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andrey2020 [161]

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

QveST [7]

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

$3x^2-3x+1=0$

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

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

$i=\sqrt{-1}$

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

The roots are

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

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

6 0

2 years ago