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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...
$$\mathrm{let}\:{x}={r}\mathrm{e}^{\mathrm{i}\theta} \wedge{r}>\mathrm{0}\wedge\mathrm{0}\leqslant\theta<\mathrm{2}\pi \\ $$$${r}\mathrm{e}^{\mathrm{i}\theta} +\frac{\mathrm{3}}{\:{r}^{\mathrm{1}/\mathrm{2}} \mathrm{e}^{\mathrm{i}\theta/\mathrm{2}} }=\mathrm{0} \\ $$$${r}^{\mathrm{3}/\mathrm{2}} \mathrm{e}^{\mathrm{3i}\theta/\mathrm{2}} +\mathrm{3}=\mathrm{0} \\ $$$${r}^{\mathrm{3}/\mathrm{2}} \mathrm{e}^{\mathrm{3i}\theta/\mathrm{2}} =\mathrm{3e}^{\mathrm{i}\pi} \\ $$$$\Rightarrow \\ $$$${r}=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \wedge\theta=\mathrm{2}\pi/\mathrm{3}\:\Rightarrow\:{x}=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \mathrm{e}^{\mathrm{2i}\pi/\mathrm{3}} \vee{x}=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \mathrm{e}^{\mathrm{4i}\pi/\mathrm{3}} \\ $$$${x}\sqrt{{x}}−\mathrm{2}{x}−\mathrm{6}= \\ $$$$=−\mathrm{12}−\mathrm{2}×\mathrm{3}^{\mathrm{2}/\mathrm{3}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)= \\ $$$$=−\mathrm{12}+\mathrm{3}^{\mathrm{2}/\mathrm{3}} \pm\mathrm{3}^{\mathrm{7}/\mathrm{6}} \mathrm{i} \\ $$$$\mathrm{well}\:\mathrm{well}\:\mathrm{well}… \\ $$
Commented by MJS_new last updated on 12/Jun/21
corrected a typo
$$\mathrm{corrected}\:\mathrm{a}\:\mathrm{typo} \\ $$
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)
$${x}\sqrt{{x}}=−\mathrm{3} \\ $$$${x}^{\mathrm{3}/\mathrm{2}} =−\mathrm{3} \\ $$$${x}=\left(−\mathrm{3}\right)^{\mathrm{2}/\mathrm{3}} =\mathrm{3}^{\mathrm{2}/\mathrm{3}} \\ $$$${x}\sqrt{{x}}−\mathrm{2}{x}−\mathrm{6}=−\mathrm{3}−\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}/\mathrm{3}} −\mathrm{6} \\ $$$$\:\:\:\:\:=−\mathrm{9}−\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}/\mathrm{3}} \\ $$
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 (!)
$${x}=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \:\left(?\right) \\ $$$$\mathrm{3}^{\mathrm{2}/\mathrm{3}} +\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }}=\mathrm{3}^{\mathrm{2}/\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} }=\mathrm{2}×\mathrm{3}^{\mathrm{2}/\mathrm{3}} \neq\mathrm{0}\:\left(!\right) \\ $$
Commented by aliibrahim1 last updated on 12/Jun/21
thank you good sir mjs
$${thank}\:{you}\:{good}\:{sir}\:{mjs} \\ $$
Commented by aliibrahim1 last updated on 12/Jun/21
thank you good sir rasheed
$${thank}\:{you}\:{good}\:{sir}\:{rasheed} \\ $$
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!
$${Sorry}\:\:\boldsymbol{{mjs}}\:\boldsymbol{{si}\mathrm{r}}\:,{I}'{ve}\:{used}\:{an}\:{extraneous} \\ $$$${root}\:\left(\mathrm{3}^{\mathrm{2}/\mathrm{3}} \right)\:{to}\:{evaluate}\:\:{given}\:{expression}! \\ $$
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) )
$${x}+\frac{\mathrm{3}}{\:\sqrt{{x}}}=\mathrm{0}\Rightarrow{x}\sqrt{{x}}=−\mathrm{3} \\ $$$${x}^{\mathrm{3}/\mathrm{2}} =−\mathrm{3} \\ $$$${x}^{\mathrm{3}} =\mathrm{9} \\ $$$${x}=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \:,\:\mathrm{3}^{\mathrm{2}/\mathrm{3}} \omega\:,\:\mathrm{3}^{\mathrm{2}/\mathrm{3}} \omega^{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \:,\:\mathrm{3}^{\mathrm{2}/\mathrm{3}} \left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:,\:\mathrm{3}^{\mathrm{2}/\mathrm{3}} \left(\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:\:=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \:,\:\mathrm{3}^{\mathrm{2}/\mathrm{3}} \left(\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} +{i}\left(\mathrm{3}^{\mathrm{2}/\mathrm{3}} \right)\left(\mathrm{3}\right)^{\mathrm{1}/\mathrm{2}} }{\mathrm{2}}\right)\:,\:\mathrm{3}^{\mathrm{2}/\mathrm{3}} \left(\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} −{i}\left(\mathrm{3}^{\mathrm{2}/\mathrm{3}} \right)\left(\mathrm{3}\right)^{\mathrm{1}/\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\:\:\:=\mathrm{3}^{\mathrm{2}/\mathrm{3}} \:,\:\left(\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} +{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right)}{\mathrm{2}}\right)\:,\:\left(\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} −{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right)}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\:^{\bullet} {x}=\mathrm{3}^{\mathrm{2}/\mathrm{3}} :\:{x}+\frac{\mathrm{3}}{\:\sqrt{{x}}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{2}/\mathrm{3}} +\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }}\neq\mathrm{0} \\ $$$${x}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=\mathrm{0}\:{doesn}'{t}\:{satisfy} \\ $$$$\:^{\bullet} {x}=\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} +{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right)}{\mathrm{2}}: \\ $$$${Assuming}\:\:{x}+\frac{\mathrm{3}}{\:\sqrt{{x}}}=\mathrm{0}\:{is}\:{satisfied} \\ $$$${x}\sqrt{{x}}−\mathrm{2}{x}−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{3}−\mathrm{2}\left(\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} +{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right)}{\mathrm{2}}\right)−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:=−\mathrm{9}+\mathrm{3}^{\mathrm{2}/\mathrm{3}} −{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right) \\ $$$$\:^{\bullet} {x}=\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} −{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right)}{\mathrm{2}} \\ $$$${Assuming}\:\:{x}+\frac{\mathrm{3}}{\:\sqrt{{x}}}=\mathrm{0}\:{is}\:{satisfied} \\ $$$${x}\sqrt{{x}}−\mathrm{2}{x}−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{3}−\mathrm{2}\left(\frac{−\mathrm{3}^{\mathrm{2}/\mathrm{3}} −{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right)}{\mathrm{2}}\right)−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{9}+\mathrm{3}^{\mathrm{2}/\mathrm{3}} +{i}\left(\mathrm{3}^{\mathrm{7}/\mathrm{6}} \right) \\ $$
Commented by aliibrahim1 last updated on 19/Jun/21
sorry was offline for awhile thank you alot boss really appreciate it
$${sorry}\:{was}\:{offline}\:{for}\:{awhile}\: \\ $$$${thank}\:{you}\:{alot}\:{boss}\:{really}\:{appreciate}\:{it} \\ $$
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)
$${x}\sqrt{{x}}=−\mathrm{3} \\ $$$${v}=−\mathrm{3}−\mathrm{2}{x}−\mathrm{6} \\ $$$$\mathrm{2}{x}=−{v}−\mathrm{9} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} =\mathrm{72} \\ $$$$\left({v}+\mathrm{9}\right)^{\mathrm{3}} =−\mathrm{72} \\ $$$${v}=−\mathrm{9}+\left(−\mathrm{72}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$ \\ $$
Commented by ajfour last updated on 19/Jun/21
This isn′t wrong though.
$${This}\:{isn}'{t}\:{wrong}\:{though}. \\ $$

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