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

 x  = (3 − i√3)/6 and (3 + i√3)/6

Step-by-step explanation: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

-2

7 - 9 = -2

reversed:

-2 + 9 = 7

step-by-step explanation: