Answer: 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
Answer:
[tex]x=\frac{1}{2} +\sqrt{\frac{1}{9} } i , \frac{1}{2} -\sqrt{\frac{1}{9} } i[/tex]
Step-by-step explanation:
Just another way to write the answer on top of me.
