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Question Number 124988 by bemath last updated on 07/Dec/20

Commented by bemath last updated on 07/Dec/20

x = ((−1±(√(1−4)))/2) = ((−1±i(√3))/2)  x_1  = −(1/2)+((i(√3))/2) ⇒x_1 = cos (((2π)/3))+ isin (((2π)/3))  x_1 ^(2007)  = cos (((4014π)/3))+isin (((4014π)/3))  x_1 ^(−2007)  = cos (((4014π)/3))−isin (((4014π)/3))  x^(2007) +(1/x^(2007) ) = 2cos (((4014π)/3))                            = 2cos (1338π)                            = 2cos (0)                            = 2×(1)=2

x=1±142=1±i32x1=12+i32x1=cos(2π3)+isin(2π3)x12007=cos(4014π3)+isin(4014π3)x12007=cos(4014π3)isin(4014π3)x2007+1x2007=2cos(4014π3)=2cos(1338π)=2cos(0)=2×(1)=2

Commented by floor(10²Eta[1]) last updated on 07/Dec/20

desnecessary long result

desnecessarylongresult

Commented by bemath last updated on 08/Dec/20

i like a long thing

ilikealongthing

Answered by floor(10²Eta[1]) last updated on 07/Dec/20

x^2 =−x−1  x^3 =−x^2 −x=x+1−x=1  x^3 =1  x^(2007) =(x^3 )^(669) =1  x^(2007) +(1/x^(2007) )=1+1=2

x2=x1x3=x2x=x+1x=1x3=1x2007=(x3)669=1x2007+1x2007=1+1=2

Commented by talminator2856791 last updated on 07/Dec/20

 please explain −x^2 −x = x+1−x

pleaseexplainx2x=x+1x

Commented by floor(10²Eta[1]) last updated on 07/Dec/20

x^2 =−x−1

x2=x1

Commented by talminator2856791 last updated on 07/Dec/20

 ooooooooooooo wow

ooooooooooooowow

Commented by talminator2856791 last updated on 07/Dec/20

 can you please look at my answer for post 124978

canyoupleaselookatmyanswerforpost124978

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