If `alpha` and `beta` are the zeros of the quadratic polynomial `f(x) = 3x^(2) – 5x - 2`, then evaluate
(i) `alpha^(2) + beta^(2)`, (ii) `alpha^(3) + beta^(3)`, (iii) `alpha^(2)/beta + beta^(2)/alpha`

Answer

To solve the problem, we need to evaluate the expressions given the zeros of the quadratic polynomial \( f(x) = 3x^2 - 5x - 2 \). Let's go through the steps one by one.

### Step 1: Identify the coefficients
The given polynomial is \( f(x) = 3x^2 - 5x - 2 \). Comparing it with the standard form \( ax^2 + bx + c \), we get:
- \( a = 3 \)
- \( b = -5 \)
- \( c = -2 \)

### Step 2: Find the zeros (roots) of the polynomial
The zeros of the polynomial can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substitute the values of \( a \), \( b \), and \( c \):
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \]
\[ x = \frac{5 \pm \sqrt{25 + 24}}{6} \]
\[ x = \frac{5 \pm \sqrt{49}}{6} \]
\[ x = \frac{5 \pm 7}{6} \]

Thus, the roots are:
\[ x = \frac{5 + 7}{6} = 2 \]
\[ x = \frac{5 - 7}{6} = -\frac{1}{3} \]

So, \( \alpha = 2 \) and \( \beta = -\frac{1}{3} \).

### Step 3: Evaluate \( \alpha^2 + \beta^2 \)
We know:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

First, find \( \alpha + \beta \) and \( \alpha\beta \):
\[ \alpha + \beta = 2 + \left(-\frac{1}{3}\right) = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \]
\[ \alpha\beta = 2 \cdot \left(-\frac{1}{3}\right) = -\frac{2}{3} \]

Now, calculate \( \alpha^2 + \beta^2 \):
\[ \alpha^2 + \beta^2 = \left(\frac{5}{3}\right)^2 - 2 \cdot \left(-\frac{2}{3}\right) \]
\[ \alpha^2 + \beta^2 = \frac{25}{9} + \frac{4}{3} \]
\[ \alpha^2 + \beta^2 = \frac{25}{9} + \frac{12}{9} \]
\[ \alpha^2 + \beta^2 = \frac{37}{9} \]

### Step 4: Evaluate \( \alpha^3 + \beta^3 \)
We know:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) \]

Using the values calculated:
\[ \alpha^3 + \beta^3 = \left(\frac{5}{3}\right) \left(\frac{37}{9} + \frac{2}{3}\right) \]
\[ \alpha^3 + \beta^3 = \left(\frac{5}{3}\right) \left(\frac{37}{9} + \frac{6}{9}\right) \]
\[ \alpha^3 + \beta^3 = \left(\frac{5}{3}\right) \left(\frac{43}{9}\right) \]
\[ \alpha^3 + \beta^3 = \frac{215}{27} \]

### Step 5: Evaluate \( \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} \)
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{2^2}{-\frac{1}{3}} + \frac{\left(-\frac{1}{3}\right)^2}{2} \]
\[ = \frac{4}{-\frac{1}{3}} + \frac{\frac{1}{9}}{2} \]
\[ = 4 \cdot \left(-3\right) + \frac{1}{18} \]
\[ = -12 + \frac{1}{18} \]
\[ = -12 + \frac{1}{18} \]
\[ = \frac{-216}{18} + \frac{1}{18} \]
\[ = \frac{-215}{18} \]

### Summary of Results
(i) \( \alpha^2 + \beta^2 = \frac{37}{9} \)

(ii) \( \alpha^3 + \beta^3 = \frac{215}{27} \)

(iii) \( \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{-215}{18} \)