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Question Number 188660 by mr W last updated on 04/Mar/23
solve x^4 +4x=1
$${solve}\:{x}^{\mathrm{4}} +\mathrm{4}{x}=\mathrm{1} \\ $$
Answered by aba last updated on 04/Mar/23
•x^4 =(x^2 +1)^2 −2x^2 −1 x^4 +4x−1=0 ⇒ (x^2 +1)^2 −2x^2 −1+4x−1=0 ⇒(x^2 −1)^2 −2(x^2 −2x+1)=0 ⇒(x^2 +1)^2 −2(x−1)^2 =0 ⇒x^2 +1=±(x−1)(√2) ⇒x^2 −(√2)x+(1+(√2))=0 ∨ x^2 +(√2)x+(1−(√2))=0 x^2 −(√2)x+(1+(√2))=0 Δ=−2−4(√2)=−2(2(√2)−1)<0 x^2 +(√2)x+(1−(√2))=0 Δ=−2+4(√2)=2(2(√2)−1)>0 x=((−(√2)±(√(2(2(√2)−1))))/2)
$$\bullet\mathrm{x}^{\mathrm{4}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{4x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}} −\mathrm{1}+\mathrm{4x}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{1}=\pm\left(\mathrm{x}−\mathrm{1}\right)\sqrt{\mathrm{2}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)=\mathrm{0}\:\vee\:\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{x}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\:\Delta=−\mathrm{2}−\mathrm{4}\sqrt{\mathrm{2}}=−\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)<\mathrm{0} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{x}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Delta=−\mathrm{2}+\mathrm{4}\sqrt{\mathrm{2}}=\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)>\mathrm{0} \\ $$$$\mathrm{x}=\frac{−\sqrt{\mathrm{2}}\pm\sqrt{\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by BaliramKumar last updated on 04/Mar/23
answer is right but last 3rd and 4th step error (take ± value )
$${answer}\:{is}\:{right}\:{but} \\ $$$${last}\:\mathrm{3}{rd}\:{and}\:\mathrm{4}{th}\:{step}\:{error}\:\left({take}\:\pm\:{value}\:\right) \\ $$
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 −((√2)x−(√2))^2 =0 (x^2 +(√2)x+1−(√2))(x^2 −(√2)x+1+(√2))=0
$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{1}=\mathrm{1} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{x}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}−\sqrt{\mathrm{2}}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}+\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$ \\ $$

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