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Question Number 7054 by Tawakalitu. last updated on 08/Aug/16

Solve for x in the equation below  x^4  + 2x^3  − x^2  + 2x + 1 = 0

$${Solve}\:{for}\:{x}\:{in}\:{the}\:{equation}\:{below} \\ $$$${x}^{\mathrm{4}} \:+\:\mathrm{2}{x}^{\mathrm{3}} \:−\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$ \\ $$

Commented by prakash jain last updated on 08/Aug/16

(x^2 +x+1)^2 =x^4 +x^2 +1+2x^3 +2x+2x^2   x^4 +2x^3 +3x^2 +2x+1  given expr−(x^2 +x+1)^2 =−4x^2 +1    (x^2 −x+1)^2 =x^4 +x^2 +1−2x^3 −2x+2x^2   =x^4 −2x^3 +3x^2 −2x+1  given expr−(x^2 −x+1)^2   =−4x^3 −4x^2 −4x=−4x(x^2 +x+1)  hence given expr  =(x^2 +x+1)^2 −4x(x^2 +x+1)  =(x^2 +x+1)(x^2 −3x+1)  can be solved as quadratic

$$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$${given}\:{expr}−\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} =−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$ \\ $$$$\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1} \\ $$$${given}\:{expr}−\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=−\mathrm{4}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}=−\mathrm{4}{x}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$${hence}\:{given}\:{expr} \\ $$$$=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{x}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}\right) \\ $$$${can}\:{be}\:{solved}\:{as}\:{quadratic} \\ $$

Commented by Tawakalitu. last updated on 08/Aug/16

Thanks so much. i appreciate

$${Thanks}\:{so}\:{much}.\:{i}\:{appreciate} \\ $$

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