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Question Number 68835 by MJS last updated on 16/Sep/19
reposting this because it′s not yet solved  correctly. I criticize that most of you people  do not test if your solutions fit the given  equations in many cases    x^2 +1−(√(x^3 +x))=6x    please determine  (1) how many real solutions  (2) how many complex solutions  we can expect  (3) solve it  (4) test your solutions
repostingthisbecauseitsnotyetsolvedcorrectly.Icriticizethatmostofyoupeopledonottestifyoursolutionsfitthegivenequationsinmanycasesx2+1x3+x=6xpleasedetermine(1)howmanyrealsolutions(2)howmanycomplexsolutionswecanexpect(3)solveit(4)testyoursolutions
Commented by Prithwish sen last updated on 16/Sep/19
x^2 −6x+1=(√(x^3 +x))  Squaring both sides we get  x^4 −13x^3 +38x^2 −13x+1=0  and this is the reciprocal equation of 1^(st)  type   ∴ (x^4 +1)−13x(x^2 +1)+38x^2 =0  Dividing the equation by x^2  we get  (x^2 +(1/x^2 ))−13(x+(1/x))+38 =0  (x+(1/x))^2 −13(x+(1/x))+36=0  putting x+(1/x) = y we get  y^2 −13y+36 =0  (y−4)(y−9)=0   and i.e  x^2 −4x+1= 0  and  x^2 −9x +1= 0  Solving we get     x = 2± (√3)  , ((9±(√(77)))/2)    now 2−(√3) and 2+(√3) are reciprocal  on the other hand ((9+(√(77)))/2)  and ((9−(√(77)))/2) are reciprocal.  now 2+(√(3 ))∽ 3.73  and 2−(√3) ∽ 0.268  ((9−(√(77)))/2) ∽ 0.11  and ((9+(√(77)))/2) ∽ 8.89  the equation satisfies only on 0.11 and 8.89  MJS Sir, please share your valuable comments.
x26x+1=x3+xSquaringbothsidesweget\boldsymbolx413\boldsymbolx3+38\boldsymbolx213\boldsymbolx+1=0\boldsymboland\boldsymbolthis\boldsymbolis\boldsymbolthe\boldsymbolreciprocal\boldsymbolequation\boldsymbolof1\boldsymbolst\boldsymboltype(x4+1)13x(x2+1)+38x2=0Dividingtheequationbyx2weget(x2+1x2)13(x+1x)+38=0(x+1x)213(x+1x)+36=0puttingx+1x=\boldsymbolywegety213y+36=0(y4)(y9)=0andi.ex24x+1=0andx29x+1=0Solvingweget\boldsymbolx=2±3,9±772\boldsymbolnow23\boldsymboland2+3\boldsymbolare\boldsymbolreciprocal\boldsymbolon\boldsymbolthe\boldsymbolother\boldsymbolhand9+772\boldsymboland9772\boldsymbolare\boldsymbolreciprocal.\boldsymbolnow2+33.73\boldsymboland230.26897720.11\boldsymboland9+7728.89\boldsymbolthe\boldsymbolequation\boldsymbolsatisfies\boldsymbolonly\boldsymbolon0.11\boldsymboland8.89\boldsymbolMJS\boldsymbolSir,\boldsymbolplease\boldsymbolshare\boldsymbolyour\boldsymbolvaluable\boldsymbolcomments.
Commented by MJS last updated on 16/Sep/19
good, thank you! we have to test, because  with squaring we get false solutions
good,thankyou!wehavetotest,becausewithsquaringwegetfalsesolutions
Commented by Prithwish sen last updated on 16/Sep/19
Thank you sir.
Thankyousir.
Commented by Rasheed.Sindhi last updated on 16/Sep/19
Slightly different(?) way  x^2 +1−(√(x^3 +x))=6x  x(x+(1/x))−x(√(x+(1/x)))−6x=0  x{(x+(1/x))−(√(x+(1/x)))−6}=0  x=0 (Not valid) ∣ (x+(1/x))−(√(x+(1/x)))−6=0  Let (√(x+(1/x)))=y  y^2 −y−6=0  (y−3)(y+2)=0  y=3  ∣   y=−2  (√(x+(1/x)))=3  ∣  (√(x+(1/x)))=−2 (false)  x+(1/x)=9  x^2 −9x+1=0  x=((9±(√(81−4)))/2)=((9±(√(77)))/2)  Substituting in original equation we  can see that these both are valid.
Slightlydifferent(?)wayx2+1x3+x=6xx(x+1x)xx+1x6x=0x{(x+1x)x+1x6}=0x=0(Notvalid)(x+1x)x+1x6=0Letx+1x=yy2y6=0(y3)(y+2)=0y=3y=2x+1x=3x+1x=2(false)x+1x=9x29x+1=0x=9±8142=9±772Substitutinginoriginalequationwecanseethatthesebotharevalid.
Answered by ajfour last updated on 16/Sep/19
(x^2 +1−6x)^2 =x^3 +x  dividing by x^2   (x+(1/x)−6)^2 =x+(1/x)  if we let  x+(1/x)=t     t^2 −13t+36=0     t=((13)/2)±(√(((169)/4)−36))     t=((13)/2)±(5/2)   ⇒   x+(1/x)=4, 9       x^2 −4x+1=0  ⇒         x=2±(√(4−1)) = 2±(√3)  >0  and from  x^2 −9x+1=0       x=(9/2)±(√(((81)/4)−1))         =(9/2)±((√(77))/2)    >0  x^3 −4x^2 +x=0   ⇒ x^3 +x=4x^2   or  x^3 −9x^2 +x=0 ⇒ x^3 +x=9x^2   testing       x^2 +1−6x=4∣x∣   and as all    found roots are >0  ⇒ x^2 −10x+1=0  ⇒  x=5±(√(25−1)) = 5±2(√6)      ⇒  x=2±(√3)  cannot be true      solutions.     x^2 +1−6x=3∣x∣   gives for x>0     x^2 −9x+1=0  (same eq.)  ⇒so x=((9±(√(77)))/2) are true solutions.
(x2+16x)2=x3+xdividingbyx2(x+1x6)2=x+1xifweletx+1x=tt213t+36=0t=132±169436t=132±52x+1x=4,9x24x+1=0x=2±41=2±3>0andfromx29x+1=0x=92±8141=92±772>0x34x2+x=0x3+x=4x2orx39x2+x=0x3+x=9x2testingx2+16x=4xandasallfoundrootsare>0x210x+1=0x=5±251=5±26x=2±3cannotbetruesolutions.x2+16x=3xgivesforx>0x29x+1=0(sameeq.)sox=9±772aretruesolutions.
Commented by MJS last updated on 16/Sep/19
good, thank you!
good,thankyou!good,thankyou!

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