Q2(ii):

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (ii) $\sqrt{2}, \frac{1}{3}$

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $\sqrt{2}$

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $\frac{1}{3}$

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 ($\alpha$ and $\beta$) 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 as follows:

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

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

Step 3: Substituting the Given Values

Substitute the given values into the formula:

$p(x) = k[x^2 - (\sqrt{2})x + (\frac{1}{3})]$

Step 4: Simplifying the Expression

To express the polynomial with integer coefficients, we can choose a value for $k$ that clears the denominator. Let $k = 3$:

$p(x) = 3[x^2 - \sqrt{2}x + \frac{1}{3}]$

Distribute the $3$ across the terms inside the brackets:

$p(x) = 3 \cdot x^2 - 3 \cdot \sqrt{2}x + 3 \cdot \frac{1}{3}$

$p(x) = 3x^2 - 3\sqrt{2}x + 1$

Step 5: Verification (Optional but Recommended)

For the polynomial $3x^2 - 3\sqrt{2}x + 1$:

Sum of zeroes = $-\frac{b}{a} = -\frac{-3\sqrt{2}}{3} = \sqrt{2}$ [Matches the given sum]

Product of zeroes = $\frac{c}{a} = \frac{1}{3}$ [Matches the given product]

Final Answer: The required quadratic polynomial is $3x^2 - 3\sqrt{2}x + 1$.