Question

find a quadratic polynomial whose zeroes are 3+√2 and 3-√2

Answer 1

$\bf{Answer\colon}$

$Zeros\:of\:polynomial\\=3+\sqrt{2}\:and\:3-\sqrt{2}\\Sum\:of\:zeros\\=3+\cancel{\sqrt{2}}+3-\cancel{\sqrt{2}}\\=6\\Product\:of\:zeros\\=(3+\sqrt{2})(3-\sqrt{2})\\={(3)}^{2}-{(\sqrt{2})}^{2}\\=9-2\\=7\\Now,\\Quadratic\:polynomial\\=>{x}^{2}-Sx+P=0\\=>{x}^{2}-6x+7=0$

$Therefore,required\:polynomial\\=>{x}^{2}-6x+7=0$

Answer 2

Answer:

Zeros : 3 + √2 ; 3 - √2

General Form of a Quadratic Equation: ax² + bx + c

Here b is the sum of roots, c is the product of roots.

Sum of roots is taken as negative value.

That is,

=> x² - ( sum of roots ) x + product of roots.

According to your question,

Sum of roots = 3 + √2 + 3 - √2

=> Sum of roots = 6

Product of roots = ( 3 + √2 ) ( 3 - √2 )

=> Product of roots = 3² - ( √2 )²

=> Product of Roots = 9 - 2 = 7

Hence on substituting the values we get,

=> x² - ( 6 ) x + 7

=> x² - 6x + 7

This is the required equation.