Question

Find a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2.​

Answer 1

Answer :

The required quadratic polynomial is :

x² - 6x + 7

Given :

• The zeroes of a quadratic polynomial are : 3 + √2 and 3 - √2

To Find :

• The quadratic polynomial having the given zeroes

Formula to be used :

If sum and product of zeroes of a quadratic polynomial are given then the polynomial can be expressed as :

$\rm x^{2} -(Sum \: of \: the \: zeroes)x + Product \: of \: the \: zeroes$

Solution :

Given the zeroes 3 + √2 and 3 - √2

$\rm \: Sum \: of \: the \: zeroes = 3 + \sqrt{2} + 3 - \sqrt{2} \\\\ \rm \implies Sum \: of \: the \: zeroes = 6$

Again ,

$\rm Product \: of \: the \: zeroes = (3+\sqrt{2})(3-\sqrt{2}) \\\\ \rm \implies Product \: of \: the \: zeroes = 3^{2} -(\sqrt{2})^{2} \\\\ \rm \implies Product \: of \: the \: zeroes = 9 - 2 \\\\ \rm \implies Product \: of \: the \: zeroes= 7$

Thus the quadratic equation is :

$\rm \dashrightarrow x^{2} - 6x + 7$

Answer 2

$\sf\large\underline{Given:}$

• $\sf{The\: zeroes\:of\: polynomial=3+\sqrt{2}\:and\:3-\sqrt{2}}$

$\sf\large\underline{To\: Find:}$

• $\sf{The\: Quardric\: polynomial=?}$

$\sf\large\underline{Solution:}$

• At first we have to find the sum of two zeroes and the product of two zeroes

$\rm{\implies Sum\:of\:two\: zeroes}$

$\rm{\implies Sum=3+\sqrt{2}+3-\sqrt{2}}$

$\rm{\implies Sum=3+3}$

$\rm{\implies Sum=6}$

$\rm{\implies Now,\: product\:of\:two\: zeroes}$

$\rm{\implies Product=(3+\sqrt{2})*(3-\sqrt{2})}$

$\rm{\implies Product=9-3\sqrt{2}+3\sqrt{2}-(\sqrt{2})^2}$

$\rm{\implies Product=9-2}$

$\rm{\implies Product=7}$

• Now, calculate quardric polynomial

$\sf\large\underline{Formula\: used:}$

$\rm{\implies x^2-(sum\:of\: zeroes)x+(product\:of\:zero)}$

$\rm{\implies Polynomial=x^2-6x+7}$

Hence,

$\rm{\implies Quardric\: polynomial=x^2-6x+7}$