Question

Solve the quadratic equation 6x2−3x+6=0
6
𝑥
2
−
3
𝑥
+
6
=
0
. Which of the following expresses its solutions in the form a±bi
𝑎
±
𝑏
𝑖

?(1 point)
Responses

−14±15√4i
−
1
4
±
15
4
𝑖
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
1
2
±
15
2
𝑖
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
4
±
17
4
𝑖
Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 17 End Root over 4 End Fraction i

14±15√4i

Answer 1

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

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

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

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

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

Since the discriminant is negative, we expect complex solutions.

2. Substitute back into the quadratic formula:

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

3. Simplify \( \sqrt{135} \):

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

4. Substitute back into the expression for \( x \):

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

Thus, the solutions can be expressed as:

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

The solutions in the form \( a \pm bi \) are:

\[
a = \frac{1}{4}, \quad b = \frac{\sqrt{15}}{4}
\]

Looking at the provided options, the correct answer is:

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

Which corresponds to:

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