Question

If α and ß are the zeroes of x2+6x +9, find a polynomial whose zeroes are
(i) 2 α and 2ß
(ii) (α+l) and (ß+l)

PLS HELP GUYS

Answer 1

Answer:

alpha =_3 and beta=_3 or alpha =3 and beta=3

Step-by-step explanation:

x2+6x+9=0

x2 +2(3)(x)+(3)square =0

(x+3)square

x+3=0 and x+3=0

x=-3,-3

Alpha =-3,beta =-3

x2 - x(-alpha +(-beta) +(-alpha) (-beta) =0

x2-x(+3+3=+(+3+3)=0

x2+6x+9=0

Hence proved

Answer 2

Step-by-step explanation:

If alpha and ß are zeroes of polynomial x²+6x+9

then alpha * ß(product of zeroes) =c/a =9/1 =9=>alpha * ß=9 ---(1)

alpha * ß(sum of zeroes) =-b/a =-6/1 =-6 =>alpha + ß= -6 ----(2)

Now zeroes of new quadratic equation are -alpha, -ß.

sum of zeroes = -alpha+ -ß = -(alpha + ß)

from (2) = --6 =6 ---(3)

product of zeroes = -alpha * -ß = -alpha * -ß =alpha * ß

from (1) = 9 ----(4)

Quadratic equation if of the form x² -(sum of zeroes)x +product of zeroes

=x² -(6)x + 9

=x² -6x + 9