Question Number 104905 by mohammad17 last updated on 24/Jul/20

Answered by OlafThorendsen last updated on 24/Jul/20

x^4 +x^3 +x^2 +x+1 = 0  ((1−x^5 )/(1−x)) = 0  x^5  = 1  x = e^((2ikπ)/5) , k∈{1,2,3,4}

Answered by ajfour last updated on 24/Jul/20

x^2 +(1/x^2 )+x+(1/x)+1=0   (x+(1/x))^2 +(x+(1/x))−1=0  ⇒   x+(1/x) = −(1/2)±((√5)/2) = −a, b  a=(((√5)+1)/2)  ,    b=(((√5)−1)/2)  x^2 +ax+1=0  x_1 , x_2  =−(a/2)±i(((√(4−a^2 ))/2))  x^2 −bx+1=0  ⇒  x_3  , x_4  =(b/2)±i(((√(4−b^2 ))/2)) .

Answered by 1549442205PVT last updated on 25/Jul/20

⇔x^2 (x^2 +x+(1/4))+((3x^2 )/4)+x+1=0  ⇔x^2 (x+(1/2))^2 +(x^2 /2)+((x/2)+1)^2 =0  This equation has no real solutions  because LHS is  sum of the  non−negative terms and they isn′t  equal to zero simultaneously

Commented by1549442205PVT last updated on 25/Jul/20

but  that steps isn′t key .

Commented byRasheed.Sindhi last updated on 25/Jul/20

Good but too many steps are left  out.

Commented byRasheed.Sindhi last updated on 25/Jul/20

Yet an important step.  If you were tronsforming from  x^2 (x^2 +x+(1/4))+((3x^2 )/4)+x+1=0 to  x^4 +x^3 +x^2 +x+1 no step were  necessary.  But if you do oppositely I think  some steps are necessary to insert.