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Question-189090




Question Number 189090 by sonukgindia last updated on 12/Mar/23
Answered by HeferH last updated on 12/Mar/23
 case 1:   (x^2 −2x) = 1   x^2 −2x−1 +2 = 2   (x−1)^2  = 2   ∣x−1∣ = (√2)   x −1 = −(√2)  ∨ x −1=(√2)    x_1  = 1+(√2) ; x_2  = 1−(√2)   case 2:   (x^2 +x−6) = 0   (x −2)(x + 3) = 0   x_3  = 2; x_4  = −3   S = Σ_n  x_n  = 2 + (−3) + (1+(√2))+(1−(√2))=1
$$\:\mathrm{case}\:\mathrm{1}:\: \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}{x}\right)\:=\:\mathrm{1} \\ $$$$\:{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\:+\mathrm{2}\:=\:\mathrm{2} \\ $$$$\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:\mid{x}−\mathrm{1}\mid\:=\:\sqrt{\mathrm{2}} \\ $$$$\:{x}\:−\mathrm{1}\:=\:−\sqrt{\mathrm{2}}\:\:\vee\:{x}\:−\mathrm{1}=\sqrt{\mathrm{2}}\: \\ $$$$\:{x}_{\mathrm{1}} \:=\:\mathrm{1}+\sqrt{\mathrm{2}}\:;\:{x}_{\mathrm{2}} \:=\:\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\:\mathrm{case}\:\mathrm{2}: \\ $$$$\:\left({x}^{\mathrm{2}} +{x}−\mathrm{6}\right)\:=\:\mathrm{0} \\ $$$$\:\left({x}\:−\mathrm{2}\right)\left({x}\:+\:\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\:{x}_{\mathrm{3}} \:=\:\mathrm{2};\:{x}_{\mathrm{4}} \:=\:−\mathrm{3} \\ $$$$\:{S}\:=\:\underset{{n}} {\sum}\:{x}_{{n}} \:=\:\mathrm{2}\:+\:\left(−\mathrm{3}\right)\:+\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)=\mathrm{1} \\ $$
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+(√(2   ))         x_2 =1−(√2)  2     x_3 =1          x_4 =1  3      x_5 = 2           x_6 =−3  Σ_6 x_n = 3
$$\mathrm{1}.\:{x}^{\mathrm{2}} −\mathrm{2}{x}=\mathrm{1} \\ $$$$\mathrm{2}.\:{x}^{\mathrm{2}} −\mathrm{2}{x}=−\mathrm{1}\:\:\:\:\:\:\:\left(−\mathrm{1}\right)^{−\mathrm{6}} =\mathrm{1} \\ $$$$\mathrm{3}.{x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0} \\ $$$${n}=\mathrm{6} \\ $$$$\mathrm{1}.\:\:\:{x}_{\mathrm{1}} =\mathrm{1}+\sqrt{\mathrm{2}\:\:\:} \\ $$$$\:\:\:\:\:\:\:{x}_{\mathrm{2}} =\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}\:\:\:\:\:{x}_{\mathrm{3}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{x}_{\mathrm{4}} =\mathrm{1} \\ $$$$\mathrm{3}\:\:\:\:\:\:{x}_{\mathrm{5}} =\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:{x}_{\mathrm{6}} =−\mathrm{3} \\ $$$$\sum_{\mathrm{6}} {x}_{{n}} =\:\mathrm{3} \\ $$

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