Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively
(i) `(1)/(4),(1)/(2)` (ii) `2,(1)/(3)`

Answer

To find a quadratic polynomial given the sum and product of its zeroes, we can use the standard form of a quadratic polynomial, which is given by:

\[ P(x) = x^2 - (sum \ of \ zeroes) \cdot x + (product \ of \ zeroes) \]

Let's solve the problem step by step for both cases.

### Case (i): Sum of zeroes = \( \frac{1}{4} \), Product of zeroes = \( \frac{1}{2} \)

1. **Identify the sum and product of zeroes**:
- Sum of zeroes, \( \alpha + \beta = \frac{1}{4} \)
- Product of zeroes, \( \alpha \cdot \beta = \frac{1}{2} \)

2. **Substitute into the polynomial formula**:
\[ P(x) = x^2 - (sum \ of \ zeroes) \cdot x + (product \ of \ zeroes) \]
\[ P(x) = x^2 - \left(\frac{1}{4}\right)x + \left(\frac{1}{2}\right) \]

3. **Simplify the polynomial**:
\[ P(x) = x^2 - \frac{1}{4}x + \frac{1}{2} \]

4. **To express it in a standard form, we can multiply through by 4 to eliminate the fractions**:
\[ P(x) = 4x^2 - x + 2 \]

### Case (ii): Sum of zeroes = \( 2 \), Product of zeroes = \( \frac{1}{3} \)

1. **Identify the sum and product of zeroes**:
- Sum of zeroes, \( \alpha + \beta = 2 \)
- Product of zeroes, \( \alpha \cdot \beta = \frac{1}{3} \)

2. **Substitute into the polynomial formula**:
\[ P(x) = x^2 - (sum \ of \ zeroes) \cdot x + (product \ of \ zeroes) \]
\[ P(x) = x^2 - (2)x + \left(\frac{1}{3}\right) \]

3. **Simplify the polynomial**:
\[ P(x) = x^2 - 2x + \frac{1}{3} \]

4. **To express it in a standard form, we can multiply through by 3 to eliminate the fractions**:
\[ P(x) = 3x^2 - 6x + 1 \]

### Final Answers:
- For case (i): \( P(x) = 4x^2 - x + 2 \)
- For case (ii): \( P(x) = 3x^2 - 6x + 1 \)