Question

What are the roots of the equation x, squared, plus, 12, x, plus, 61, equals, 0x
2
+12x+61=0 in simplest a, plus, b, ia+bi form?

Answer 1

To find the roots of the equation \( x^2 + 12x + 61 = 0 \), we can use the quadratic formula:

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

In this equation, \( a = 1 \), \( b = 12 \), and \( c = 61 \).

Now, let's calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = 12^2 = 144
\]
\[
4ac = 4 \cdot 1 \cdot 61 = 244
\]
\[
b^2 - 4ac = 144 - 244 = -100
\]

Since the discriminant is negative, the roots are complex. Now we can substitute back into the quadratic formula:

\[
x = \frac{-12 \pm \sqrt{-100}}{2 \cdot 1}
\]

We know that \( \sqrt{-100} = 10i \), so we can continue:

\[
x = \frac{-12 \pm 10i}{2}
\]
\[
x = \frac{-12}{2} \pm \frac{10i}{2}
\]
\[
x = -6 \pm 5i
\]

Thus, the roots in the form \( a + bi \) are:

\[
-6 + 5i \quad \text{and} \quad -6 - 5i
\]

Therefore, the roots of the equation \( x^2 + 12x + 61 = 0 \) are:

\[
-6 + 5i \quad \text{and} \quad -6 - 5i
\]