Question

which quadratic equation has roots
$-3 + 6i$ and $-3 - 6i$
?
answer
$\bigcirc$ $x^2 + 6x + 45 = 0$
$\bigcirc$ $x^2 + 6x + 18 = 0$
$\bigcirc$ $x^2 - 6x + 18 = 0$
$\bigcirc$ $x^2 - 6x + 45 = 0$
submit answer

Explanation:

Step1: Recall Vieta's formulas for quadratic equations. For a quadratic equation \(ax^2 + bx + c = 0\) with roots \(r_1\) and \(r_2\), we have \(r_1 + r_2 = -\frac{b}{a}\) and \(r_1 \cdot r_2 = \frac{c}{a}\). Here, let's take \(a = 1\) (since we can compare with the given options which have leading coefficient 1).

The roots are \(r_1 = -3 + 6i\) and \(r_2 = -3 - 6i\).

Step2: Calculate the sum of the roots.

\(r_1 + r_2 = (-3 + 6i) + (-3 - 6i) = -3 - 3 + 6i - 6i = -6\).
From Vieta's formula, \(r_1 + r_2 = -\frac{b}{a}\). Since \(a = 1\), we have \(-6 = -b\), so \(b = 6\).

Step3: Calculate the product of the roots.

\(r_1 \cdot r_2 = (-3 + 6i)(-3 - 6i)\). Using the difference of squares formula \((a + b)(a - b) = a^2 - b^2\), where \(a = -3\) and \(b = 6i\), we get:
\((-3)^2 - (6i)^2 = 9 - 36i^2\). Since \(i^2 = -1\), this becomes \(9 - 36(-1) = 9 + 36 = 45\).
From Vieta's formula, \(r_1 \cdot r_2 = \frac{c}{a}\). Since \(a = 1\), \(c = 45\).

Step4: Form the quadratic equation.

With \(a = 1\), \(b = 6\), and \(c = 45\), the quadratic equation is \(x^2 + 6x + 45 = 0\).

Answer:

\(x^2 + 6x + 45 = 0\) (corresponding to the first option: \(x^2 + 6x + 45 = 0\))