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

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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)

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