Question

What are the roots of the equation x, squared, plus, 12, x, plus, 45, equals, 0x
2
+12x+45=0 in simplest a, plus, b, ia+bi form?

Answer 1

To solve the quadratic equation \( x^2 + 12x + 45 = 0 \), we will use the quadratic formula:

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

For the equation \( x^2 + 12x + 45 = 0 \), we have:
- \( a = 1 \)
- \( b = 12 \)
- \( c = 45 \)

First, we calculate the discriminant:

\[
b^2 - 4ac = 12^2 - 4 \cdot 1 \cdot 45 = 144 - 180 = -36
\]

Since the discriminant is negative, the roots will be complex. Now we will substitute the values into the quadratic formula:

\[
x = \frac{-12 \pm \sqrt{-36}}{2 \cdot 1}
\]

We know that \( \sqrt{-36} = 6i \), so now we can substitute that into the formula:

\[
x = \frac{-12 \pm 6i}{2}
\]

Dividing both the real and imaginary components by 2:

\[
x = -6 \pm 3i
\]

Thus, the roots of the equation \( x^2 + 12x + 45 = 0 \) in simplest \( a + bi \) form are:

\[
x = -6 + 3i \quad \text{and} \quad x = -6 - 3i
\]