Question

What are the roots of the equation x, squared, minus, 10, x, plus, 41, equals, 0x
2
−10x+41=0 in simplest a+bi form?

Answer 1

To find the roots of the equation

\[
x^2 - 10x + 41 = 0,
\]

we will use the quadratic formula, which is given by:

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

where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).

In this case, we have:

- \( a = 1 \)
- \( b = -10 \)
- \( c = 41 \)

Let's first calculate the discriminant (\( D \)):

\[
D = b^2 - 4ac = (-10)^2 - 4(1)(41) = 100 - 164 = -64.
\]

Since the discriminant is negative, the roots will be complex.

Now we can substitute the values into the quadratic formula:

\[
x = \frac{-(-10) \pm \sqrt{-64}}{2 \cdot 1} = \frac{10 \pm \sqrt{-64}}{2}.
\]

We know that \(\sqrt{-64} = 8i\) (since \(i\) is the imaginary unit, defined as \(i = \sqrt{-1}\)). Therefore, we have:

\[
x = \frac{10 \pm 8i}{2}.
\]

Now, simplifying the expression:

\[
x = 5 \pm 4i.
\]

Thus, the roots of the equation \( x^2 - 10x + 41 = 0 \) in simplest \( a + bi \) form are:

\[
x = 5 + 4i \quad \text{and} \quad x = 5 - 4i.
\]