Question

Algebra 2A Semester Exam
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Question
Solve the equation 2x2=6x−5
. Write your conjugate pair solution(s) as two values separated by a ±
sign.(1 point)
x=

Answer 1

To solve the equation \(2x^2 = 6x - 5\), we first rearrange it into standard quadratic form:

\[
2x^2 - 6x + 5 = 0
\]

Next, we can use the quadratic formula to find the solutions. The quadratic formula is:

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

For the equation \(2x^2 - 6x + 5 = 0\), the coefficients are:
- \(a = 2\)
- \(b = -6\)
- \(c = 5\)

Let's calculate the discriminant (\(b^2 - 4ac\)):

\[
b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4
\]

Since the discriminant is negative, we have two complex conjugate solutions. Now we can substitute into the quadratic formula:

\[
x = \frac{-(-6) \pm \sqrt{-4}}{2(2)} = \frac{6 \pm \sqrt{-4}}{4}
\]

We know that \(\sqrt{-4} = 2i\). Therefore, we have:

\[
x = \frac{6 \pm 2i}{4}
\]

This simplifies to:

\[
x = \frac{6}{4} \pm \frac{2i}{4} = \frac{3}{2} \pm \frac{i}{2}
\]

Thus, the solutions are:

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

So, the final answer is:

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