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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

$$\mathrm{let}x=r{\mathrm{e}}^{\mathrm{i}\theta }\wedge r>0\wedge 0⩽\theta <2\pi$$$$r{\mathrm{e}}^{\mathrm{i}\theta }+\frac{3}{r^{1/2}{\mathrm{e}}^{\mathrm{i}\theta /2}}=0$$$$r^{3/2}{\mathrm{e}}^{3\mathrm{i}\theta /2}+3=0$$$$r^{3/2}{\mathrm{e}}^{3\mathrm{i}\theta /2}={3\mathrm{e}}^{\mathrm{i}\pi }$$$$\Rightarrow$$$$r=3^{2/3}\wedge \theta =2\pi /3\Rightarrow x=3^{2/3}{\mathrm{e}}^{2\mathrm{i}\pi /3}\vee x=3^{2/3}{\mathrm{e}}^{4\mathrm{i}\pi /3}$$$$x\sqrt{x}-2x-6=$$$$=-12-2\times 3^{2/3}(-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i})=$$$$=-12+3^{2/3}\pm 3^{7/6}\mathrm{i}$$$$\mathrm{well}\mathrm{well}\mathrm{well}...$$

Commented by MJS_new last updated on 12/Jun/21

$$\mathrm{corrected}\mathrm{a}\mathrm{typo}$$

Answered by Rasheed.Sindhi last updated on 12/Jun/21

$$x\sqrt{x}=-3$$$$x^{3/2}=-3$$$$x={(-3)}^{2/3}=3^{2/3}$$$$x\sqrt{x}-2x-6=-3-2{\left(3\right)}^{2/3}-6$$$$=-9-2{\left(3\right)}^{2/3}$$

Commented by MJS_new last updated on 12/Jun/21

$$x=3^{2/3}(?)$$$$3^{2/3}+\frac{3}{\sqrt{3^{2/3}}}=3^{2/3}+\frac{3}{3^{1/3}}=2\times 3^{2/3}\ne 0(!)$$

Commented by aliibrahim1 last updated on 12/Jun/21

$$thankyougoodsirmjs$$

Commented by aliibrahim1 last updated on 12/Jun/21

$$thankyougoodsirrasheed$$

Commented by Rasheed.Sindhi last updated on 12/Jun/21

$$Sorry\mathit{m}\mathit{j}\mathit{s}\mathit{s}\mathit{i}\mathbf{r},I^{\prime }veusedanextraneous$$$$root\left(3^{2/3}\right)toevaluategivenexpression!$$

Answered by Rasheed.Sindhi last updated on 12/Jun/21

$$x+\frac{3}{\sqrt{x}}=0\Rightarrow x\sqrt{x}=-3$$$$x^{3/2}=-3$$$$x^3=9$$$$x=3^{2/3},3^{2/3}\omega ,3^{2/3}{\omega }^2$$$$=3^{2/3},3^{2/3}\left(\frac{-1+i\sqrt{3}}{2}\right),3^{2/3}\left(\frac{-1-i\sqrt{3}}{2}\right)$$$$=3^{2/3},3^{2/3}\left(\frac{-3^{2/3}+i\left(3^{2/3}\right){\left(3\right)}^{1/2}}{2}\right),3^{2/3}\left(\frac{-3^{2/3}-i\left(3^{2/3}\right){\left(3\right)}^{1/2}}{2}\right)$$$$=3^{2/3},\left(\frac{-3^{2/3}+i\left(3^{7/6}\right)}{2}\right),\left(\frac{-3^{2/3}-i\left(3^{7/6}\right)}{2}\right)$$$$$$$${}^{\bullet }x=3^{2/3}:x+\frac{3}{\sqrt{x}}=0$$$$3^{2/3}+\frac{3}{\sqrt{3^{2/3}}}\ne 0$$$$x+\frac{1}{\sqrt{x}}=0{doesn}^{\prime }tsatisfy$$$${}^{\bullet }x=\frac{-3^{2/3}+i\left(3^{7/6}\right)}{2}:$$$$Assumingx+\frac{3}{\sqrt{x}}=0issatisfied$$$$x\sqrt{x}-2x-6$$$$=-3-2\left(\frac{-3^{2/3}+i\left(3^{7/6}\right)}{2}\right)-6$$$$=-9+3^{2/3}-i\left(3^{7/6}\right)$$$${}^{\bullet }x=\frac{-3^{2/3}-i\left(3^{7/6}\right)}{2}$$$$Assumingx+\frac{3}{\sqrt{x}}=0issatisfied$$$$x\sqrt{x}-2x-6$$$$=-3-2\left(\frac{-3^{2/3}-i\left(3^{7/6}\right)}{2}\right)-6$$$$=-9+3^{2/3}+i\left(3^{7/6}\right)$$

Commented by aliibrahim1 last updated on 19/Jun/21

$$sorrywasofflineforawhile$$$$thankyoualotbossreallyappreciateit$$

Answered by ajfour last updated on 19/Jun/21

$$x\sqrt{x}=-3$$$$v=-3-2x-6$$$$2x=-v-9$$$$8x^3=72$$$${(v+9)}^3=-72$$$$v=-9+{(-72)}^{1/3}$$$$$$

Commented by ajfour last updated on 19/Jun/21

$$This{isn}^{\prime }twrongthough.$$

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