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Question Number 218473 by Nicholas666 last updated on 10/Apr/25
Solve the following equation for the real x value; x^4 −4x^3 +6x^2 −4x+2 = (√(2x^4 −8x^3 +12x^2 −8x+5))
$$ \\ $$$$\:{Solve}\:{the}\:{following}\:{equation}\:{for}\:{the}\:{real}\:\boldsymbol{{x}}\:{value}; \\ $$$$\:{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\:=\:\sqrt{\mathrm{2}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{5}}\:\:\:\: \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 10/Apr/25
x^4 −4x^3 +6x^2 −4x+2 = (√(2x^4 −8x^3 +12x^2 −8x+5)) x^4 −4x^3 +6x^2 −4x+2 =(√2) (√(x^4 −4x^3 +6x^2 −4x+5/2)) (√(x^4 −4x^3 +6x^2 −4x+5/2)) =y x^4 −4x^3 +6x^2 −4x+5/2 =y^2 x^4 −4x^3 +6x^2 −4x+2+(1/2) =y^2 x^4 −4x^3 +6x^2 −4x+2 =y^2 −(1/2) y^2 −(1/2)=(√2) y 2y^2 −2(√2) y−1=0 y=((2(√2) ±(√(8+8)))/4) y=(((√2) ±2)/2) x^4 −4x^3 +6x^2 −4x+5/2=((((√2) ±2)/2))^2
$$\:{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\:=\:\sqrt{\mathrm{2}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{5}}\:\:\:\: \\ $$$$\:{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\:=\sqrt{\mathrm{2}}\:\sqrt{{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}/\mathrm{2}}\:\:\:\: \\ $$$$\sqrt{{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}/\mathrm{2}}\:\:={y} \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}/\mathrm{2}\:\:={y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\:\:={y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\:\:={y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}=\sqrt{\mathrm{2}}\:{y} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\:{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{8}+\mathrm{8}}}{\mathrm{4}} \\ $$$${y}=\frac{\sqrt{\mathrm{2}}\:\pm\mathrm{2}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}/\mathrm{2}=\left(\frac{\sqrt{\mathrm{2}}\:\pm\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$
Answered by Ghisom last updated on 10/Apr/25
let t=(x−1)^4 t+1=(√(2t+3)) t=(√2) (x−1)^4 =(√2) x=1±2^(1/8) ∨x=1±2^(1/8) i
$$\mathrm{let}\:{t}=\left({x}−\mathrm{1}\right)^{\mathrm{4}} \\ $$$${t}+\mathrm{1}=\sqrt{\mathrm{2}{t}+\mathrm{3}} \\ $$$${t}=\sqrt{\mathrm{2}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{4}} =\sqrt{\mathrm{2}} \\ $$$${x}=\mathrm{1}\pm\mathrm{2}^{\mathrm{1}/\mathrm{8}} \vee{x}=\mathrm{1}\pm\mathrm{2}^{\mathrm{1}/\mathrm{8}} \mathrm{i} \\ $$

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