Question

What are the roots of the equation 9, x, squared, minus, 24, x, plus, 17, equals, 09x
2
−24x+17=0 in simplest a, plus, b, ia+bi form?

Answer 1

To find the roots of the quadratic equation \( 9x^2 - 24x + 17 = 0 \), we can use the quadratic formula:

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

In this equation, the coefficients are:
- \( a = 9 \)
- \( b = -24 \)
- \( c = 17 \)

Now, we will calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = (-24)^2 = 576
\]

\[
4ac = 4 \cdot 9 \cdot 17 = 612
\]

Now we can find the discriminant:

\[
b^2 - 4ac = 576 - 612 = -36
\]

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

Now applying the quadratic formula:

\[
x = \frac{-(-24) \pm \sqrt{-36}}{2 \cdot 9}
\]

This simplifies to:

\[
x = \frac{24 \pm \sqrt{-36}}{18}
\]

Since \( \sqrt{-36} = 6i \), we substitute that into the equation:

\[
x = \frac{24 \pm 6i}{18}
\]

This can be simplified further:

\[
x = \frac{24}{18} \pm \frac{6i}{18}
\]

\[
x = \frac{4}{3} \pm \frac{i}{3}
\]

Thus, the roots of the equation \( 9x^2 - 24x + 17 = 0 \) in simplest \( a + bi \) form are:

\[
x = \frac{4}{3} + \frac{i}{3} \quad \text{and} \quad x = \frac{4}{3} - \frac{i}{3}
\]