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Question Number 121423 by Jamshidbek2311 last updated on 08/Nov/20

$$x^4-2\sqrt{2}{x}^2-x+2-\sqrt{2}=0x=?$$

Answered by 2004 last updated on 08/Nov/20

$$\sqrt{2}=\mathrm{t}$$$${\mathrm{t}}^2-\mathrm{t}({2\mathrm{x}}^2+1)-\mathrm{x}+{\mathrm{x}}^4=0$$$$\mathrm{D}={4\mathrm{x}}^4+{4\mathrm{x}}^2+1-{4\mathrm{x}}^4+4\mathrm{x}={(2\mathrm{x}+1)}^2$$$$\sqrt{2}=\mathrm{t}=\frac{{2\mathrm{x}}^2+1\mp (2\mathrm{x}+1)}{2}$$$$0={\mathrm{x}}^2-\mathrm{x}-\sqrt{2}0={\mathrm{x}}^2+\mathrm{x}+1-\sqrt{2}$$$$\mathrm{x}=\frac{1\mp \sqrt{1+4\sqrt{2}}}{2}\mathrm{x}=\frac{-1\mp \sqrt{4\sqrt{2}-3}}{2}$$

Commented by Jamshidbek2311 last updated on 08/Nov/20

$$thankyou$$

Commented by 2004 last updated on 08/Nov/20

$$\mathit{W}\mathit{e}\mathit{l}\mathit{c}\mathit{o}\mathit{m}\mathit{e}\mathit{b}\mathit{r}\mathit{o}(\mathit{a}\mathit{r}\mathit{z}\mathit{i}\mathit{m}\mathit{a}\mathit{y}\mathit{d}\mathit{i}:))$$

Commented by Jamshidbek2311 last updated on 08/Nov/20

$$AreyouUzbek?$$

Commented by pooooop last updated on 10/Nov/20

$$\mathit{H}\mathit{a}.\mathit{U}zbek$$

Answered by mathmax by abdo last updated on 08/Nov/20

$$\mathrm{e}\Rightarrow {\mathrm{x}}^4-\sqrt{2}({2\mathrm{x}}^2+1)-\mathrm{x}+2=0\mathrm{let}\sqrt{2}=\mathrm{t}$$$$\mathrm{e}\Rightarrow {\mathrm{x}}^4-\mathrm{t}({2\mathrm{x}}^2+1)-\mathrm{x}+{\mathrm{t}}^2=0\Rightarrow {\mathrm{t}}^2-({2\mathrm{x}}^2+1)\mathrm{t}+{\mathrm{x}}^4-\mathrm{x}=0$$$$\mathrm{\Delta }={({2\mathrm{x}}^2+1)}^2-4({\mathrm{x}}^4-\mathrm{x})={4\mathrm{x}}^4+{4\mathrm{x}}^2+1-{4\mathrm{x}}^4+4\mathrm{x}={(2\mathrm{x}+1)}^2$$$$\Rightarrow \sqrt{2}=\frac{{2\mathrm{x}}^2+1+2\mathrm{x}+1}{2}={\mathrm{x}}^2+\mathrm{x}+1\Rightarrow {\mathrm{x}}^2+\mathrm{x}+1-\sqrt{2}=0$$$$\mathrm{\Delta }=1-4(1-\sqrt{2})=-3+4\sqrt{2}>0\Rightarrow {\mathrm{x}}_1=\frac{-1+\sqrt{4\sqrt{2}-3}}{2}$$$$\mathrm{and}{\mathrm{x}}_2=\frac{-1-\sqrt{4\sqrt{2}-3}}{2}$$

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