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Question Number 67903 by rajesh4661kumar@gmail.com last updated on 02/Sep/19

Answered by $@ty@m123 last updated on 02/Sep/19

$$Let\sqrt{x}=y$$$$3y^2+\frac{2}{y}=1$$$$3y^3+2=y$$$$3y^3-y+2=0$$$$3y^3+3y^2-3y^2-3y+2y+2=0$$$$3y^2(y+1)-3y(y+1)+2(y+1)=0$$$$(y+1)(3y^2-3y+2)=0$$$$y=-1\mid 3y^2-3y+2=0$$$$\mid 3(y^2-y)+2=0$$$$\mid y^2-y=\frac{-2}{3}$$

Commented by MJS last updated on 02/Sep/19

$$y=-1\Rightarrow \sqrt{x}=-1\mathrm{and}\mathrm{this}\mathrm{has}\mathrm{no}\mathrm{solution}$$$$\mathrm{you}\mathrm{get}\mathrm{the}\mathrm{right}\mathrm{result}\mathrm{but}\mathrm{the}\mathrm{path}\mathrm{is}\mathrm{wrong}$$

Commented by Prithwish sen last updated on 02/Sep/19

$$\mathrm{I}\mathrm{think}\sqrt{\mathrm{x}}=-1\nRightarrow \mathbf{x}=\mathbf{i}\mathbf{as}\sqrt{\mathbf{x}}=-1\Rightarrow \mathbf{x}=1\mathbf{only}$$$$\mathbf{That}\mathbf{is}\mathbf{why}\mathbf{the}\mathbf{matter}\mathbf{of}\mathbf{complex}\mathbf{number}$$$$\mathbf{is}\mathbf{irrelvant}.\mathbf{Mjs}\mathbf{sir}\mathbf{waiting}\mathbf{for}\mathbf{your}\mathbf{opinion}.$$

Commented by MJS last updated on 02/Sep/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{a}\mathrm{matter}\mathrm{of}\mathrm{being}\mathrm{broad}\mathrm{or}\mathrm{narrow}$$$$\mathrm{minded}.$$$$x^2=-1$$$$x^2+1=0$$$$(x-\mathrm{i})(x+\mathrm{i})=0$$$$\Rightarrow x=\pm \mathrm{i}$$$${(-\mathrm{i})}^2=-1$$$${(+\mathrm{i})}^2=-1$$$$x^3=-1$$$$x^3+1=0$$$$(x+1)(x^2-x+1)=0$$$$\Rightarrow x=-1\vee x=\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}$$$${(-1)}^3=-1$$$${(\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})}^3=-1$$$${(\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})}^3=-1$$$$...$$$$$$$$x^n=-1$$$$\mathrm{the}\mathrm{solutions}\mathrm{form}\mathrm{a}\mathrm{cyclic}n-\mathrm{gon}$$$$$$$$x^{\frac{1}{2}}=-1$$$$\left(1\right){\mathrm{what}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{cyclic}\frac{1}{2}-\mathrm{gon}?$$$$\left(2\right)x=r{\mathrm{e}}^{\mathrm{i}\theta };-1={\mathrm{e}}^{\mathrm{i}\pi }$$$$\sqrt{r}{\mathrm{e}}^{\mathrm{i}\frac{\theta }{2}}={1\mathrm{e}}^{\mathrm{i}\pi }$$$$\stackrel{?}{\Rightarrow }r=1\wedge \theta =2\pi \iff x={\mathrm{e}}^{2\pi \mathrm{i}}=1$$$$\mathrm{but}:\sqrt{1}=1$$$$\mathrm{the}\mathrm{problem}\mathrm{is},\mathrm{if}\mathrm{we}\mathrm{square}\mathrm{an}\mathrm{equation}\mathrm{we}$$$$\mathrm{might}\mathrm{get}\mathrm{false}\mathrm{solutions}\Rightarrow \mathrm{we}\mathrm{have}\mathrm{to}\mathrm{try}$$$$\mathrm{these}\mathrm{solutions}\mathrm{in}\mathrm{the}\mathrm{original}\mathrm{equation}$$$$$$$$\mathrm{another}\mathrm{problem}:\mathrm{if}\sqrt{1}=\pm 1\mathrm{we}\mathrm{can}\mathrm{forget}$$$$\mathrm{the}\mathrm{uniqueness}\mathrm{of}\mathrm{terms}\mathrm{like}$$$$\sqrt{\frac{7}{16}-\frac{\sqrt{3}}{4}}$$$$\mathrm{this}\mathrm{would}\mathrm{have}4\mathrm{different}\mathrm{values}$$$$$$$$\mathrm{usually}\mathrm{all}n-\mathrm{roots}\mathrm{are}\mathrm{defined}\mathrm{as}\mathrm{the}‘‘{\mathrm{natural}}^{″}$$$$\mathrm{roots}:$$$$\sqrt{4}=2$$$$\sqrt{-4}=2\mathrm{i}$$$$\sqrt[3]{8}=2$$$$\sqrt[3]{-8}=-2(\mathrm{this}\mathrm{violates}\mathrm{other}\mathrm{laws},\mathrm{electronic}$$$$\mathrm{calculators}\mathrm{give}\sqrt[3]{-8}=1+\sqrt{3}\mathrm{i}\mathrm{because}\mathrm{the}$$$$‘‘{\mathrm{natural}}^{″}{3}^{\mathrm{rd}}\mathrm{root}\mathrm{of}{8\mathrm{e}}^{\mathrm{i}\pi }\mathrm{is}\sqrt[3]{8}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{3}})$$$$...$$$$\mathrm{this}\mathrm{also}\mathrm{makes}\mathrm{it}\mathrm{complicated}\mathrm{to}\mathrm{calculate}$$$${(-1)}^{\frac{2}{5}}:\sqrt[5]{{(-1)}^2}\ne {\left(\sqrt[5]{-1}\right)}^2\mathrm{in}\mathrm{machine}\mathrm{logic}$$

Commented by Prithwish sen last updated on 02/Sep/19

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{for}\mathrm{your}\mathrm{effort}.$$

Answered by MJS last updated on 02/Sep/19

$$3x+\frac{2}{\sqrt{x}}=1$$$$3x\sqrt{x}-\sqrt{x}+2=0$$$$x=t^2$$$$3t^2\mid t\mid -\mid t\mid +2=0$$$$\mathrm{case}1:t<0$$$$-3t^3+t+2=0$$$$-(t-1)(3t^2+3t+2)=0$$$$\Rightarrow t=1\mathrm{but}t<0\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{in}\mathbb{R}$$$$\mathrm{case}2:t>0$$$$3t^3-t+2=0$$$$(t+1)(3t^2-3t+2)=0$$$$\Rightarrow t=-1\mathrm{but}t>0\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{in}\mathbb{R}$$$$$$$$\mathrm{but}\mathrm{in}\mathrm{both}\mathrm{cases}\mathrm{we}\mathrm{get}2\mathrm{complex}\mathrm{solutions}$$$$\mathrm{and}\mathrm{with}x=t^2\mathrm{we}\mathrm{have}$$$$x=-\frac{1}{6}\pm \frac{\sqrt{15}}{6}\mathrm{i}$$$$\Rightarrow x-\sqrt{x}=-\frac{2}{3}$$

Commented by MJS last updated on 02/Sep/19

$$x=\mathrm{i}\mathrm{is}\mathrm{wrong}\mathrm{anyway}$$$$\sqrt{\mathrm{i}}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}$$

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