Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve the problem, we need to utilize the sum and product of the roots of the given polynomial [tex]\(2x^2 - 4x + 5\)[/tex]. Here, [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the zeroes of the polynomial.
Sum and Product of roots:
For a polynomial [tex]\(ax^2 + bx + c = 0\)[/tex]:
- The sum of the roots, [tex]\(\alpha + \beta\)[/tex], is given by [tex]\(-\frac{b}{a}\)[/tex].
- The product of the roots, [tex]\(\alpha \beta\)[/tex], is given by [tex]\(\frac{c}{a}\)[/tex].
Given the polynomial [tex]\(2x^2 - 4x + 5\)[/tex]:
- [tex]\(a = 2\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 5\)[/tex].
Step-by-step solutions:
### Part (a): [tex]\(\alpha^2 + \beta^2\)[/tex]
We use the identity:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \][/tex]
1. Calculate the sum of the roots:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{-4}{2} = 2 \][/tex]
2. Calculate the product of the roots:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{5}{2} \][/tex]
3. Substitute the values into the identity:
[tex]\[ \alpha^2 + \beta^2 = (2)^2 - 2 \left(\frac{5}{2}\right) \][/tex]
[tex]\[ \alpha^2 + \beta^2 = 4 - 2 \times \frac{5}{2} \][/tex]
[tex]\[ \alpha^2 + \beta^2 = 4 - 5 \][/tex]
[tex]\[ \alpha^2 + \beta^2 = -1 \][/tex]
Thus, [tex]\(\alpha^2 + \beta^2 = -1\)[/tex].
### Part (b): [tex]\(\frac{1}{\alpha} + \frac{1}{\beta}\)[/tex]
We use the identity:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \][/tex]
1. Substitute the values of the sum and product of the roots:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{2}{\frac{5}{2}} \][/tex]
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = 2 \times \frac{2}{5} \][/tex]
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{4}{5} \][/tex]
Thus, [tex]\(\frac{1}{\alpha} + \frac{1}{\beta} = 0.8\)[/tex].
### Part (c): [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]
We use the identity:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} \][/tex]
1. Substitute the values of [tex]\(\alpha^2 + \beta^2\)[/tex] and [tex]\((\alpha \beta)^2\)[/tex]:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{-1}{\left(\frac{5}{2}\right)^2} \][/tex]
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{-1}{\frac{25}{4}} \][/tex]
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = -1 \times \frac{4}{25} \][/tex]
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = -\frac{4}{25} \][/tex]
Thus, [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = -0.16\)[/tex].
Final Results:
a) [tex]\(\alpha^2 + \beta^2 = -1\)[/tex]
b) [tex]\(\frac{1}{\alpha} + \frac{1}{\beta} = 0.8\)[/tex]
c) [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = -0.16\)[/tex]
Sum and Product of roots:
For a polynomial [tex]\(ax^2 + bx + c = 0\)[/tex]:
- The sum of the roots, [tex]\(\alpha + \beta\)[/tex], is given by [tex]\(-\frac{b}{a}\)[/tex].
- The product of the roots, [tex]\(\alpha \beta\)[/tex], is given by [tex]\(\frac{c}{a}\)[/tex].
Given the polynomial [tex]\(2x^2 - 4x + 5\)[/tex]:
- [tex]\(a = 2\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 5\)[/tex].
Step-by-step solutions:
### Part (a): [tex]\(\alpha^2 + \beta^2\)[/tex]
We use the identity:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \][/tex]
1. Calculate the sum of the roots:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{-4}{2} = 2 \][/tex]
2. Calculate the product of the roots:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{5}{2} \][/tex]
3. Substitute the values into the identity:
[tex]\[ \alpha^2 + \beta^2 = (2)^2 - 2 \left(\frac{5}{2}\right) \][/tex]
[tex]\[ \alpha^2 + \beta^2 = 4 - 2 \times \frac{5}{2} \][/tex]
[tex]\[ \alpha^2 + \beta^2 = 4 - 5 \][/tex]
[tex]\[ \alpha^2 + \beta^2 = -1 \][/tex]
Thus, [tex]\(\alpha^2 + \beta^2 = -1\)[/tex].
### Part (b): [tex]\(\frac{1}{\alpha} + \frac{1}{\beta}\)[/tex]
We use the identity:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \][/tex]
1. Substitute the values of the sum and product of the roots:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{2}{\frac{5}{2}} \][/tex]
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = 2 \times \frac{2}{5} \][/tex]
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{4}{5} \][/tex]
Thus, [tex]\(\frac{1}{\alpha} + \frac{1}{\beta} = 0.8\)[/tex].
### Part (c): [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]
We use the identity:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} \][/tex]
1. Substitute the values of [tex]\(\alpha^2 + \beta^2\)[/tex] and [tex]\((\alpha \beta)^2\)[/tex]:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{-1}{\left(\frac{5}{2}\right)^2} \][/tex]
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{-1}{\frac{25}{4}} \][/tex]
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = -1 \times \frac{4}{25} \][/tex]
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = -\frac{4}{25} \][/tex]
Thus, [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = -0.16\)[/tex].
Final Results:
a) [tex]\(\alpha^2 + \beta^2 = -1\)[/tex]
b) [tex]\(\frac{1}{\alpha} + \frac{1}{\beta} = 0.8\)[/tex]
c) [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = -0.16\)[/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
