Question

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

Answer 1

Hii friend,

(✓3+✓5) and (✓5-✓3) are the zeros of the polynomial.

Let Alpha = (✓3+✓5) and Beta = (✓5-✓3)

Sum of zeros = (Alpha+Beta) = (✓3+✓5+✓5-✓3) = 2✓5.


Product of zeros = (Alpha × Beta) = (✓3+✓5)(✓5-✓3) = ✓8 -3 + 5 -✓8 = 2.

Therefore,

The required polynomial = X²-(Alpha + Beta)X + Alpha × Beta.

=> X²-(2✓5)X+2

=> X²-2✓5X +2..


Hence,

The polynomial = X²-2✓5X+2


HOPE IT WILL HELP YOU..... :-)

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

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