Question

which answer shows the conjugate pair solutions to the equation x^2−3x+8=−2 ?

Answer 1

To solve the equation \( x^2 - 3x + 8 = -2 \), we first rearrange it into standard quadratic form:

\[
x^2 - 3x + 8 + 2 = 0
\]
This simplifies to:

\[
x^2 - 3x + 10 = 0
\]

Next, we can use the quadratic formula to find the solutions to this equation. The quadratic formula is given by:

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

For our equation \( x^2 - 3x + 10 = 0 \), the coefficients are:

- \( a = 1 \)
- \( b = -3 \)
- \( c = 10 \)

Now we can substitute these values into the quadratic formula:

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}
\]
This simplifies to:

\[
x = \frac{3 \pm \sqrt{9 - 40}}{2}
\]
\[
x = \frac{3 \pm \sqrt{-31}}{2}
\]
Since the discriminant \( -31 \) is negative, the solutions will be complex. We can express the square root of a negative number as a complex number:

\[
\sqrt{-31} = i\sqrt{31}
\]

Substituting this back into our formula gives:

\[
x = \frac{3 \pm i\sqrt{31}}{2}
\]

Thus, the conjugate pair solutions to the equation are:

\[
x = \frac{3}{2} + \frac{i\sqrt{31}}{2} \quad \text{and} \quad x = \frac{3}{2} - \frac{i\sqrt{31}}{2}
\]