Q2(i):

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

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $\frac{1}{4}$

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $-1$

To Find:

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

Step 1: Understanding the Relationship between Zeroes and Coefficients

A quadratic polynomial in the variable $x$ is generally expressed in the form $p(x) = ax^2 + bx + c$, where $a \neq 0$.

According to the relationship between the zeroes and coefficients of a quadratic polynomial:

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

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

Step 2: Formulating the Polynomial

Any quadratic polynomial can be expressed in terms of its sum and product of 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 into the identity:

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

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

Step 4: Simplifying the Expression

To obtain a polynomial with integer coefficients, we can choose a suitable value for the constant $k$. Let $k = 4$ (the denominator of the fraction):

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

Distribute the $4$ across the terms inside the bracket:

$p(x) = 4(x^2) - 4(\frac{1}{4}x) - 4(1)$

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

$p(x) = 4x^2 - x - 4$

Step 5: Verification (Optional but Recommended)

For $p(x) = 4x^2 - x - 4$:

Sum of zeroes = $-\frac{b}{a} = -(\frac{-1}{4}) = \frac{1}{4}$ [Matches the given sum]

Product of zeroes = $\frac{c}{a} = \frac{-4}{4} = -1$ [Matches the given product]

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