Answer :
To find the roots of the quadratic equation [tex]\(2x^2 + 11x + 15 = 0\)[/tex], we need to identify the values of [tex]\(x\)[/tex] that satisfy the equation. Here is a detailed step-by-step process to determine these roots:
1. Write down the quadratic equation:
[tex]\[ 2x^2 + 11x + 15 = 0 \][/tex]
2. Identify the coefficients:
The general form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex].
[tex]\[ a = 2, \quad b = 11, \quad c = 15 \][/tex]
3. Use the quadratic formula to find the roots:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
4. Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[ b^2 - 4ac = 11^2 - 4(2)(15) = 121 - 120 = 1 \][/tex]
5. Calculate the roots using the quadratic formula:
[tex]\[ x = \frac{-11 \pm \sqrt{1}}{2 \cdot 2} \][/tex]
Simplify further:
[tex]\[ x = \frac{-11 \pm 1}{4} \][/tex]
6. Consider the two possible solutions for the roots:
- For the positive square root:
[tex]\[ x = \frac{-11 + 1}{4} = \frac{-10}{4} = -\frac{5}{2} \][/tex]
- For the negative square root:
[tex]\[ x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3 \][/tex]
Hence, the roots of the quadratic equation [tex]\(2x^2 + 11x + 15 = 0\)[/tex] are [tex]\(x = -\frac{5}{2}\)[/tex] and [tex]\(x = -3\)[/tex].
Given the options:
- A. [tex]\(x = -\frac{5}{2}\)[/tex]
- B. [tex]\(x = -6\)[/tex]
- C. [tex]\(x = -3\)[/tex]
- D. [tex]\(x = -5\)[/tex]
The correct answers are:
- A. [tex]\(x = -\frac{5}{2}\)[/tex]
- C. [tex]\(x = -3\)[/tex]
1. Write down the quadratic equation:
[tex]\[ 2x^2 + 11x + 15 = 0 \][/tex]
2. Identify the coefficients:
The general form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex].
[tex]\[ a = 2, \quad b = 11, \quad c = 15 \][/tex]
3. Use the quadratic formula to find the roots:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
4. Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[ b^2 - 4ac = 11^2 - 4(2)(15) = 121 - 120 = 1 \][/tex]
5. Calculate the roots using the quadratic formula:
[tex]\[ x = \frac{-11 \pm \sqrt{1}}{2 \cdot 2} \][/tex]
Simplify further:
[tex]\[ x = \frac{-11 \pm 1}{4} \][/tex]
6. Consider the two possible solutions for the roots:
- For the positive square root:
[tex]\[ x = \frac{-11 + 1}{4} = \frac{-10}{4} = -\frac{5}{2} \][/tex]
- For the negative square root:
[tex]\[ x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3 \][/tex]
Hence, the roots of the quadratic equation [tex]\(2x^2 + 11x + 15 = 0\)[/tex] are [tex]\(x = -\frac{5}{2}\)[/tex] and [tex]\(x = -3\)[/tex].
Given the options:
- A. [tex]\(x = -\frac{5}{2}\)[/tex]
- B. [tex]\(x = -6\)[/tex]
- C. [tex]\(x = -3\)[/tex]
- D. [tex]\(x = -5\)[/tex]
The correct answers are:
- A. [tex]\(x = -\frac{5}{2}\)[/tex]
- C. [tex]\(x = -3\)[/tex]
