# Roots.Given that 4x^3-16x^2+21x-27 has the solution 3 find the conjugate complex pair of roots.

justaguide | Certified Educator

The equation 4x^3-16x^2+21x-27=0 has a real root equal to 3.

We can write the equation as:

4x^3 - 16x^2 + 21x - 27 = 0

=> 4x^3 - 12x^2 - 4x^2 + 12x + 9x - 27 = 0

=> 4x^2(x - 3) - 4x(x - 3) + 9(x - 3) = 0

=> (4x^2 - 4x + 9)(x - 3) = 0

The roots of 4x^2 - 4x + 9 = 0 are

x1 = 4/8 + sqrt(16 - 144)/8

=> 1/2 + i*sqrt 2

x2 = 1/2 - i*sqrt 2

The complex roots of the equation 4x^3-16x^2+21x-27=0 are 1/2 + i*sqrt 2 and 1/2 - i*sqrt 2

giorgiana1976 | Student

We'll use Viete's relations for finding the other 2 conjugate roots, which always come in pairs.

3 + x2 + x3 = 16/4

3 + x2 + x3 = 4

We'll subtract 3:

x2 + x3 = 4-3

x2 + x3 = 1

3*x2*x3 = 27/4

We'll divide by 3:

x2*x3 = 9/4

We'll write the equation of the quadratic when knowing the sum and the product of the roots:

x^2 - x + 9/4 = 0

We'll multiply by 4:

4x^2 - 4x + 9 = 0

x2 = [4 + sqrt(16 - 144)]/8

x2 = (4 + 8sqrt2)/8

x2 = (1+2isqrt2)/2

x3 = (1-2isqrt2)/2

The pair of complex roots of the equation are :{(1-2isqrt2)/2 ; (1+2isqrt2)/2}.