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Question Number 189090 by sonukgindia last updated on 12/Mar/23

Answered by HeferH last updated on 12/Mar/23

$$\mathrm{case}1:$$$$(x^2-2x)=1$$$$x^2-2x-1+2=2$$$${(x-1)}^2=2$$$$\mid x-1\mid =\sqrt{2}$$$$x-1=-\sqrt{2}\vee x-1=\sqrt{2}$$$$x_1=1+\sqrt{2};x_2=1-\sqrt{2}$$$$\mathrm{case}2:$$$$(x^2+x-6)=0$$$$(x-2)(x+3)=0$$$$x_3=2;x_4=-3$$$$S=\sum _n{x}_n=2+(-3)+(1+\sqrt{2})+(1-\sqrt{2})=1$$

Answered by manxsol last updated on 12/Mar/23

$$1.x^2-2x=1$$$$2.x^2-2x=-1{(-1)}^{-6}=1$$$$3.x^2+x-6=0$$$$n=6$$$$1.x_1=1+\sqrt{2}$$$$x_2=1-\sqrt{2}$$$$2x_3=1$$$$x_4=1$$$$3x_5=2$$$$x_6=-3$$$$\sum _6{x}_n=3$$

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