Question

find quadratic polynomial if zeroes are
-√5/4 , √3/3

Answer 1

We know that a quadratic equation is given by k [ x² - (sum of the zeroes)x + (product of the zeroes) ], where k = any real number.
Here, the zeroes are -√5/4 and √3/3
Sum of the zeroes = -√5/4 + √3/3
=( 4√3-3√5)/12
Product of the zeroes = -√5/4 × √3/3
= -√15/12

So, the quadratic equation is k [ x² - (4√3-3√5/12)× - √15/12]

Answer 2

The quadratic polynomial whose zeroes are,

$5 \sqrt{3} ,5 - \sqrt{3}$

$\alpha , \beta \: is \: f(x) = k[ {x}^{2} - ( \alpha + \beta )x + \alpha \times \beta ]$

where k is any non-zero real no.

THE QUADRATIC POLY POLYNOMIAL WHOSE ZEROES ARE

$5 \sqrt{3} ,5 - \sqrt{3}$

$f(x) = k[ {x}^{2} - ( \alpha + \beta )x + \alpha \times \beta ]$

$f(x) = k[ {x}^{2} - ( 5 \cancel{ + \sqrt{3}} + 5 \cancel{ - \sqrt{3}} )x + (5 + \sqrt{3} ) (5 - \sqrt{3} ) ]$

$f(x) = k[ {x}^{2} -10x + ( {5)}^{2} - ({ \sqrt{3} )}^{2} ]$

$f(x) = k[ {x}^{2} -10x + (25 - 3)]$

$f(x) = k[ {x}^{2} -10x + 22]$

so, the QUADRATIC polynomial is

$f(x) = k[ {x}^{2} -10x + 22]$