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pychu [463]
3 years ago
8

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

Mathematics
1 answer:
QveST [7]3 years ago
6 0

Answer:

The roots are

x=\frac{1}{6} [3+ i\sqrt{3}]

x=\frac{1}{6} [3- i\sqrt{3}]

Step-by-step explanation:

we have

3x^2-3x+1

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

3x^2-3x+1=0  

so

a=3\\b=-3\\c=1

substitute in the formula

x=\frac{-(-3)\pm\sqrt{-3^{2}-4(3)(1)}} {2(3)}

x=\frac{3\pm\sqrt{-3}} {6}

Remember that

i=\sqrt{-1}

so

x=\frac{1}{6} [3\pm i\sqrt{3}]

The roots are

x=\frac{1}{6} [3+ i\sqrt{3}]

x=\frac{1}{6} [3- i\sqrt{3}]

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