Question

What are the roots of the equation x, squared, minus, 10, x, plus, 50, equals, 0x
2
−10x+50=0 in simplest a, plus, b, ia+bi form?

Answer 1

To solve the quadratic equation \( x^2 - 10x + 50 = 0 \), we can use the quadratic formula:

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

In our equation, the coefficients are:
- \( a = 1 \)
- \( b = -10 \)
- \( c = 50 \)

Now we can substitute these values into the quadratic formula:

1. Calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = (-10)^2 - 4(1)(50) = 100 - 200 = -100
\]

2. The discriminant is negative (\(-100\)), which indicates that there are two complex roots.

3. Now, substitute back into the quadratic formula:

\[
x = \frac{-(-10) \pm \sqrt{-100}}{2(1)} = \frac{10 \pm \sqrt{-100}}{2}
\]

4. Since \( \sqrt{-100} = 10i \) (where \( i \) is the imaginary unit), we have:

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

5. This simplifies to:

\[
x = 5 \pm 5i
\]

Thus, the roots of the equation \( x^2 - 10x + 50 = 0 \) in simplest form \( a + bi \) are:

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