At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.