Question

13. Find a quadratic polynomial whose sum and product of zeroes are V2 and 3 respectively.​

Answer 1

Correct Question:-

Find a Quadratic Polynomial whose sum & product of zeroes are $\sf\sqrt{2}$ and 3 respectively.

AnswEr:-

Polynomial is $\sf\; x^2 -\sqrt{2} + 3$

Given:-

Sum of Zeroes = $\sf\sqrt{2}$

Product of zeroes = 3

By using Formula :-

$:\implies\large\boxed{\sf{\red {x^2 - (\alpha\;+\;\beta) + (\alpha\;\beta)}}}$

$\rule{150}2$

Sum of Zeroes = $\sf\;(\alpha \; + \; \beta)$

And, Product of zeroes = $\sf\; (\alpha\; \beta)$

Sum of Zeroes = $\sf\sqrt{2}$

Product of zeroes = 3

$:\implies\sf\; x^2 - \sqrt{2} + 3$

$:\implies\large\boxed{\sf{\blue{x^2 - \sqrt{2} + 3}}}$

$\rule{150}3$

Answer 2

Step-by-step explanation:

Given :

$\sqrt{2} \: \: and \: 3 \: are \: the \: sum \: and \: product \\ \: of \: a \: quadratic \: polynomial.$

To Find:

A quadratic polynomial whose sum and product of zeroes are V2 and 3 respectively.

Solution :

As we know that a quadratic polynomial has the degree 2 and is in the form ${x} ^{2}-(\alpha+\beta) +\alpha\times\beta.$

$And \: if \: the \: roots \: of \\ \: the \: polynomial \: are \: \\ \alpha \: and \: \beta \: then$

$Sum \: of \: the \: zeros \: of \: the \: polynomial \\ \: =( \alpha + \beta ) = \sqrt{2}$

$product \: of \: the \: polynomial \: \\ = ( \alpha \times \beta ) = 3$

$So \: the \: required \: polynomial \\ \: is \: \: \: {x}^{2} - \sqrt{2} x + 3.$