Question

Solve the quadratic equation 3x2+3x+12=10 . Express its solutions in the form a±bi .(1 point)

Answer 1

First, let's start by rewriting the quadratic equation \( 3x^2 + 3x + 12 = 10 \) in standard form. We can do this by moving 10 to the left side:

\[
3x^2 + 3x + 12 - 10 = 0
\]

This simplifies to:

\[
3x^2 + 3x + 2 = 0
\]

Next, we can use the quadratic formula to find the solutions. The quadratic formula is given by:

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

In our equation, \( a = 3 \), \( b = 3 \), and \( c = 2 \). Now, we will calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 3^2 - 4 \cdot 3 \cdot 2 = 9 - 24 = -15
\]

Since the discriminant is negative, the solutions will involve imaginary numbers. Now, we can substitute the values into the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{-15}}{2 \cdot 3}
\]

This simplifies to:

\[
x = \frac{-3 \pm i\sqrt{15}}{6}
\]

We can further simplify this expression:

\[
x = \frac{-3}{6} \pm \frac{i\sqrt{15}}{6}
\]

This simplifies to:

\[
x = -\frac{1}{2} \pm \frac{i\sqrt{15}}{6}
\]

Thus, the solutions of the quadratic equation \( 3x^2 + 3x + 12 = 10 \) in the form \( a \pm bi \) are:

\[
x = -\frac{1}{2} \pm \frac{i\sqrt{15}}{6}
\]