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Question-143251

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Question Number 143251 by aliibrahim1 last updated on 12/Jun/21
Answered by MJS_new last updated on 12/Jun/21
let x=re^(iθ) ∧r>0∧0≤θ<2π re^(iθ) +(3/( r^(1/2) e^(iθ/2) ))=0 r^(3/2) e^(3iθ/2) +3=0 r^(3/2) e^(3iθ/2) =3e^(iπ) ⇒ r=3^(2/3) ∧θ=2π/3 ⇒ x=3^(2/3) e^(2iπ/3) ∨x=3^(2/3) e^(4iπ/3) x(√x)−2x−6= =−12−2×3^(2/3) (−(1/2)±((√3)/2)i)= =−12+3^(2/3) ±3^(7/6) i well well well...
letx=reiθ∧r>0∧0⩽θ<2πreiθ+3r1/2eiθ/2=0r3/2e3iθ/2+3=0r3/2e3iθ/2=3eiπ⇒r=32/3∧θ=2π/3⇒x=32/3e2iπ/3∨x=32/3e4iπ/3xx−2x−6==−12−2×32/3(−12±32i)==−12+32/3±37/6iwellwellwell…
Commented by MJS_new last updated on 12/Jun/21
corrected a typo
correctedatypo
Answered by Rasheed.Sindhi last updated on 12/Jun/21
x(√x)=−3 x^(3/2) =−3 x=(−3)^(2/3) =3^(2/3) x(√x)−2x−6=−3−2(3)^(2/3) −6 =−9−2(3)^(2/3)
xx=−3x3/2=−3x=(−3)2/3=32/3xx−2x−6=−3−2(3)2/3−6=−9−2(3)2/3
Commented by MJS_new last updated on 12/Jun/21
x=3^(2/3) (?) 3^(2/3) +(3/( (√3^(2/3) )))=3^(2/3) +(3/3^(1/3) )=2×3^(2/3) ≠0 (!)
x=32/3(?)32/3+332/3=32/3+331/3=2×32/3≠0(!)
Commented by aliibrahim1 last updated on 12/Jun/21
thank you good sir mjs
thankyougoodsirmjs
Commented by aliibrahim1 last updated on 12/Jun/21
thank you good sir rasheed
thankyougoodsirrasheed
Commented by Rasheed.Sindhi last updated on 12/Jun/21
Sorry mjs sir ,I′ve used an extraneous root (3^(2/3) ) to evaluate given expression!
Sorrymjssir,I′veusedanextraneousroot(32/3)toevaluategivenexpression!
Answered by Rasheed.Sindhi last updated on 12/Jun/21
x+(3/( (√x)))=0⇒x(√x)=−3 x^(3/2) =−3 x^3 =9 x=3^(2/3) , 3^(2/3) ω , 3^(2/3) ω^2 =3^(2/3) , 3^(2/3) (((−1+i(√3))/2)) , 3^(2/3) (((−1−i(√3))/2)) =3^(2/3) , 3^(2/3) (((−3^(2/3) +i(3^(2/3) )(3)^(1/2) )/2)) , 3^(2/3) (((−3^(2/3) −i(3^(2/3) )(3)^(1/2) )/2)) =3^(2/3) , (((−3^(2/3) +i(3^(7/6) ))/2)) , (((−3^(2/3) −i(3^(7/6) ))/2)) ^• x=3^(2/3) : x+(3/( (√x)))=0 3^(2/3) +(3/( (√3^(2/3) )))≠0 x+(1/( (√x)))=0 doesn′t satisfy ^• x=((−3^(2/3) +i(3^(7/6) ))/2): Assuming x+(3/( (√x)))=0 is satisfied x(√x)−2x−6 =−3−2(((−3^(2/3) +i(3^(7/6) ))/2))−6 =−9+3^(2/3) −i(3^(7/6) ) ^• x=((−3^(2/3) −i(3^(7/6) ))/2) Assuming x+(3/( (√x)))=0 is satisfied x(√x)−2x−6 =−3−2(((−3^(2/3) −i(3^(7/6) ))/2))−6 =−9+3^(2/3) +i(3^(7/6) )
x+3x=0⇒xx=−3x3/2=−3x3=9x=32/3,32/3ω,32/3ω2=32/3,32/3(−1+i32),32/3(−1−i32)=32/3,32/3(−32/3+i(32/3)(3)1/22),32/3(−32/3−i(32/3)(3)1/22)=32/3,(−32/3+i(37/6)2),(−32/3−i(37/6)2)∙x=32/3:x+3x=032/3+332/3≠0x+1x=0doesn′tsatisfy∙x=−32/3+i(37/6)2:Assumingx+3x=0issatisfiedxx−2x−6=−3−2(−32/3+i(37/6)2)−6=−9+32/3−i(37/6)∙x=−32/3−i(37/6)2Assumingx+3x=0issatisfiedxx−2x−6=−3−2(−32/3−i(37/6)2)−6=−9+32/3+i(37/6)
Commented by aliibrahim1 last updated on 19/Jun/21
sorry was offline for awhile thank you alot boss really appreciate it
sorrywasofflineforawhilethankyoualotbossreallyappreciateit
Answered by ajfour last updated on 19/Jun/21
x(√x)=−3 v=−3−2x−6 2x=−v−9 8x^3 =72 (v+9)^3 =−72 v=−9+(−72)^(1/3)
xx=−3v=−3−2x−62x=−v−98x3=72(v+9)3=−72v=−9+(−72)1/3
Commented by ajfour last updated on 19/Jun/21
This isn′t wrong though.
Thisisn′twrongthough.

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