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]
