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x-2-x-a-x-a-a-R-solve-for-x-




Question Number 90194 by behi83417@gmail.com last updated on 21/Apr/20
(x^2 /(x+a))+(√x)=a     (a∈R)  solve for:  x  .
x2x+a+x=a(aR)solvefor:x.
Answered by ajfour last updated on 22/Apr/20
let (√x)=t  t^4 =(a−t)(t^2 +a)  t^4 +t^3 −at^2 +at−a^2 =0  let  (t^2 +pt+q)(t^2 +rt+m)=0  ⇒  t^4 +(p+r)t^3 +(m+pr+q)t^2     +(pm+qr)t+qm=0  ⇒  p+r=1  ,  pr+m+q=−a,    pm+qr =a ,  qm=−a^2   ⇒  pm−((a^2 r)/m)=a  pm^2 −am−a^2 r=0  m=(a/(2p))±(√((a^2 /(4p^2 ))+((a^2 r)/p)))     m= (a/(2p)){1±(√(1+4pr))}    q=−(a^2 /m) = (2ap)(({1∓(√(1+4pr))})/(4pr))    q= (a/(2r)){1∓(√(1+4pr))}  q+m=(a/(2p))+(a/(2r))±(√(1+4pr))((a/(2p))−(a/(2r)))  −a−pr=(a/2)(((p+r)/(pr)))±(a/2)(((r−p)/(pr)))(√(1+4pr))  let  pr=z  −a−z=(a/(2z))±(a/(2z))(√(1−16z^2 ))  ⇒  ((a/(2z))+z+a)^2 =(a^2 /(4z^2 ))(1−16z^2 )  z^2 +a^2 +a+2az+(a^2 /z)+4a^2 =0  z^3 +2az^2 +a(5a+1)z+a^2 =0  let  z=y−((2a)/3)  y^3 −2ay^2 +((4a^2 y)/3)−((8a^3 )/(27))      +2ay^2 −((8a^2 y)/3)+((8a^3 )/9)     +5a^2 y+ay−((10a^3 )/3)−((2a^2 )/3)+a^2 =0  y^3 +(((11a^2 )/3)+a)y−((70a^3 )/(27))+(a^2 /3)=0  .........
letx=tt4=(at)(t2+a)t4+t3at2+ata2=0let(t2+pt+q)(t2+rt+m)=0t4+(p+r)t3+(m+pr+q)t2+(pm+qr)t+qm=0p+r=1,pr+m+q=a,pm+qr=a,qm=a2pma2rm=apm2ama2r=0m=a2p±a24p2+a2rpm=a2p{1±1+4pr}q=a2m=(2ap){11+4pr}4prq=a2r{11+4pr}q+m=a2p+a2r±1+4pr(a2pa2r)apr=a2(p+rpr)±a2(rppr)1+4prletpr=zaz=a2z±a2z116z2(a2z+z+a)2=a24z2(116z2)z2+a2+a+2az+a2z+4a2=0z3+2az2+a(5a+1)z+a2=0letz=y2a3y32ay2+4a2y38a327+2ay28a2y3+8a39+5a2y+ay10a332a23+a2=0y3+(11a23+a)y70a327+a23=0
Commented by behi83417@gmail.com last updated on 22/Apr/20
thank you very much dear Ajfour.
thankyouverymuchdearAjfour.thankyouverymuchdearAjfour.
Commented by ajfour last updated on 22/Apr/20
i will try to finish..
iwilltrytofinish..iwilltrytofinish..

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