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Question Number 189748 by normans last updated on 21/Mar/23

$$$$$$\mathit{x}-\sqrt{\mathit{x}}+1=0$$$$\mathit{x}=??$$

Commented by mr W last updated on 21/Mar/23

$$x>0$$$$\sqrt{x}=x+1⩾2\sqrt{x}$$$$1⩾2\Rightarrow impossible$$

Commented by Frix last updated on 21/Mar/23

$$\mathrm{Are}\mathrm{we}\mathrm{suddenly}\mathrm{limited}\mathrm{to}\mathbb{R}?$$

Answered by TUN last updated on 21/Mar/23

$$NOROOT!$$

Answered by Frix last updated on 21/Mar/23

$$\mathrm{Let}a,b\in \mathbb{R}\wedge \sqrt{x}=a+b\mathrm{i}\Rightarrow x=a^2-b^2+2ab\mathrm{i}$$$$x-\sqrt{x}+1=0$$$$(a^2-b^2-a+1)+(2ab-b)\mathrm{i}=0$$$$\{\begin{array}{l}{a}^2-b^2-a+1=0\\ (2a-1)b=0\Rightarrow a=\frac{1}{2}\vee b=0\end{array}$$$$b=0\Rightarrow a^2-a+1=0\Rightarrow a\notin \mathbb{R}\Rightarrow \mathrm{no}\mathrm{solution}$$$$a=\frac{1}{2}\Rightarrow b=\pm \frac{\sqrt{3}}{2}\Rightarrow$$$$\sqrt{x}=\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}$$$$x=\frac{1}{4}-\frac{3}{4}\pm 2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}\mathrm{i}=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}$$

Commented by MJS_new last updated on 21/Mar/23

$$\mathrm{why}\mathrm{not}\mathrm{simply}\mathrm{square}\mathrm{it}?$$$$x-\sqrt{x}+1=0$$$$x+1=\sqrt{x}$$$${(x+1)}^2=x$$$$x^2+x+1=0$$$$x=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}$$

Commented by mathocean1 last updated on 21/Mar/23

$$S=\varnothing (becauseof\sqrt{x},x\in {\mathbb{R}}^{+})$$$$$$

Commented by MJS_new last updated on 21/Mar/23

$$\mathrm{the}\mathrm{question}\mathrm{does}\mathrm{not}\mathrm{say}‘‘x\in {\mathbb{R}}^{″}$$$$\mathrm{obviously}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{it}\mathrm{in}\mathbb{C}$$

Commented by talminator2856792 last updated on 22/Mar/23

$$$$$$k=\sqrt{x}$$$$k^2-k+1=0$$$${(k-\frac{1}{2})}^2+\frac{3}{4}=0$$$${(k-\frac{1}{2})}^2=-\frac{3}{4}$$$$\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{positive}\mathrm{square}$$$$\Rightarrow \mathrm{not}\mathrm{solvable}\mathrm{in}\mathbb{R}$$

Commented by MJS_new last updated on 22/Mar/23

$$\mathrm{yes}.\mathrm{but}\mathrm{remember}\mathrm{Descartes}\mathrm{introduced}$$$$\sqrt{-1}=\mathrm{i}\mathrm{back}\mathrm{in}1637$$

Commented by Frix last updated on 22/Mar/23

$$\mathrm{Be}\mathrm{patient},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{been}\mathrm{only}386\mathrm{years}...$$

Commented by mathocean1 last updated on 27/Mar/23

$$Iwassayingthatbecause{it}^{\prime }soften$$$$saidthatthecharacter‘‘{\sqrt{}}^{″}isused$$$$onlyforpositivereal...$$

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