If `alpha, beta` are the roots of the quadratic equation `cx^2- 2bx + 4a = 0` then find the quadratic equation whose roots are:
(i) `alpha/2, beta/2`, (ii) `alpha^(2), beta^(2)`, (iii) `alpha + 1, beta+1`, (iv) `(1+alpha)/(1-alpha) (1+beta)/(1-beta)`, (v) `alpha/beta, beta/alpha`

Answer

To find the quadratic equations with the specified roots based on the quadratic equation \( cx^2 - 2bx + 4a = 0 \) with roots \( \alpha \) and \( \beta \), we will use the relationships between the roots and coefficients of the quadratic equation.

### Step-by-Step Solution:

1. **Identify the Sum and Product of Roots**:
- From the given equation \( cx^2 - 2bx + 4a = 0 \), we can express the sum and product of the roots:
- Sum of roots \( \alpha + \beta = \frac{2b}{c} \)
- Product of roots \( \alpha \beta = \frac{4a}{c} \)

2. **(i) Roots: \( \frac{\alpha}{2}, \frac{\beta}{2} \)**:
- **Sum**:
\[ \frac{\alpha}{2} + \frac{\beta}{2} = \frac{\alpha + \beta}{2} = \frac{2b}{2c} = \frac{b}{c} \]
- **Product**:
\[ \frac{\alpha}{2} \cdot \frac{\beta}{2} = \frac{\alpha \beta}{4} = \frac{4a}{4c} = \frac{a}{c} \]
- **Quadratic Equation**:
\[ x^2 - \left(\frac{b}{c}\right)x + \frac{a}{c} = 0 \implies cx^2 - bx + a = 0 \]

3. **(ii) Roots: \( \alpha^2, \beta^2 \)**:
- **Sum**:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{2b}{c}\right)^2 - 2\left(\frac{4a}{c}\right) = \frac{4b^2}{c^2} - \frac{8a}{c} \]
- **Product**:
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{4a}{c}\right)^2 = \frac{16a^2}{c^2} \]
- **Quadratic Equation**:
\[ x^2 - \left(\frac{4b^2 - 8ac}{c^2}\right)x + \frac{16a^2}{c^2} = 0 \implies c^2x^2 - (4b^2 - 8ac)x + 16a^2 = 0 \]

4. **(iii) Roots: \( \alpha + 1, \beta + 1 \)**:
- **Sum**:
\[ (\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{2b}{c} + 2 = \frac{2b + 2c}{c} \]
- **Product**:
\[ (\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = \frac{4a}{c} + \frac{2b}{c} + 1 = \frac{4a + 2b + c}{c} \]
- **Quadratic Equation**:
\[ x^2 - \left(\frac{2b + 2c}{c}\right)x + \frac{4a + 2b + c}{c} = 0 \implies cx^2 - (2b + 2c)x + (4a + 2b + c) = 0 \]

5. **(iv) Roots: \( \frac{1 + \alpha}{1 - \alpha}, \frac{1 + \beta}{1 - \beta} \)**:
- **Sum**:
\[ \frac{1 + \alpha}{1 - \alpha} + \frac{1 + \beta}{1 - \beta} = \frac{(1 + \alpha)(1 - \beta) + (1 + \beta)(1 - \alpha)}{(1 - \alpha)(1 - \beta)} = \frac{2 - (\alpha + \beta) + \alpha \beta}{1 - (\alpha + \beta) + \alpha \beta} \]
- **Product**:
\[ \frac{(1 + \alpha)(1 + \beta)}{(1 - \alpha)(1 - \beta)} = \frac{1 + \alpha + \beta + \alpha\beta}{1 - (\alpha + \beta) + \alpha\beta} \]
- **Quadratic Equation**:
\[ \text{(This step requires further simplification)} \]

6. **(v) Roots: \( \frac{\alpha}{\beta}, \frac{\beta}{\alpha} \)**:
- **Sum**:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{\left(\frac{2b}{c}\right)^2 - 2\left(\frac{4a}{c}\right)}{\frac{4a}{c}} = \frac{4b^2 - 8ac}{4a} \]
- **Product**:
\[ \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1 \]
- **Quadratic Equation**:
\[ x^2 - \left(\frac{4b^2 - 8ac}{4a}\right)x + 1 = 0 \implies 4ax^2 - (4b^2 - 8ac)x + 4a = 0 \]