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solve-for-R-a-bcd-100-b-cda-100-c-dab-100-d-abc-100-

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Question Number 177112 by mr W last updated on 01/Oct/22
solve for R a+bcd=100 b+cda=100 c+dab=100 d+abc=100
solveforRa+bcd=100b+cda=100c+dab=100d+abc=100
Answered by aleks041103 last updated on 01/Oct/22
1 case:a=b=c=d a+a^3 =100⇒a=b=c=d≈4.5698 Now a^2 +abcd=100a b^2 +abcd=100b c^2 +abcd=100c d^2 +abcd=100d ⇒a^2 −b^2 =100(a−b) ⇒(a+b−100)(a−b)=0 ⇒(a+c−100)(a−c)=0 ⇒(a+d−100)(a−d)=0 ⇒(b+c−100)(b−c)=0 ⇒(b+d−100)(b−d)=0 ⇒(c+d−100)(c−d)=0 2 case:a=b≠c≠d b≠c⇒b+c=100 c≠a⇒a+c=100 a≠d⇒a+d=100 ⇒c=d 3 case:a=b=x≠y=c=d x+xy^2 =100 y+yx^2 =100 ⇒(x+y−100)(x−y)=0 ⇒x+y=100 ⇒x=100−y ⇒x(1+y^2 )=100 (100−y)(1+y^2 )=100 100−y+100y^2 −y^3 =100 ⇒y^3 −100y^2 +y=0 ⇒y=0(obv. wrong) and y=((100±(√(100^2 −4)))/2)=50±(√(2499)) ⇒a=b=50±(√(2499)) and c=d=50∓(√(2499)) answer: a=b=c=d=x≈4.5698 (x^3 +x−100=0) and a=b=50±(√(2499)) and c=d=50∓(√(2499)) (with all of its permutations)
1case:a=b=c=da+a3=100⇒a=b=c=d≈4.5698Nowa2+abcd=100ab2+abcd=100bc2+abcd=100cd2+abcd=100d⇒a2−b2=100(a−b)⇒(a+b−100)(a−b)=0⇒(a+c−100)(a−c)=0⇒(a+d−100)(a−d)=0⇒(b+c−100)(b−c)=0⇒(b+d−100)(b−d)=0⇒(c+d−100)(c−d)=02case:a=b≠c≠db≠c⇒b+c=100c≠a⇒a+c=100a≠d⇒a+d=100⇒c=d3case:a=b=x≠y=c=dx+xy2=100y+yx2=100⇒(x+y−100)(x−y)=0⇒x+y=100⇒x=100−y⇒x(1+y2)=100(100−y)(1+y2)=100100−y+100y2−y3=100⇒y3−100y2+y=0⇒y=0(obv.wrong)andy=100±1002−42=50±2499⇒a=b=50±2499andc=d=50∓2499answer:a=b=c=d=x≈4.5698(x3+x−100=0)anda=b=50±2499andc=d=50∓2499(withallofitspermutations)
Commented by mr W last updated on 01/Oct/22
thanks sir!
thankssir!
Commented by Tawa11 last updated on 02/Oct/22
Great sirs
Greatsirs
Answered by Rasheed.Sindhi last updated on 02/Oct/22
solve for R ;a+bcd=100,b+cda=100, c+dab=100,d+abc=100_ (i)−(ii): a−cda−b+bcd=0 a(1−cd)−b(1−cd)=0 (a−b)(1−cd)=0 a=b ∣ cd=1 (i)+(ii): (a+b)(1+cd)=200 (a=b ∨ cd=1)∧ (a+b)(1+cd)=200 2a(1+1)=200 a=50 Similar way: (a=c ∨ bd=1) a=d ∣ bc=1 b=c ∣ ad=1 b=d ∣ ac=1 c=d ∣ ab=1 Similar way a=c ∣ bd=1...... a=d ∣ bc=1 b=c ∣ ad=1 b=d ∣ ac=1 c=d ∣ ab=1 a=b=c=d ....
solveforR;a+bcd=100,b+cda=100,c+dab=100,d+abc=100―(i)−(ii):a−cda−b+bcd=0a(1−cd)−b(1−cd)=0(a−b)(1−cd)=0a=b∣cd=1(i)+(ii):(a+b)(1+cd)=200(a=b∨cd=1)∧(a+b)(1+cd)=2002a(1+1)=200a=50Similarway:(a=c∨bd=1)a=d∣bc=1b=c∣ad=1b=d∣ac=1c=d∣ab=1Similarwaya=c∣bd=1……a=d∣bc=1b=c∣ad=1b=d∣ac=1c=d∣ab=1a=b=c=d….
Commented by mr W last updated on 01/Oct/22
thanks sir!
thankssir!
Answered by mr W last updated on 01/Oct/22
a^2 −100a+abcd=0 b^2 −100b+abcd=0 c^2 −100c+abcd=0 d^2 −100d+abcd=0 let k=abcd we see a,b,c,d are the roots of the uadratic e uation x^2 −100x+k=0. since a quadratic equation can have at most two different roots, we can only have following cases: case 1: two roots are equal, ⇒a,b,c,d are equal. a=b=c=d a^3 +a−100=0 ⇒a=((((√(202503))/9)+50))^(1/3) −((((√(202503))/9)−50))^(1/3) ≈4.56978 ⇒(a,b,c,d)=(4.56978,4.56978,4.56978,4.56978) case 2: two roots are different, a≠b=c=d (as example) a+b^3 =100 ...(i) b+ab^2 =100 ...(ii) (i)−(ii): (a−b)(1−b^2 )=0 ⇒1−b^2 =0 ⇒b=±1 ⇒a=100−b^3 =100∓1 i.e. b=1, a=99 or b=−1, a=101 ⇒(a,b,c,d)=(99,1,1,1) or (101,−1,−1,−1) and we can exchange a with b,c,d. case 3: two roots are different, a=b≠c=d (as example) a+ac^2 =100 ...(iii) c+a^2 c=100 ...(iv) (iii)−(iv): (a−c)(1−ac)=0 ⇒1−ac=0 ⇒ac=1 ⇒a+c=100 ⇒a,c are roots of z^2 −100z+1=0 a,c=50±(√(50^2 −1))=50±7(√(51)) ⇒(a,b,c,d)=(50+7(√(51)),50+7(√(51)),50−7(√(51)),50−7(√(51))) or (50−7(√(51)),50−7(√(51)),50+7(√(51)),50+7(√(51))) and we can exchange a, c with a,b or a,d.
a2−100a+abcd=0b2−100b+abcd=0c2−100c+abcd=0d2−100d+abcd=0letk=abcdweseea,b,c,daretherootsoftheuadraticeuationx2−100x+k=0.sinceaquadraticequationcanhaveatmosttwodifferentroots,wecanonlyhavefollowingcases:case1:tworootsareequal,⇒a,b,c,dareequal.a=b=c=da3+a−100=0⇒a=2025039+503−2025039−503≈4.56978⇒(a,b,c,d)=(4.56978,4.56978,4.56978,4.56978)case2:tworootsaredifferent,a≠b=c=d(asexample)a+b3=100…(i)b+ab2=100…(ii)(i)−(ii):(a−b)(1−b2)=0⇒1−b2=0⇒b=±1⇒a=100−b3=100∓1i.e.b=1,a=99orb=−1,a=101⇒(a,b,c,d)=(99,1,1,1)or(101,−1,−1,−1)andwecanexchangeawithb,c,d.case3:tworootsaredifferent,a=b≠c=d(asexample)a+ac2=100…(iii)c+a2c=100…(iv)(iii)−(iv):(a−c)(1−ac)=0⇒1−ac=0⇒ac=1⇒a+c=100⇒a,carerootsofz2−100z+1=0a,c=50±502−1=50±751⇒(a,b,c,d)=(50+751,50+751,50−751,50−751)or(50−751,50−751,50+751,50+751)andwecanexchangea,cwitha,bora,d.

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