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:

⚡ Using what you learned: quadratic formula and its applications

Step 1: Identify the given solutions and the quadratic formula

The given solutions are:

$$ x = \frac{-3 \pm \sqrt{3}i}{2} $$

The quadratic formula for an equation of the form \(ax^2 + bx + c = 0\) is:

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

Step 2: Compare the denominators to find \(a\)

Comparing the denominators:

$$ 2a = 2 \implies a = 1 $$

Step 3: Compare the linear terms to find \(b\)

Comparing the first term in the numerator:

$$ -b = -3 \implies b = 3 $$

Step 4: Use the discriminant to find \(c\)

Comparing the terms under the square root:

$$ \sqrt{b^2 - 4ac} = \sqrt{3}i $$

Since \(i = \sqrt{-1}\), we can rewrite the right side:

$$ \sqrt{3}i = \sqrt{-3} $$

Now, equate the expressions inside the square roots:

$$ b^2 - 4ac = -3 $$

Substitute \(a = 1\) and \(b = 3\):

$$ 3^2 - 4(1)(c) = -3 $$

$$ 9 - 4c = -3 $$

$$ -4c = -12 $$

$$ c = 3 $$

Step 5: Write the quadratic equation

Using \(a = 1\), \(b = 3\), and \(c = 3\):

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

Answer:

\( x^2 + 3x + 3 = 0 \) (the third option)