Q2(v):

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (v) $-\frac{1}{4}, \frac{1}{4}$

Given: The sum of the zeroes ($\alpha + \beta$) is $-\frac{1}{4}$ and the product of the zeroes ($\alpha \cdot \beta$) is $\frac{1}{4}$.

To Find: A quadratic polynomial $p(x)$ that satisfies the given conditions.

Step 1: Understanding the Relationship between Zeroes and Coefficients

For any quadratic polynomial of the form $ax^2 + bx + c$, where $a \neq 0$, the relationship between the zeroes ($\alpha, \beta$) and the coefficients is given by the following formulas:

Sum of zeroes: $\alpha + \beta = -\frac{b}{a}$

Product of zeroes: $\alpha \cdot \beta = \frac{c}{a}$

Step 2: Formulating the General Quadratic Polynomial

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes using the following identity:

$p(x) = k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})]$

where $k$ is a non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values $\alpha + \beta = -\frac{1}{4}$ and $\alpha \cdot \beta = \frac{1}{4}$ into the identity:

$p(x) = k\left[x^2 - \left(-\frac{1}{4}\right)x + \left(\frac{1}{4}\right)\right]$

[Simplifying the signs: $-(- \frac{1}{4}) = +\frac{1}{4}$]

$p(x) = k\left[x^2 + \frac{1}{4}x + \frac{1}{4}\right]$

Step 4: Simplifying the Expression

To obtain a polynomial with integer coefficients, we can choose $k = 4$ (or any multiple of the denominator):

$p(x) = 4\left[x^2 + \frac{1}{4}x + \frac{1}{4}\right]$

Distributing the $4$ across the terms inside the brackets:

$p(x) = 4(x^2) + 4\left(\frac{1}{4}x\right) + 4\left(\frac{1}{4}\right)$

$p(x) = 4x^2 + x + 1$

Final Answer: The required quadratic polynomial is $4x^2 + x + 1$.