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Question Number 188660 by mr W last updated on 04/Mar/23

$$solve{x}^4+4x=1$$

Answered by aba last updated on 04/Mar/23

$$\bullet {\mathrm{x}}^4={({\mathrm{x}}^2+1)}^2-{2\mathrm{x}}^2-1$$$${\mathrm{x}}^4+4\mathrm{x}-1=0\Rightarrow {({\mathrm{x}}^2+1)}^2-{2\mathrm{x}}^2-1+4\mathrm{x}-1=0$$$$\Rightarrow {({\mathrm{x}}^2-1)}^2-2({\mathrm{x}}^2-2\mathrm{x}+1)=0$$$$\Rightarrow {({\mathrm{x}}^2+1)}^2-2{(\mathrm{x}-1)}^2=0$$$$\Rightarrow {\mathrm{x}}^2+1=\pm (\mathrm{x}-1)\sqrt{2}$$$$\Rightarrow {\mathrm{x}}^2-\sqrt{2}\mathrm{x}+(1+\sqrt{2})=0\vee {\mathrm{x}}^2+\sqrt{2}\mathrm{x}+(1-\sqrt{2})=0$$$${\mathrm{x}}^2-\sqrt{2}\mathrm{x}+(1+\sqrt{2})=0$$$$\mathrm{\Delta }=-2-4\sqrt{2}=-2(2\sqrt{2}-1)<0$$$$$$$${\mathrm{x}}^2+\sqrt{2}\mathrm{x}+(1-\sqrt{2})=0$$$$\mathrm{\Delta }=-2+4\sqrt{2}=2(2\sqrt{2}-1)>0$$$$\mathrm{x}=\frac{-\sqrt{2}\pm \sqrt{2(2\sqrt{2}-1)}}{2}$$$$$$

Commented by BaliramKumar last updated on 04/Mar/23

$$answerisrightbut$$$$last3rdand4thsteperror(take\pm value)$$

Answered by manxsol last updated on 05/Mar/23

$$x^4+2x^2+1-2x^2+4x-1=1$$$${(x^2+1)}^2-2(x^2-2x+1)=0$$$${(x^2+1)}^2-{(\sqrt{2}x-\sqrt{2})}^2=0$$$$(x^2+\sqrt{2}x+1-\sqrt{2})(x^2-\sqrt{2}x+1+\sqrt{2})=0$$$$$$

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