If `alpha` and `beta` are the roots of the quation `ax^(2)+bx+c=0`, then find the equation whose roots are given by
(i) `alpha+1/(beta), beta+1/(alpha)`
(ii) `alpha^(2)+2,beta^(2)+2`

Answer

To solve the given problem, we will find the equations whose roots are specified in the question. We will tackle each part separately.

### Part (i): Roots are \( \alpha + \frac{1}{\beta} \) and \( \beta + \frac{1}{\alpha} \)

1. **Identify the roots**: The roots are given as \( r_1 = \alpha + \frac{1}{\beta} \) and \( r_2 = \beta + \frac{1}{\alpha} \).

2. **Simplify the roots**:
- We can express \( r_1 \) and \( r_2 \) in a common form:
\[ r_1 = \alpha + \frac{1}{\beta} = \alpha + \frac{\alpha}{\alpha \beta} = \frac{\alpha^2 + 1}{\beta} \]
\[ r_2 = \beta + \frac{1}{\alpha} = \beta + \frac{\beta}{\alpha \beta} = \frac{\beta^2 + 1}{\alpha} \]

3. **Use Vieta's formulas**:
- From the original quadratic equation \( ax^2 + bx + c = 0 \), we know:
- \( \alpha + \beta = -\frac{b}{a} \)
- \( \alpha \beta = \frac{c}{a} \)

4. **Calculate the sum of the roots**:
\[ r_1 + r_2 = \left( \alpha + \frac{1}{\beta} \right) + \left( \beta + \frac{1}{\alpha} \right) = (\alpha + \beta) + \left( \frac{1}{\beta} + \frac{1}{\alpha} \right) \]
\[ = -\frac{b}{a} + \frac{\alpha + \beta}{\alpha \beta} = -\frac{b}{a} + \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{a} - \frac{b}{c} \]
\[ = -\frac{b(c + a)}{ac} \]

5. **Calculate the product of the roots**:
\[ r_1 \cdot r_2 = \left( \alpha + \frac{1}{\beta} \right) \left( \beta + \frac{1}{\alpha} \right) = \alpha \beta + \frac{\alpha}{\alpha} + \frac{\beta}{\beta} + \frac{1}{\alpha \beta} \]
\[ = \alpha \beta + 1 + 1 + \frac{1}{\alpha \beta} = \frac{c}{a} + 2 + \frac{a}{c} \]

6. **Form the new quadratic equation**:
- The new quadratic equation can be formed using the sum and product of the roots:
\[ x^2 - \left( -\frac{b(c + a)}{ac} \right)x + \left( \frac{c}{a} + 2 + \frac{a}{c} \right) = 0 \]

### Part (ii): Roots are \( \alpha^2 + 2 \) and \( \beta^2 + 2 \)

1. **Identify the roots**: The roots are \( r_1 = \alpha^2 + 2 \) and \( r_2 = \beta^2 + 2 \).

2. **Calculate the sum of the roots**:
\[ r_1 + r_2 = (\alpha^2 + 2) + (\beta^2 + 2) = \alpha^2 + \beta^2 + 4 \]
- Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\[ = \left( -\frac{b}{a} \right)^2 - 2 \cdot \frac{c}{a} + 4 = \frac{b^2}{a^2} - \frac{2c}{a} + 4 \]

3. **Calculate the product of the roots**:
\[ r_1 \cdot r_2 = (\alpha^2 + 2)(\beta^2 + 2) = \alpha^2\beta^2 + 2\alpha^2 + 2\beta^2 + 4 \]
- Using \( \alpha^2\beta^2 = (\alpha\beta)^2 = \left( \frac{c}{a} \right)^2 \):
\[ = \left( \frac{c}{a} \right)^2 + 2(\alpha^2 + \beta^2) + 4 = \left( \frac{c}{a} \right)^2 + 2\left( \frac{b^2}{a^2} - \frac{2c}{a} \right) + 4 \]

4. **Form the new quadratic equation**:
- The new quadratic equation can be formed using the sum and product of the roots:
\[ x^2 - \left( \frac{b^2}{a^2} - \frac{2c}{a} + 4 \right)x + \left( \left( \frac{c}{a} \right)^2 + 2\left( \frac{b^2}{a^2} - \frac{2c}{a} \right) + 4 \right) = 0 \]