To find the zeros of the function [tex]\( f(x) = 3x^2 - x + 6 \)[/tex], we need to solve the equation [tex]\( f(x) = 0 \)[/tex]. Thus, we set:
[tex]\[ 3x^2 - x + 6 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = 6 \)[/tex].
To solve for [tex]\( x \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-1)^2 - 4 \cdot 3 \cdot 6 \][/tex]
[tex]\[ \Delta = 1 - 72 \][/tex]
[tex]\[ \Delta = -71 \][/tex]
Since the discriminant is negative, the solutions will be complex numbers. We continue by using the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{-71}}{2 \cdot 3} \][/tex]
This simplifies to:
[tex]\[ x = \frac{1 \pm \sqrt{-71}}{6} \][/tex]
Recall that [tex]\( \sqrt{-71} \)[/tex] can be written as [tex]\( \sqrt{71} \, i \)[/tex] where [tex]\( i \)[/tex] is the imaginary unit. Therefore:
[tex]\[ x = \frac{1 \pm \sqrt{71} \, i}{6} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{1 - \sqrt{71} \, i}{6} \][/tex]
[tex]\[ x = \frac{1 + \sqrt{71} \, i}{6} \][/tex]
Thus, the zeros of the function are:
[tex]\[ \boxed{\frac{1 - \sqrt{71} \, i}{6}, \frac{1 + \sqrt{71} \, i}{6}} \][/tex]
