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x-1-0-find-x-Mastermind-




Question Number 169396 by Mastermind last updated on 29/Apr/22
(√x)+1=0  find x    Mastermind
x+1=0findxMastermind
Commented by mr W last updated on 29/Apr/22
i^4 =1  (√i^4 )+1=2≠0
i4=1i4+1=20
Commented by infinityaction last updated on 29/Apr/22
i^4  where i is a complex number
i4whereiisacomplexnumber
Commented by mr W last updated on 29/Apr/22
(√x)+1=0 has no solution even for  x∈C.
x+1=0hasnosolutionevenforxC.
Commented by Mastermind last updated on 29/Apr/22
Wow!  thanks man
Wow!thanksman
Commented by Mastermind last updated on 29/Apr/22
i love the way you answered it
ilovethewayyouansweredit
Commented by Mastermind last updated on 29/Apr/22
Thank you too, for attempting it
Thankyoutoo,forattemptingit
Commented by Mastermind last updated on 29/Apr/22
But why didn′t you consider law of  exponential  (√((x^n )))=x^(n/2)   Now, x=i^4   (√((i^4 )))=i^(4/2) ⇒i^2 =−1    I want you to check this @mr W
Butwhydidntyouconsiderlawofexponential(xn)=xn2Now,x=i4(i4)=i42i2=1Iwantyoutocheckthis@mrW
Commented by mr W last updated on 30/Apr/22
i just say: be careful! otherwise you  will get things like this:  1=(1)^(3/2) =((−1)^2 )^(3/2) =(−1)^(2×(3/2)) =(−1)^3 =−1
ijustsay:becareful!otherwiseyouwillgetthingslikethis:1=(1)32=((1)2)32=(1)2×32=(1)3=1
Commented by infinityaction last updated on 30/Apr/22
i^4  = 1   (√x) = −1  i^2   = −1  (√x) = i^2   x = i^4
i4=1x=1i2=1x=i2x=i4
Commented by mr W last updated on 30/Apr/22
(√x)=i^2 ⇎x=i^4   be careful again!  when you square an equation, you  get a new equation which may contain   roots which are not the roots of the   original equation. so you must check  if the roots fulfill the original  equation. in current case you′ll see  that x=i^4  doesn′t fulfill the original  equation.
x=i2x=i4becarefulagain!whenyousquareanequation,yougetanewequationwhichmaycontainrootswhicharenottherootsoftheoriginalequation.soyoumustcheckiftherootsfulfilltheoriginalequation.incurrentcaseyoullseethatx=i4doesntfulfilltheoriginalequation.
Commented by mr W last updated on 30/Apr/22
you may insist on that x=i^4  is the  root of (√x)+1=0. that′s your right.  but nevertheless it is wrong!
youmayinsistonthatx=i4istherootofx+1=0.thatsyourright.butneverthelessitiswrong!
Commented by mr W last updated on 30/Apr/22
Commented by infinityaction last updated on 30/Apr/22
got it sir  thank you sir
gotitsirthankyousir
Commented by MJS_new last updated on 30/Apr/22
steps we use to solve i.e.  x^3 =−1; x∈C  x^3 =e^(iπ)   we must search for x=e^(iθ)  with e^(3iθ) =e^(iπ)   ⇒ 3θ=π ⇔ θ=(π/3) ⇒ x=e^(i(π/3)) =−(1/2)+((√3)/2)i  but there′s more:  x_k ^3 =e^(i(π+2πn)) ; n=0, 1, 2∧k=n+1  this “trick” on the rhs leads to 3 solutions.  we did nothing on the lhs  x_1 ^3 =e^(iπ)  ⇒ x_1 =e^(i(π/3)) =−(1/2)+((√3)/2)i  x_2 ^3 =e^(3iπ)  ⇒ x_2 =e^(iπ) =−1  x_3 ^3 =e^(5iπ)  ⇒ x_3 =e^(i((5π)/3)) =−(1/2)−((√3)/2)i  tbese 3 solutions represent 3 different points  in the complex plane  now test the solutions  (e^(i(π/3)) )^3 =e^(iπ) =−1 true  (e^(iπ) )^3 =e^(3iπ) =e^(iπ) =−1 true  (e^(i((5π)/3)) )^3 =e^(5iπ) =e^(iπ) =−1 true    now let′s try  x^(1/3) =−1; x∈C  x^(1/3) =e^(iπ)   obviously our ”trick” on the rhs doesn′t  work (n∈N because we need a countable  number of solutions)  we must search for x=e^(iθ)  with e^(i(θ/3)) =e^(iπ)   ⇒ (θ/3)=π ⇔ θ=3π ⇒ x=e^(3iπ) =e^(iπ) =−1  but here we did something on the lhs  test: (e^(iπ) )^(1/3) =e^(i(π/3)) ≠−1  so here′s a totally different case  the rule x^(1/3) =−(−x)^(1/3)  is only valid for x∈R  strictly following the rules of C x^(1/3) =−1 has  no solution    now try  x^(1/2) =−1; x∈C  x^(1/2) =e^(iπ)   x=e^(iθ)  with e^(i(θ/2)) =e^(iπ)   ⇒ (θ/2)=π ⇔ θ=2π ⇒ x=e^(2iπ) =e^(0iπ) =e^0 =1  again we did something on the lhs  test: (e^0 )^(1/2) =e^0 =1≠−1  again no solution in C and obviously none  in R (because (√1)=1; also f(x)=(√x) is only  a half parabola)    ⇒ we need a different approach for solving  equations in C
stepsweusetosolvei.e.x3=1;xCx3=eiπwemustsearchforx=eiθwithe3iθ=eiπ3θ=πθ=π3x=eiπ3=12+32ibuttheresmore:xk3=ei(π+2πn);n=0,1,2k=n+1thistrickontherhsleadsto3solutions.wedidnothingonthelhsx13=eiπx1=eiπ3=12+32ix23=e3iπx2=eiπ=1x33=e5iπx3=ei5π3=1232itbese3solutionsrepresent3differentpointsinthecomplexplanenowtestthesolutions(eiπ3)3=eiπ=1true(eiπ)3=e3iπ=eiπ=1true(ei5π3)3=e5iπ=eiπ=1truenowletstryx1/3=1;xCx1/3=eiπobviouslyourtrickontherhsdoesntwork(nNbecauseweneedacountablenumberofsolutions)wemustsearchforx=eiθwitheiθ3=eiπθ3=πθ=3πx=e3iπ=eiπ=1butherewedidsomethingonthelhstest:(eiπ)1/3=eiπ31soheresatotallydifferentcasetherulex1/3=(x)1/3isonlyvalidforxRstrictlyfollowingtherulesofCx1/3=1hasnosolutionnowtryx1/2=1;xCx1/2=eiπx=eiθwitheiθ2=eiπθ2=πθ=2πx=e2iπ=e0iπ=e0=1againwedidsomethingonthelhstest:(e0)1/2=e0=11againnosolutioninCandobviouslynoneinR(because1=1;alsof(x)=xisonlyahalfparabola)weneedadifferentapproachforsolvingequationsinC

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