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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}

x4+x3+x2+x+1=01−x51−x=0x5=1x=e2ikπ5,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)) .

x2+1x2+x+1x+1=0(x+1x)2+(x+1x)−1=0⇒x+1x=−12±52=−a,ba=5+12,b=5−12x2+ax+1=0x1,x2=−a2±i(4−a22)x2−bx+1=0⇒x3,x4=b2±i(4−b22).

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

⇔x2(x2+x+14)+3x24+x+1=0⇔x2(x+12)2+x22+(x2+1)2=0ThisequationhasnorealsolutionsbecauseLHSissumofthenon−negativetermsandtheyisn′tequaltozerosimultaneously

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

but that steps isn′t key .

butthatstepsisn′tkey.

Commented by Rasheed.Sindhi last updated on 25/Jul/20

Good but too many steps are left out.

Goodbuttoomanystepsareleftout.

Commented by Rasheed.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.

Yetanimportantstep.Ifyouweretronsformingfromx2(x2+x+14)+3x24+x+1=0tox4+x3+x2+x+1nostepwerenecessary.ButifyoudooppositelyIthinksomestepsarenecessarytoinsert.

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