Q2(vi):

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (vi) $4, 1$

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $4$.

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

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 standard 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 any non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values into the identity:

Sum of zeroes = $4$

Product of zeroes = $1$

$p(x) = k[x^2 - (4)x + (1)]$

Step 4: Simplifying the Expression

By choosing the simplest case where $k = 1$, we obtain the polynomial:

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

Justification:

If we verify the zeroes of $p(x) = x^2 - 4x + 1$:

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

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

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