To solve for the roots of the quadratic equation given as [tex]\(2x^2 - 3x - 6 = 0\)[/tex], we need to find the values of [tex]\(x\)[/tex] that satisfy this equation. Let's denote these roots by [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
### Step-by-Step Solution:
1. Quadratic Equation:
[tex]\[
2x^2 - 3x - 6 = 0
\][/tex]
2. Roots of the Quadratic Equation:
The roots of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] can be found using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For our equation, [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -6\)[/tex].
Substituting these values into the quadratic formula:
[tex]\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}
\][/tex]
3. Calculating the Discriminant:
[tex]\[
\text{Discriminant} = (-3)^2 - 4 \cdot 2 \cdot (-6) = 9 + 48 = 57
\][/tex]
4. Finding the Roots:
[tex]\[
x = \frac{3 \pm \sqrt{57}}{4}
\][/tex]
Hence, the roots are:
[tex]\[
x = \frac{3 - \sqrt{57}}{4} \quad \text{and} \quad x = \frac{3 + \sqrt{57}}{4}
\][/tex]
Therefore, [tex]\(\alpha = \frac{3 - \sqrt{57}}{4}\)[/tex] and [tex]\(\beta = \frac{3 + \sqrt{57}}{4}\)[/tex].
5. Formation of the New Quadratic Equation:
The problem statement asks for the equation whose roots are [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]. However, we see that the roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are exactly the roots of the given equation [tex]\(2x^2 - 3x - 6 = 0\)[/tex].
Given the provided options:
- Option A: [tex]\(2x^2 + 3x - 6 = 0\)[/tex]
- Option B: [tex]\(6x^2 - 4x + 8 = 0\)[/tex]
- Option C: [tex]\(4x^2 - 5x - 8 = 0\)[/tex]
- Option D: Duplicate of Option C with [tex]\(4x^2 - 5x - 8 = 0\)[/tex]
- Option E: None
Since none of the options match our original equation [tex]\(2x^2 - 3x - 6 = 0\)[/tex] directly, the correct answer is:
\[
\boxed{\text{E. none}}
