Solve the quadratic equation 12x2−12x+4=−3x

12
𝑥
2

12
𝑥
+
4
=

3
𝑥
. Which of the following expresses its solutions in the form a±bi
𝑎
±
𝑏
𝑖

?(1 point)
Responses

38+111√24i
3
8
+
111
24
𝑖
Start Fraction 3 over 8 End Fraction plus Start Fraction Start Root 111 End Root over 24 End Fraction i

38±111√24i
3
8
±
111
24
𝑖
Start Fraction 3 over 8 End Fraction plus or minus Start Fraction Start Root 111 End Root over 24 End Fraction i

38±273√24i
3
8
±
273
24
𝑖
Start Fraction 3 over 8 End Fraction plus or minus Start Fraction Start Root 273 End Root over 24 End Fraction i

−38±111√24i

1 answer

To solve the quadratic equation \( 12x^2 - 12x + 4 = -3x \), we first rearrange the equation to standard form:

\[ 12x^2 - 12x + 4 + 3x = 0 \] \[ 12x^2 - 9x + 4 = 0 \]

Next, we can use the quadratic formula, which is given by:

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

In our case, \( a = 12 \), \( b = -9 \), and \( c = 4 \). First, we calculate the discriminant \( b^2 - 4ac \):

\[ b^2 = (-9)^2 = 81 \] \[ 4ac = 4 \cdot 12 \cdot 4 = 192 \] \[ b^2 - 4ac = 81 - 192 = -111 \]

Since the discriminant is negative, we can conclude that the solutions will be complex. Now, substituting the values into the quadratic formula:

\[ x = \frac{-(-9) \pm \sqrt{-111}}{2(12)} = \frac{9 \pm \sqrt{-111}}{24} \]

We can write \(\sqrt{-111}\) as \(i\sqrt{111}\):

\[ x = \frac{9 \pm i\sqrt{111}}{24} \]

Now we can split the expression into its real and imaginary parts. The real part is:

\[ \frac{9}{24} = \frac{3}{8} \]

And the imaginary part is:

\[ \frac{\sqrt{111}}{24} \]

So, we can express it in the form \( a \pm bi \):

\[ x = \frac{3}{8} \pm \frac{\sqrt{111}}{24} i \]

Thus, the solutions can be expressed as:

\[ \frac{3}{8} \pm \frac{\sqrt{111}}{24} i \]

This corresponds to the second option in your provided responses:

\( \frac{3}{8} \pm \frac{\sqrt{111}}{24} i \).