Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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}}
### 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}}
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.