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Given-a-gt-b-gt-0-a-amp-b-real-number-such-that-a-2-ab-b-2-7-and-a-ab-b-1-find-the-value-of-a-2-b-2-




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&brealnumbersuchthata2ab+b2=7andaab+b=1.findthevalueofa2b2
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{a2ab+b2=7(1)aab+b=1(2).aab+b=1b=a+1a1(3).Replaceinto(1)wegeta2+(a+1a1)2a2+aa1=7a2+a2+2a+1a22a+1a2+aa1=7a42a3+a2+a2+2a+1(a3a)=7a214a+7a43a35a2+17a6=0(a3)(a2)(a2+2a1)=0a{2,3,1+2}.Replaceinto(1)wegetb{3,2,12}butsincea>b>0,wegetb=2Thus,a2b2=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>0whya2b2negativesir?
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=ab1(a+b)2=3ab+73ab+7=(ab1)2(ab)25ab6=0a,b>0ab=6a2+b2=13a+b=5a2+b2=(ab)2+2ab=(ab)2+12(ab)2=a2+b212=1ab=1a2b2=(ab)(a+b)=5
Commented by bemath last updated on 07/Oct/20
thank you prof
thankyouprof

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