Question

Which two values of x are roots of the polynomial below? 3x2 - 3x + 1

Answer 1

The roots of a polynomial are where the polynomial equals zero

3x² − 3x + 1 = 0

∴ 3x² − 3x = -1

∴ x² − x = -1/3

∴ x² − x + (-1/2)² = (-1/2)² − 1/3

given x² + bx + (b/2)² = (x + b/2)²

∴ (x − 1/2)² = 1/4 − 1/3

∴ (x − 1/2)² = 3/12 − 4/12

∴ (x − 1/2)² = -1/12

∴ x − 1/2 = ±√(-1/12)

This tells us there are no real roots and if you need real number solutions we stop here

∴ x − 1/2 = ±i/(2√3)

∴ x − 1/2 = ±i√(3)/6

∴ x = 1/2 ± i√(3)/6

∴ x = 3/6 ± i√(3)/6

∴ x = (3 − i√3)/6 and (3 + i√3)/6 <= the 2 complex roots

Answer 2

For this case we have the following polynomial:

`3x^2 - 3x + 1`

Applying the formula of the resolvent we have:

`x=(-b+/-√(b^2-4ac))/(2a)`

Substituting values we have:

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

Rewriting we have:

`x=(3+/-√(9-12))/(6)`

`x=(3+/-√(-3))/(6)`

`x=(3+/-i√(3))/(6)`

Therefore, the roots of the polynomial are given by:

`x=(3+i√(3))/(6)`

`x=(3-i√(3))/(6)`

Answer:

The values of x that are roots of the polynomial are:

`x=(3+i√(3))/(6)`

`x=(3-i√(3))/(6)`