Advertisement Remove all ads

Solve X 4 − X 3 + X 2 − X + 1 = 0 . - Applied Mathematics 1

Sum

Solve  `x^4-x^3+x^2-x+1=0.`

Advertisement Remove all ads

Solution

`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)`.

Concept: D’Moivre’S Theorem
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×