Solve the quadratic equation 6x2−3x+6=0 . Which of the following expresses its solutions in the form a±bi ?(1 point)

Question

Responses

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
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

14±17√4i

Answer 1

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

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

where $a=6$, $b=-3$, and $c=6$.

1. First, we calculate $b^2-4ac$:

$$b^2=(-3{)}^2=9$$ $$4ac=4\times 6\times 6=144$$ $$b^2-4ac=9-144=-135$$

2. Since the determinant $b^2-4ac$ is negative, there will be complex solutions.

3. Now we substitute $a$, $b$, and the value of the determinant into the quadratic formula:

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

4. We can simplify $\sqrt{-135}$:

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

5. Substitute this back into the equation:

$$x=\frac{3\pm 3\sqrt{15}\cdot i}{12}$$

6. We can simplify $\frac{3}{12}$ and $\frac{3\sqrt{15}}{12}$:

$$x=\frac{1}{4}\pm \frac{\sqrt{15}}{4}\cdot i$$

Thus, the solutions can be expressed in the form $a\pm bi$ as:

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

Therefore, the correct response from your options is:

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