Answer :
To determine which equation has the solutions [tex]\( x = \frac{-3 \pm \sqrt{3}i}{2} \)[/tex], let's test each given equation by substituting these complex solutions and checking if the equation holds true.
First, let’s denote the solutions:
[tex]\[ x_1 = \frac{-3 + \sqrt{3}i}{2} \][/tex]
[tex]\[ x_2 = \frac{-3 - \sqrt{3}i}{2} \][/tex]
We'll evaluate each equation at both values.
1. Equation: [tex]\( 2x^2 + 6x + 9 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ 2\left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 + \sqrt{3}i}{2}\right) + 9 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ 2\left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 - \sqrt{3}i}{2}\right) + 9 = 0 \][/tex]
Both these evaluations need to equal zero in order for [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] to be solutions. After performing the above substitutions and calculations we would verify if they indeed equal zero.
2. Equation: [tex]\( x^2 + 3x + 12 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ \left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 + \sqrt{3}i}{2}\right) + 12 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ \left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 - \sqrt{3}i}{2}\right) + 12 = 0 \][/tex]
Both must evaluate to zero.
3. Equation: [tex]\( x^2 + 3x + 3 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ \left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 + \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ \left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 - \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Again, both must be zero.
4. Equation: [tex]\( 2x^2 + 6x + 3 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ 2\left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 + \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ 2\left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 - \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Both need to simplify to zero.
Evaluating these, we discover that the correct equation where both solutions [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] satisfy the equation is:
[tex]\[ x^2 + 3x + 3 = 0 \][/tex]
Therefore, the correct equation is:
[tex]\[ x^2 + 3x + 3 = 0 \][/tex]
First, let’s denote the solutions:
[tex]\[ x_1 = \frac{-3 + \sqrt{3}i}{2} \][/tex]
[tex]\[ x_2 = \frac{-3 - \sqrt{3}i}{2} \][/tex]
We'll evaluate each equation at both values.
1. Equation: [tex]\( 2x^2 + 6x + 9 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ 2\left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 + \sqrt{3}i}{2}\right) + 9 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ 2\left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 - \sqrt{3}i}{2}\right) + 9 = 0 \][/tex]
Both these evaluations need to equal zero in order for [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] to be solutions. After performing the above substitutions and calculations we would verify if they indeed equal zero.
2. Equation: [tex]\( x^2 + 3x + 12 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ \left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 + \sqrt{3}i}{2}\right) + 12 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ \left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 - \sqrt{3}i}{2}\right) + 12 = 0 \][/tex]
Both must evaluate to zero.
3. Equation: [tex]\( x^2 + 3x + 3 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ \left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 + \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ \left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 3\left(\frac{-3 - \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Again, both must be zero.
4. Equation: [tex]\( 2x^2 + 6x + 3 = 0 \)[/tex]
Substituting [tex]\( x_1 \)[/tex]:
[tex]\[ 2\left(\frac{-3 + \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 + \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Substituting [tex]\( x_2 \)[/tex]:
[tex]\[ 2\left(\frac{-3 - \sqrt{3}i}{2}\right)^2 + 6\left(\frac{-3 - \sqrt{3}i}{2}\right) + 3 = 0 \][/tex]
Both need to simplify to zero.
Evaluating these, we discover that the correct equation where both solutions [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] satisfy the equation is:
[tex]\[ x^2 + 3x + 3 = 0 \][/tex]
Therefore, the correct equation is:
[tex]\[ x^2 + 3x + 3 = 0 \][/tex]
