The given expression is:

\[
2x^2 - 6x + 8
\]

We can analyze it step by step:

### 1. Simplification:
The expression is already simplified, so there's no need to perform additional simplifications.

### 2. Factoring (if possible):
To factor this quadratic expression, we can first check if there's a common factor among all terms. In this case, we can factor out a **2**:

\[
2(x^2 - 3x + 4)
\]

Now we try to factor \( x^2 - 3x + 4 \) further. However, the discriminant (\( b^2 - 4ac \)) of this quadratic is:

\[
\Delta = (-3)^2 - 4(1)(4) = 9 - 16 = -7
\]

Since the discriminant is negative, this quadratic expression has no real roots, meaning it cannot be factored further over the real numbers. It only has complex solutions.

### 3. Solving (if required):
If you wanted to solve \( 2x^2 - 6x + 8 = 0 \), you would use the quadratic formula:

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

For \( x^2 - 3x + 4 = 0 \), apply the quadratic formula:

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(4)}}{2(1)} = \frac{3 \pm \sqrt{9 - 16}}{2} = \frac{3 \pm \sqrt{-7}}{2}
\]

Since \( \sqrt{-7} = i\sqrt{7} \), the solutions are:

\[
x = \frac{3 \pm i\sqrt{7}}{2}
\]

Thus, the two complex solutions are:

\[
x = \frac{3 + i\sqrt{7}}{2} \quad \text{or} \quad x = \frac{3 - i\sqrt{7}}{2}
\]

Let me know if you'd like more details or further explanation!

---

Here are five related questions:
1. How do you calculate the discriminant of a quadratic equation?
2. What are the steps to solve a quadratic equation using the quadratic formula?
3. How do complex solutions arise from negative discriminants?
4. Can you factor quadratics with complex roots?
5. What is the significance of factoring in solving quadratic equations?

**Tip**: When the discriminant is negative, the quadratic equation has no real solutions but two complex conjugate solutions!

Explore how to solve the quadratic equation 2x^2 - 6x + 8, factoring and applying the quadratic formula to find complex solutions. This problem highlights key algebraic concepts, including discriminants and complex numbers. Suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation