Question

find the quadratic polynomial whose zeroes are √3+5 and √3-5​

Answer 1

Hey dude ....

hre is ur answer....!!!

Given ->

3 - √5 and 3 + √5 are zeros of a polynomial

let p(x) be required polynomial

====> x - (3 + √5) and x - (3 - √5) are factors of p(x)

====> [x - (3 + √5)] [x - (3 - √5) is required polynomial

= [x - 3 - √5] [x - 3 + √5]

= x - 3x + √5x - 3x + 9 - 3 √5 - √5x +

3 √5 - 5

= x² - 6x + 4

or...

zeroes = 3+root5 and 3-root5

product of zeroes= (3+root5)(3-root5)

= 9-5

=4

sum of zeroes=3+root5 + 3-root5

=6

we know p(x) = k(x2^ - sum of zeroes (x) + product of zeroes

= k(x2^ -6x +4)

hope it will.help u..

thanks. !

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

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