Question

which equation has the solutions $x = \frac{-3 \pm \sqrt{3}i}{2}$?\
\bigcirc $2x^2 + 6x + 9 = 0$\
\bigcirc $x^2 + 3x + 12 = 0$\
\bigcirc $x^2 + 3x + 3 = 0$\
\bigcirc $2x^2 + 6x + 3 = 0$

Explanation:

Step1: Recall quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Step2: Match given solution to formula

Given $x=\frac{-3\pm\sqrt{3}i}{2}$, rewrite $\sqrt{3}i$ as $\sqrt{-3}$. So $x=\frac{-3\pm\sqrt{-3}}{2}$. Compare to $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$: set $2a=2\Rightarrow a=1$, $b=3$.

Step3: Solve for $c$

Use discriminant: $b^2-4ac=-3$. Substitute $a=1,b=3$:
$3^2-4(1)c=-3$
$9-4c=-3$
$-4c=-12$
$c=3$

Step4: Form the equation

With $a=1,b=3,c=3$, equation is $x^2+3x+3=0$.
(Verify: $x=\frac{-3\pm\sqrt{9-12}}{2}=\frac{-3\pm\sqrt{-3}}{2}=\frac{-3\pm\sqrt{3}i}{2}$)

Answer:

C. $x^2 + 3x + 3 = 0$