Question

Solve  `x^4-x^3+x^2-x+1=0.`

Answer 1

`x^4-x^3+x^2-x+1=0`
Multiply the given eqn by (x+1),

`(x+1)(x^4-x^3+x^2-x+1)=0`

`x^5=(-1)`

But -1 = cos 𝝅+𝒊 𝒔𝒊𝒏 𝝅

`therefore x = [cospi+isinpi]^(1/5)`

But By De Moivres theorem ,

(𝐜𝐨𝐬 𝐱 + 𝐢 𝐬𝐢𝐧 𝐱)^𝒏= cos nx + i sin nx

`therefore x=cos  pi/5+isin  pi/5`

Add period 2k𝝅 ,

`therefore x=cos(1+2k)(pi/5)+isin(1+2k)(pi/5)`

Where k = 0,1,2,3,4.

The roots of given eqn is given by ,

Put k = 0 `x_0=cos  pi/5+isin  pi/5=e^(pi/5)`

k = 1 `x_1=cos  (3pi)/5+isin  (3pi)/5=e^((3pi)/5)`

k = 2 `x_2=cos  pi/1+isin  pi/1=e^(pi/1)`

k = 3 `x_3=cos  (7pi)/5+isin  (7pi)/5=e^((7pi)/5)`

k = 4 `x_4=cos  (9pi)/5+isin  (9pi)/5=e^((9pi)/5)`

The roots of eqn are : `e^(pi/5),e^((3pi)/5),e^(pi/1),e^((7pi)/5),e^((9pi)/5)`.