Question

which equation has the solutions $x = \frac{-3 \pm \sqrt{3}i}{2}$?

\\( 2x^2 + 6x + 9 = 0 \\)
\\( x^2 + 3x + 12 = 0 \\)
\\( x^2 + 3x + 3 = 0 \\)
\\( 2x^2 + 6x + 3 = 0 \\)

Explanation:

Step1: Recall the quadratic formula

The solutions of a quadratic equation \(ax^2 + bx + c = 0\) are given by \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). We are given the solutions \(x = \frac{-3\pm\sqrt{3}i}{2}\), so we can equate this to the quadratic formula and find \(a\), \(b\), and \(c\).

Step2: Compare with quadratic formula

From \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}=\frac{-3\pm\sqrt{3}i}{2}\), we can see that \(2a = 2\) (so \(a = 1\)) or \(2a=2\) (we can also check other possibilities, but let's first assume \(a = 1\)). Then \(-b=-3\), so \(b = 3\). Now, the discriminant \(b^2-4ac=( \sqrt{3}i)^2\) (since the square root part is \(\sqrt{3}i\)). Let's compute the discriminant: \((\sqrt{3}i)^2=3i^2=- 3\). So \(b^2-4ac=-3\). Substituting \(a = 1\) and \(b = 3\) into \(b^2-4ac\), we get \(9-4(1)c=-3\). Solving for \(c\): \(9 + 3=4c\), \(12 = 4c\), \(c = 3\). Wait, but let's check the first option. Wait, maybe \(a = 2\). Let's try \(a = 2\). Then \(2a=2\) implies \(a = 1\)? No, wait \(2a\) is the denominator. Wait, if \(x=\frac{-3\pm\sqrt{3}i}{2}\), then \(2x=-3\pm\sqrt{3}i\), so \(2x + 3=\pm\sqrt{3}i\). Squaring both sides: \((2x + 3)^2=(\pm\sqrt{3}i)^2\). \((2x + 3)^2=4x^2+12x + 9\) and \((\sqrt{3}i)^2=-3\). So \(4x^2+12x + 9=-3\), \(4x^2+12x+12 = 0\), divide by 2: \(2x^2+6x + 6=0\)? No, that's not one of the options. Wait, maybe I made a mistake. Wait, let's check the discriminant for each option.

Step3: Check each option's discriminant

• Option 1: \(2x^2+6x + 9 = 0\). Here, \(a = 2\), \(b = 6\), \(c = 9\). Discriminant \(D=b^2-4ac=36-4\times2\times9=36 - 72=-36\). Then solutions are \(x=\frac{-6\pm\sqrt{-36}}{4}=\frac{-6\pm6i}{4}=\frac{-3\pm3i}{2}\). Not matching.
• Option 2: \(x^2+3x + 12 = 0\). \(a = 1\), \(b = 3\), \(c = 12\). Discriminant \(D = 9-48=-39\). Solutions \(x=\frac{-3\pm\sqrt{-39}}{2}=\frac{-3\pm\sqrt{39}i}{2}\). Not matching.
• Option 3: \(x^2+3x + 3 = 0\). \(a = 1\), \(b = 3\), \(c = 3\). Discriminant \(D=9 - 12=-3\). Solutions \(x=\frac{-3\pm\sqrt{-3}}{2}=\frac{-3\pm\sqrt{3}i}{2}\). This matches the given solutions.
• Option 4: \(2x^2+6x + 3 = 0\). \(a = 2\), \(b = 6\), \(c = 3\). Discriminant \(D = 36-24 = 12\). Solutions \(x=\frac{-6\pm\sqrt{12}}{4}=\frac{-6\pm2\sqrt{3}}{4}=\frac{-3\pm\sqrt{3}}{2}\). Real solutions, not matching.

Answer:

The equation with the given solutions is \(x^2 + 3x + 3 = 0\), which is the third option: C. \(x^2 + 3x + 3 = 0\)