Questions: Find the zeros of the function algebraically. f(x)=3 x^2-x+3 The zeros are (Simplify your answer. Type an exact answer, using radicals and i as needed. Use integers or fractions for an

Solution Steps

To find the zeros of the function \( f(x) = 3x^2 - x + 3 \), we need to solve the quadratic equation \( 3x^2 - x + 3 = 0 \). This can be done using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -1 \), and \( c = 3 \). We will calculate the discriminant \( b^2 - 4ac \) to determine the nature of the roots and then apply the formula to find the exact zeros.

We start with the quadratic function given by \[ f(x) = 3x^2 - x + 3. \] To find the zeros of this function, we set it equal to zero: \[ 3x^2 - x + 3 = 0. \]

Using the coefficients \( a = 3 \), \( b = -1 \), and \( c = 3 \), we calculate the discriminant \( D \) using the formula: \[ D = b^2 - 4ac. \] Substituting the values, we find: \[ D = (-1)^2 - 4 \cdot 3 \cdot 3 = 1 - 36 = -35. \]

Since the discriminant \( D = -35 \) is negative, this indicates that the quadratic equation has two complex conjugate roots.

We use the quadratic formula to find the roots: \[ x = \frac{-b \pm \sqrt{D}}{2a}. \] Substituting the values, we have: \[ x = \frac{-(-1) \pm \sqrt{-35}}{2 \cdot 3} = \frac{1 \pm \sqrt{35}i}{6}. \] This simplifies to: \[ x_1 = \frac{1}{6} + \frac{\sqrt{35}}{6}i, \] \[ x_2 = \frac{1}{6} - \frac{\sqrt{35}}{6}i. \]

Final Answer