Question

From a polynomial whose zeroes are 3+2√5, 3-2√5

Answer 1

Answer:

$k{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta }$

sum of the zeroes=3+2√5+3-2√5

=6

product of the zeroes=(3+2√5)(3-2√5)=3²-(2√5)²=9-(4×5)=9-20=-11

k[x²-(6)x+(-11)]

when k=1, the required quadratic polynomial=x²-6x-11