Sure! Let's solve the problem step-by-step.
Given the polynomial [tex]\(3x^2 + 2x - 6\)[/tex], we need to find the following:
(a) [tex]\(\alpha + \beta\)[/tex]
(b) [tex]\(\frac{1}{\alpha} + \frac{1}{\beta}\)[/tex]
(c) [tex]\(\alpha + \beta + \alpha\beta\)[/tex]
### Step 1: Finding [tex]\(\alpha + \beta\)[/tex]
For a quadratic polynomial [tex]\(ax^2 + bx + c\)[/tex], the sum of the zeroes, [tex]\(\alpha + \beta\)[/tex], is given by:
[tex]\[
\alpha + \beta = -\frac{b}{a}
\][/tex]
Here, [tex]\(a = 3\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -6\)[/tex]. Plugging in the values:
[tex]\[
\alpha + \beta = -\frac{2}{3}
\][/tex]
Thus, [tex]\(\alpha + \beta = -0.6666666666666666\)[/tex].
### Step 2: Finding [tex]\(\frac{1}{\alpha} + \frac{1}{\beta}\)[/tex]
The sum of the reciprocals of the zeroes is given by:
[tex]\[
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}
\][/tex]
We already know [tex]\(\alpha + \beta = -0.6666666666666666\)[/tex]. Next, we need to find [tex]\(\alpha\beta\)[/tex].
### Step 3: Finding [tex]\(\alpha\beta\)[/tex]
For a quadratic polynomial [tex]\(ax^2 + bx + c\)[/tex], the product of the zeroes, [tex]\(\alpha\beta\)[/tex], is given by:
[tex]\[
\alpha\beta = \frac{c}{a}
\][/tex]
Plugging in the values:
[tex]\[
\alpha\beta = \frac{-6}{3} = -2
\][/tex]
Thus, [tex]\(\alpha\beta = -2\)[/tex].
### Step 4: Computing [tex]\(\frac{1}{\alpha} + \frac{1}{\beta}\)[/tex]
Using the values obtained:
[tex]\[
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-0.6666666666666666}{-2} = 0.3333333333333333
\][/tex]
### Step 5: Finding [tex]\(\alpha + \beta + \alpha\beta\)[/tex]
This is a simple addition of the already known values of [tex]\(\alpha + \beta\)[/tex] and [tex]\(\alpha\beta\)[/tex]:
[tex]\[
\alpha + \beta + \alpha\beta = (-0.6666666666666666) + (-2) = -2.6666666666666665
\][/tex]
### Conclusion
(a) [tex]\(\alpha + \beta = -0.6666666666666666\)[/tex]
(b) [tex]\(\frac{1}{\alpha} + \frac{1}{\beta} = 0.3333333333333333\)[/tex]
(c) [tex]\(\alpha + \beta + \alpha\beta = -2.6666666666666665\)[/tex]
