Get the information you need from a community of experts on IDNLearn.com. Join our community to receive prompt and reliable responses to your questions from knowledgeable professionals.
Sagot :
To find the roots of the quadratic equation [tex]\(2x^2 + 11x + 15 = 0\)[/tex], we will use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the quadratic equation are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 11\)[/tex]
- [tex]\(c = 15\)[/tex]
First, we need to calculate the discriminant, [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 11^2 - 4 \cdot 2 \cdot 15 = 121 - 120 = 1 \][/tex]
Since the discriminant is a positive number, the quadratic equation has two real and distinct roots. Next, we use the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute the values of [tex]\(b\)[/tex], [tex]\( \Delta \)[/tex], and [tex]\(a\)[/tex]:
[tex]\[ x = \frac{-(11) \pm \sqrt{1}}{2 \cdot 2} = \frac{-11 \pm 1}{4} \][/tex]
This gives us two possible solutions:
1. Root 1:
[tex]\[ x = \frac{-11 + 1}{4} = \frac{-10}{4} = -2.5 \][/tex]
2. Root 2:
[tex]\[ x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3 \][/tex]
Now, let's compare these roots with the available choices:
A. [tex]\(x = -6\)[/tex]
B. [tex]\(x = -3\)[/tex]
C. [tex]\(x = -5\)[/tex]
D. [tex]\(x = -\frac{5}{2}\)[/tex]
The roots we found are [tex]\(x = -2.5\)[/tex] and [tex]\(x = -3\)[/tex], which correspond to the choices:
B. [tex]\(x = -3\)[/tex]
D. [tex]\(x = -\frac{5}{2}\)[/tex] (which is the same as [tex]\(x = -2.5\)[/tex])
Therefore, the correct answers are:
[tex]\[ \boxed{x = -3 \text{ and } x = -\frac{5}{2}} \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the quadratic equation are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 11\)[/tex]
- [tex]\(c = 15\)[/tex]
First, we need to calculate the discriminant, [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 11^2 - 4 \cdot 2 \cdot 15 = 121 - 120 = 1 \][/tex]
Since the discriminant is a positive number, the quadratic equation has two real and distinct roots. Next, we use the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute the values of [tex]\(b\)[/tex], [tex]\( \Delta \)[/tex], and [tex]\(a\)[/tex]:
[tex]\[ x = \frac{-(11) \pm \sqrt{1}}{2 \cdot 2} = \frac{-11 \pm 1}{4} \][/tex]
This gives us two possible solutions:
1. Root 1:
[tex]\[ x = \frac{-11 + 1}{4} = \frac{-10}{4} = -2.5 \][/tex]
2. Root 2:
[tex]\[ x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3 \][/tex]
Now, let's compare these roots with the available choices:
A. [tex]\(x = -6\)[/tex]
B. [tex]\(x = -3\)[/tex]
C. [tex]\(x = -5\)[/tex]
D. [tex]\(x = -\frac{5}{2}\)[/tex]
The roots we found are [tex]\(x = -2.5\)[/tex] and [tex]\(x = -3\)[/tex], which correspond to the choices:
B. [tex]\(x = -3\)[/tex]
D. [tex]\(x = -\frac{5}{2}\)[/tex] (which is the same as [tex]\(x = -2.5\)[/tex])
Therefore, the correct answers are:
[tex]\[ \boxed{x = -3 \text{ and } x = -\frac{5}{2}} \][/tex]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.