Student Question

Solve the equation x^3-4x^2+6x-4=0, given that l+i is a root of this equation.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

We have to solve the equation given that 1+ i is a root of the equation. As complex roots always come in pairs with their complex conjugate, 1 - i is also a root of the equation.

So we have (x - (1 + i))( x - (1 -i))(x - a) = x^3-4x^2+6x-4=0, where the final root is a

(x - (1 + i))( x - (1 -i))(x - a) = x^3-4x^2+6x-4=0

=> (x - 1- i)(x - 1 + i)(x - a) = x^3 - 4x^2 + 6x - 4

=> [(x - 1)^2 - i^2](x - a) = x^3 - 4x^2 + 6x - 4

=> (x^2 + 1 - 2x + 1)(x - a) = x^3 - 4x^2 + 6x - 4

=> (x^2 - 2x + 2)(x - a) = x^3 - 4x^2 + 6x - 4

=> x^3 - 2x^2 + 2x - ax^2 + 2ax - 2a = x^3 - 4x^2 + 6x - 4

=> - ax^2 + 2ax - 2a = -2x^2 + 4x - 4

Equate the numeric term -2a = -4

=> a = 2

The roots of the equation are (2 , 1 + i , 1 - i)

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial