Solve the quadratic equation 6x2−3x+6=0

. Which of the following expresses its solutions in the form a±bi
?(1 point)
Responses

14±15√4i
Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i

−14±15√4i
negative Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i

12±15√2i
Start Fraction 1 over 2 End Fraction plus or minus Start Fraction Start Root 15 End Root over 2 End Fraction i

14±17√4i

1 answer

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

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

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

First, we calculate the discriminant \( b^2 - 4ac \):

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

Since the discriminant is negative, the solutions will be complex numbers.

Now we substitute \( a \), \( b \), and the discriminant into the quadratic formula:

\[ x = \frac{-(-3) \pm \sqrt{-135}}{2 \cdot 6} = \frac{3 \pm \sqrt{-135}}{12} \]

We can simplify \( \sqrt{-135} \):

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

Now substituting this back into the equation gives us:

\[ x = \frac{3 \pm 3\sqrt{15} \cdot i}{12} = \frac{3}{12} \pm \frac{3\sqrt{15} \cdot i}{12} = \frac{1}{4} \pm \frac{\sqrt{15}}{4} \cdot i \]

Thus, the solutions can be expressed as:

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

Now looking at the options provided, we can determine the one that matches:

  • Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i

This option accurately reflects the calculated solutions.

Thus, the correct response is:

Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i.