Answer:
The roots are
[tex]x=\frac{1}{6} [3+ i\sqrt{3}][/tex]
[tex]x=\frac{1}{6} [3- i\sqrt{3}][/tex]
Step-by-step explanation:
we have
[tex]3x^2-3x+1[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^2-3x+1=0[/tex]
so
[tex]a=3\\b=-3\\c=1[/tex]
substitute in the formula
[tex]x=\frac{-(-3)\pm\sqrt{-3^{2}-4(3)(1)}} {2(3)}[/tex]
[tex]x=\frac{3\pm\sqrt{-3}} {6}[/tex]
Remember that
[tex]i=\sqrt{-1}[/tex]
so
[tex]x=\frac{1}{6} [3\pm i\sqrt{3}][/tex]
The roots are
[tex]x=\frac{1}{6} [3+ i\sqrt{3}][/tex]
[tex]x=\frac{1}{6} [3- i\sqrt{3}][/tex]
