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

x^4 −2(√2)x^2 −x+2−(√2)=0 x=?

$${x}^{\mathrm{4}} −\mathrm{2}\sqrt{\mathrm{2}}{x}^{\mathrm{2}} −{x}+\mathrm{2}−\sqrt{\mathrm{2}}=\mathrm{0}\:\:\:{x}=? \\ $$

Answered by 2004 last updated on 08/Nov/20

(√2)=t t^2 −t(2x^2 +1)−x+x^4 =0 D=4x^4 +4x^2 +1−4x^4 +4x=(2x+1)^2 (√2)=t=((2x^2 +1∓(2x+1))/2) 0=x^2 −x −(√2) 0=x^2 +x+1−(√2) x=((1∓(√(1+4(√2))))/2) x=((−1∓(√(4(√2)−3)))/2)

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

Commented by Jamshidbek2311 last updated on 08/Nov/20

thank you

$${thank}\:{you} \\ $$

Commented by 2004 last updated on 08/Nov/20

Welcome bro(arzimaydi:))

$$\left.\boldsymbol{{Welcome}}\:\boldsymbol{{bro}}\left(\boldsymbol{{arzimaydi}}:\right)\right) \\ $$

Commented by Jamshidbek2311 last updated on 08/Nov/20

Are you Uzbek?

$${Are}\:{you}\:{Uzbek}? \\ $$

Commented by pooooop last updated on 10/Nov/20

Ha.Uzbek

$$\boldsymbol{{Ha}}.\boldsymbol{{U}}{zbek} \\ $$

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

e⇒x^4 −(√2)(2x^2 +1)−x+2 =0 let (√2)=t e⇒x^4 −t(2x^2 +1)−x+t^2 =0 ⇒t^2 −(2x^2 +1)t+x^4 −x =0 Δ=(2x^2 +1)^2 −4(x^4 −x) =4x^4 +4x^2 +1−4x^4 +4x= (2x+1)^2 ⇒(√2)=((2x^2 +1+2x+1)/2) =x^2 +x+1 ⇒x^2 +x+1−(√2)=0 Δ=1−4(1−(√2)) =−3+4(√2) >0 ⇒x_1 =((−1+(√(4(√2)−3)))/2) and x_2 =((−1−(√(4(√2)−3)))/2)

$$\mathrm{e}\Rightarrow\mathrm{x}^{\mathrm{4}} −\sqrt{\mathrm{2}}\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{x}+\mathrm{2}\:=\mathrm{0}\:\:\mathrm{let}\:\sqrt{\mathrm{2}}=\mathrm{t} \\ $$$$\mathrm{e}\Rightarrow\mathrm{x}^{\mathrm{4}} −\mathrm{t}\left(\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}\right)−\mathrm{x}+\mathrm{t}^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow\mathrm{t}^{\mathrm{2}} −\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{t}+\mathrm{x}^{\mathrm{4}} −\mathrm{x}\:=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}\right)\:=\mathrm{4x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{4x}^{\mathrm{4}} \:+\mathrm{4x}=\:\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}=\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{1}+\mathrm{2x}+\mathrm{1}}{\mathrm{2}}\:=\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}−\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\:=−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}}\:>\mathrm{0}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{3}}}{\mathrm{2}} \\ $$

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