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Question Number 155153 by mathdanisur last updated on 26/Sep/21

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}\mathrm{in}\mathbb{R}$$$$\sqrt{2({\mathrm{x}}^2-\mathrm{x}+1)}=1+\sqrt{\mathrm{x}}-\mathrm{x}$$

Answered by MJS_new last updated on 26/Sep/21

$$\sqrt{x}\in \mathbb{R}\Rightarrow x⩾0\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}x\ne 0$$$$\mathrm{and}\mathrm{lhs}>0\Rightarrow 1+\sqrt{x}-x>0$$$$\Rightarrow$$$$0<x<\frac{3}{2}+\frac{\sqrt{5}}{2}$$$$$$$$\mathrm{let}x=t^2\wedge t>0$$$$\sqrt{2(t^4-t^2+1)}=-t^2+t+1[\Rightarrow 0<t<\frac{1}{2}+\frac{\sqrt{5}}{2}]$$$$\mathrm{squaring}$$$$2(t^4-t^2+1)={(t^2-t-1)}^2$$$$t^4+2t^3-t^2-2t+1=0$$$${(t^2+t+1)}^2=0$$$$\Rightarrow t=-\frac{1}{2}+\frac{\sqrt{5}}{2}$$$$\Rightarrow x=\frac{3}{2}-\frac{\sqrt{5}}{2}$$

Commented by mathdanisur last updated on 26/Sep/21

$$\mathrm{awesome}\mathrm{solution},\mathrm{thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

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