ECO 111: INTRODUCTION TO MATHEMATICS FOR ECONOMICS exam. Questions cover roots of quadratic equations, factorization, quadratic formula, sets, community problems, Cramer's rule, and matrices.

Step 1: Identify coefficients for the quadratic equation $5x^2 + 21x - 20 = 0$.
The standard form is $ax^2 + bx + c = 0$.
Here, $a=5$, $b=21$, $c=-20$.

a) Find $\alpha + \beta$
The sum of the roots of a quadratic equation is given by $\alpha + \beta = -\frac{b}{a}$.

$$\alpha + \beta = -\frac{21}{5}$$

The sum of the roots is $\boxed{-\frac{21}{5}}$.

b) Find $\alpha \beta$
The product of the roots of a quadratic equation is given by $\alpha \beta = \frac{c}{a}$.

$$\alpha \beta = \frac{-20}{5} = -4$$

The product of the roots is $\boxed{-4}$.

c) Find $\alpha^2 + \beta^2$
We use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.
Substitute the values from parts (a) and (b):

$$\alpha^2 + \beta^2 = \left(-\frac{21}{5}\right)^2 - 2(-4)$$ $$\alpha^2 + \beta^2 = \frac{441}{25} + 8$$ $$\alpha^2 + \beta^2 = \frac{441}{25} + \frac{8 \times 25}{25}$$ $$\alpha^2 + \beta^2 = \frac{441}{25} + \frac{200}{25}$$ $$\alpha^2 + \beta^2 = \frac{641}{25}$$

The value of $\alpha^2 + \beta^2$ is $\boxed{\frac{641}{25}}$.

d) Find $\alpha^3 + \beta^3$
We use the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.
Substitute the values from parts (a) and (b):

$$\alpha^3 + \beta^3 = \left(-\frac{21}{5}\right)^3 - 3(-4)\left(-\frac{21}{5}\right)$$ $$\alpha^3 + \beta^3 = -\frac{9261}{125} - 12\left(\frac{21}{5}\right)$$ $$\alpha^3 + \beta^3 = -\frac{9261}{125} - \frac{252}{5}$$

To combine these, find a common denominator, which is 125:

$$\alpha^3 + \beta^3 = -\frac{9261}{125} - \frac{252 \times 25}{125}$$ $$\alpha^3 + \beta^3 = -\frac{9261}{125} - \frac{6300}{125}$$ $$\alpha^3 + \beta^3 = -\frac{15561}{125}$$

The value of $\alpha^3 + \beta^3$ is $\boxed{-\frac{15561}{125}}$.

e) Find $\frac{1}{\alpha} + \frac{1}{\beta}$
Combine the fractions:

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta}$$

Substitute the values from parts (a) and (b):

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{-\frac{21}{5}}{-4}$$ $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{21}{5} \times \frac{1}{4}$$ $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{21}{20}$$

The value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $\boxed{\frac{21}{20}}$.

f) Find $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$
Combine the fractions:

$$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha \beta)^2}$$

Substitute the values from parts (b) and (c):

$$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\frac{641}{25}}{(-4)^2}$$ $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\frac{641}{25}}{16}$$ $$$$