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Q2(iii):
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (iii) $0, \sqrt{5}$

Solution :

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $0$.

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $\sqrt{5}$.

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 its zeroes ($\alpha$ and $\beta$) and its coefficients is given by:

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 standard formula:

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

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

Step 3: Substituting the Given Values

Substitute the given values into the formula:

Sum of zeroes = $0$

Product of zeroes = $\sqrt{5}$

$p(x) = k[x^2 - (0)x + (\sqrt{5})]$

Step 4: Simplifying the Expression

Perform the arithmetic operations inside the brackets:

$p(x) = k[x^2 - 0 + \sqrt{5}]$

$p(x) = k[x^2 + \sqrt{5}]$

Step 5: Determining the Polynomial

By choosing the simplest constant $k = 1$, we obtain the quadratic polynomial:

$p(x) = x^2 + \sqrt{5}$

Verification:

For $p(x) = x^2 + 0x + \sqrt{5}$:

Sum of zeroes = $-\frac{b}{a} = -\frac{0}{1} = 0$ (Matches given)

Product of zeroes = $\frac{c}{a} = \frac{\sqrt{5}}{1} = \sqrt{5}$ (Matches given)

Final Answer: The required quadratic polynomial is $x^2 + \sqrt{5}$.


More Questions from Class 10 Mathematics Polynomials EXERCISE 2.2


CBSE Solutions for Class 10 Mathematics Polynomials


Chapters in CBSE - Class 10 Mathematics


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