Question

Which 2 values of x are roots of the polynomial below?

Answer 1

Given:

The given polynomial is:

$f(x)=$ $3x^{2}-3x+1$

To find the roots of the given polynomial.

To find the roots we have to take $f(x) = 0$

So,

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

Formula

By quadratic formula, the root of the equation $ax^{2} +bx+c = 0$ is,

$x = \frac{-b+\sqrt{b^{2}-4ac } }{2a}$ and $\frac{-b-\sqrt{b^{2} -4ac} }{2a}$

Now,

Putting, $a=3, b=-3, c=1$ we get,

$x = \frac{3+\sqrt{3^{2}-4(3)(1) } }{(2)(3)}$ and $\frac{3-\sqrt{3^{2}-4(3)(1) } }{(2)(3)}$

$x = \frac{3+\sqrt{-3} }{6}$ and $x=\frac{3-\sqrt{-3} }{6}$

Hence,

The values of the roots of the given polynomial are$x=\frac{3+\sqrt{-3} }{6}$ and $x=\frac{3-\sqrt{-3} }{6}$

Hence, Option A and F are the correct answer.