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Sum-of-two-irrational-numbers-is-1-less-than-their-product-and-8-less-than-their-sum-of-squares-Find-the-larger-of-the-two-numbers-

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Question Number 199011 by necx122 last updated on 26/Oct/23
Sum of two irrational numbers is 1 less than their product, and 8 less than their sum of squares. Find the larger of the two numbers.
Sumoftwoirrationalnumbersis1lessthantheirproduct,and8lessthantheirsumofsquares.Findthelargerofthetwonumbers.
Commented by nikif99 last updated on 27/Oct/23
a+b=ab−1 ⇒b=((a+1)/(a−1)) (1) ab−1=a^2 +b^2 −8 ⇒^((1)) a((a+1)/(a−1))=a^2 +(((a+1)/(a−1)))^2 −8 ⇒ ((a(a+1)(a−1))/((a−1)^2 ))−(((a−1)^2 )/((a−1)^2 ))=((a^2 (a−1)^2 )/((a−1)^2 ))+(((a+1)^2 )/((a−1)^2 ))−((8(a−1)^2 )/((a−1)^2 )) ⇒ a(a^2 −1)+(7−a^2 )(a−1)^2 −(a+1)^2 =0 ⇒ a^4 −3a^3 −5a^2 +17a−6=0 ⇒^(Peano) a=2∨a=3, ∈N rejected ((a^4 −3a^3 −5a^2 +17a−6)/((a−2)(a−3)))=a^2 +2a−1 ⇒a=−1+(√2)∨a=−1−(√2) greatest a=−1+(√2)
a+b=ab−1⇒b=a+1a−1(1)ab−1=a2+b2−8⇒(1)aa+1a−1=a2+(a+1a−1)2−8⇒a(a+1)(a−1)(a−1)2−(a−1)2(a−1)2=a2(a−1)2(a−1)2+(a+1)2(a−1)2−8(a−1)2(a−1)2⇒a(a2−1)+(7−a2)(a−1)2−(a+1)2=0⇒a4−3a3−5a2+17a−6=0⇒Peanoa=2∨a=3,∈Nrejecteda4−3a3−5a2+17a−6(a−2)(a−3)=a2+2a−1⇒a=−1+2∨a=−1−2greatesta=−1+2
Answered by Rasheed.Sindhi last updated on 27/Oct/23
a,b are two irrational numbers a+b=ab−1 a+b=a^2 +b^2 −8 a+b=(a+b)^2 −2ab−8 (a+b)^2 −(a+b)−2ab−8=0 Let a+b=t=ab−1⇒ab=t+1 t^2 −t−2(t+1)−8=0 t^2 −3t−10=0 (t−5)(t+2)=0 t=a+b=ab−1=5 , −2 { ((a+b=5⇒ab=6⇒(a,b)=(2,3) or (3,2)^★ )),((a+b=−2⇒ab=−1^(★★) )) :} ^★ a+b=5 , ab=6 , specify integer values for a,b Hence rejected. ^(★★) a+b=−2,ab=−1: b=−(1/a) a−(1/a)=−2 a^2 +2a−1=0 a=((−2±(√(4+4)))/2)=−1±(√2) b=−(1/a)=−(1/(−1±(√2)))×((−1∓(√2))/(−1∓(√2)))=−((−1∓(√2))/(1−2))=−1∓(√2) (a,b)=(−1+(√(2 )) , −1−(√2) ),(−1−(√(2 )) , −1+(√2) ) Clearly, Larger value=−1+(√2)
a,baretwoirrationalnumbersa+b=ab−1a+b=a2+b2−8a+b=(a+b)2−2ab−8(a+b)2−(a+b)−2ab−8=0Leta+b=t=ab−1⇒ab=t+1t2−t−2(t+1)−8=0t2−3t−10=0(t−5)(t+2)=0t=a+b=ab−1=5,−2{a+b=5⇒ab=6⇒(a,b)=(2,3)or(3,2)★a+b=−2⇒ab=−1★★★a+b=5,ab=6,specifyintegervaluesfora,bHencerejected.★★a+b=−2,ab=−1:b=−1aa−1a=−2a2+2a−1=0a=−2±4+42=−1±2b=−1a=−1−1±2×−1∓2−1∓2=−−1∓21−2=−1∓2(a,b)=(−1+2,−1−2),(−1−2,−1+2)Clearly,Largervalue=−1+2
Commented by Rasheed.Sindhi last updated on 27/Oct/23
necx sir please go through my solution once more. I′ve corrected it now!
necxsirpleasegothroughmysolutiononcemore.I′vecorrecteditnow!
Commented by necx122 last updated on 27/Oct/23
how is that possible sir? by merely looking at both values it seems like 1+(√2) is greater
howisthatpossiblesir?bymerelylookingatbothvaluesitseemslike1+2isgreater
Commented by mr W last updated on 27/Oct/23
but the two numbers are −1±(√2). so −1+(√2) is the larger one and −1−(√2) is the smaller one.
butthetwonumbersare−1±2.so−1+2isthelargeroneand−1−2isthesmallerone.
Commented by necx122 last updated on 27/Oct/23
Yes, it's clear now. I think the error was just an oversight. Thank you, sir.
Commented by Rasheed.Sindhi last updated on 27/Oct/23
You′re right mr W sir. I′ve found my mistake and now corrected it!
You′reright\boldsymbolmr\boldsymbolW\boldsymbolsir.I′vefoundmymistakeandnowcorrectedit!

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