Question

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

14±17√4i
Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 17 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±15√4i

Answer 1

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

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

In this equation:

- \( a = 6 \)
- \( b = -3 \)
- \( c = 6 \)

Calculating the discriminant (\( b^2 - 4ac \)):

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

Since the discriminant is negative, the solutions will be complex. Now we can substitute into the quadratic formula:

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

To express \( \sqrt{-135} \) in terms of \( i \):

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

Now substitute back into the equation:

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

Thus, the solutions can be expressed as:

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

From the options provided, the correct response is:

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