Solve the quadratic equation 6x^2-3x+6=0. Which of the following expresses its solutions in the form a± bi ? (1 point) 1/4 ± sqrt(17)/4 i 1/2 ± sqrt(15)/2 i - 1/4 ± sqrt(15)/4 i 1/4 ± sqrt(15)/4 i

1 answer

To solve the quadratic equation \( 6x^2 - 3x + 6 = 0 \), we will use the quadratic formula:

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

Here, \( a = 6 \), \( b = -3 \), and \( c = 6 \).

  1. Calculate the discriminant \( b^2 - 4ac \):

\[ b^2 - 4ac = (-3)^2 - 4(6)(6) = 9 - 144 = -135 \]

  1. Since the discriminant is negative, we will have complex solutions. Now we compute \( \sqrt{-135} \):

\[ \sqrt{-135} = \sqrt{135} \cdot i = \sqrt{9 \cdot 15} \cdot i = 3\sqrt{15} \cdot i \]

  1. Now substitute back into the quadratic formula:

\[ x = \frac{-(-3) \pm 3\sqrt{15} i}{2 \cdot 6} = \frac{3 \pm 3\sqrt{15} i}{12} = \frac{1}{4} \pm \frac{\sqrt{15}}{4} i \]

Thus, the solutions for the equation \( 6x^2 - 3x + 6 = 0 \) are:

\[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4} i \]

The correct option that expresses its solutions in the form \( a \pm bi \) is:

\[ \boxed{\frac{1}{4} \pm \frac{\sqrt{15}}{4} i} \]