Question

Solve the equation x2−5x+10=0%0D%0A𝑥%0D%0A2%0D%0A−%0D%0A5%0D%0A𝑥%0D%0A+%0D%0A10%0D%0A=%0D%0A0%0D%0A. Write your conjugate pair solution(s) as two values separated by a ±%0D%0A±%0D%0A sign.

Answer 1

To solve the quadratic equation \(x^2 - 5x + 10 = 0\), we can use the quadratic formula:

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

In this case, the coefficients are:
- \(a = 1\)
- \(b = -5\)
- \(c = 10\)

Now, we will calculate the discriminant:

\[
b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot 10 = 25 - 40 = -15
\]

Since the discriminant is negative, this means the solutions will be complex (conjugate pairs). Now we can substitute these values into the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{-15}}{2 \cdot 1}
\]
\[
x = \frac{5 \pm \sqrt{-15}}{2}
\]

This can be simplified further. Since \(\sqrt{-1} = i\), we have:

\[
\sqrt{-15} = i\sqrt{15}
\]

Substituting this back into our formula gives:

\[
x = \frac{5 \pm i\sqrt{15}}{2}
\]

Thus, the solutions can be expressed as:

\[
x = \frac{5}{2} \pm \frac{\sqrt{15}}{2} i
\]

So the conjugate pair solution is:

\[
\frac{5}{2} \pm \frac{\sqrt{15}}{2} i
\]