^2 =ab+7 a+b=ab−1 (a+b)^2 =3ab+7 3ab+7=(ab−1)^2 (ab)^2 −5ab−6=0 a, b >0 ⇒ ab=6 a^2 +b^2 =13 a+b=5 a^2 +b^2 =(a−b)^2 +2ab=(a−b)^2 +12 (a−b)^2 =a^2 +b^2 −12=1 ⇒ a−b=1 ⇒ a^2 −b^2 =(a−b)(a+b)=5

a2+b2=ab+7a+b=ab−1(a+b)2=3ab+73ab+7=(ab−1)2(ab)2−5ab−6=0a,b>0⇒ab=6a2+b2=13a+b=5a2+b2=(a−b)2+2ab=(a−b)2+12(a−b)2=a2+b2−12=1⇒a−b=1⇒a2−b2=(a−b)(a+b)=5

Commented by bemath last updated on 07/Oct/20

thank you prof

thankyouprof

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