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Question Number 179095 by SAMIRA last updated on 24/Oct/22

Commented by HeferH last updated on 24/Oct/22

$$x_1=1;x_2=0$$$$$$

Answered by aleks041103 last updated on 24/Oct/22

$$3x^2-2x-1⩾0$$$$(x-1)(3x+1)⩾0\Rightarrow x\in (-\mathrm{\infty },-\frac{1}{3}]\cup [1,\mathrm{\infty })$$$$2x^2-x-1⩾0$$$$(x-1)(2x+1)⩾0\Rightarrow x\in (-\mathrm{\infty },-\frac{1}{2}]\cup [1,\mathrm{\infty })$$$$bothmustbesatisfied$$$$\Rightarrow x\in (-\mathrm{\infty },-\frac{1}{2}]\cup [1,\mathrm{\infty })$$$$then$$$$3x^2-2x-1=2x^2-x-1$$$$x^2-x=0\Rightarrow x=0,1$$$$\Rightarrow onlyans.x=1$$

Commented by Frix last updated on 24/Oct/22

$$\mathrm{why}3x^2-2x-1⩾0?$$$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{restriction}\mathrm{saying}\mathrm{both}\mathrm{sides}\mathrm{of}$$$$\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{must}\mathrm{be}\mathrm{real}\mathrm{numbers}$$$$\Rightarrow$$$$x=0\mathrm{is}\mathrm{a}\mathrm{solution}\mathrm{because}\sqrt{-1}=\sqrt{-1}$$

Commented by Acem last updated on 25/Oct/22

$$Idiomatically{it}^{\prime }sdealingwithrealnumbersunless$$$$otherwisestated$$$$$$

Commented by Acem last updated on 25/Oct/22

$$GoodSir!$$

Commented by MJS_new last updated on 25/Oct/22

$$\mathrm{for}\mathrm{me}\mathrm{everything}\mathrm{is}\mathrm{dealing}\mathrm{with}\mathrm{complex}$$$$\mathrm{numbers}\mathrm{unless}\mathrm{otherwise}\mathrm{stated}.$$

Answered by Acem last updated on 25/Oct/22

$$x=0\notin IRefused,\mathit{x}=1\in I;$$$$ThecommondomainisI=]-\mathrm{\infty },-\frac{1}{2}]\cup [1,+\mathrm{\infty }[$$$$$$

Commented by MJS_new last updated on 25/Oct/22

$$\mathrm{why}?\mathrm{obviously}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{in}\mathbb{C}$$

Commented by Acem last updated on 25/Oct/22

$$Yesyoucan,butontheotherhand,andin$$$$thiscase,{what}^{\prime }sthedefinitionsetof\sqrt{x}?$$

Commented by MJS_new last updated on 25/Oct/22

$$\mathrm{sorry}\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{like}\mathrm{saying}\mathrm{we}\mathrm{cannot}\mathrm{solve}$$$$2x+5=2\mathrm{because}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{generally}\mathrm{understood}$$$$\mathrm{that}x\in \mathbb{N}$$

Commented by Acem last updated on 25/Oct/22

$$Thegeneralcaseforpolynomialisx\in \mathbb{R}not\mathbb{N}$$$$unlessotherwisestated,asifx\in \mathbb{N}orx\in \mathbb{Z}..etc$$$$$$$$Thesamethingfor\sqrt{{ax}^n+....}thegeneralcaseis$$$${ax}^n+...⩾0$$

Commented by MJS_new last updated on 25/Oct/22

$$\mathrm{these}\mathrm{are}your\mathrm{rules},\mathrm{not}\mathrm{general}\mathrm{rules}.\mathrm{it}\mathrm{makes}$$$$\mathrm{no}\mathrm{sense}\mathrm{to}\mathrm{keep}\mathrm{on}\mathrm{discussing}$$

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