Question

Answer these.... :-) 100 points class 10

Answer 1

13. p(x) = ax²+bx +c
sum of zeroes = -b/a
product of zeroes = c/a
let the zeroes be p and 2p as stated in the question that one zero of the polynomial is double the other.
so
sum:
p+2p = -b/a
3p = -b/a
p = -b/3a
product:
p×2p = c/a
2p²= c/a
2(-b/3a)² = c/a
2b²/9a² = c/a
2b² = 9ac
Hence proved.

12. f(x) = ax²+bx +c
let the zeroes of the polynomial be p and q.
p+q = -b/a
pq = -c/a
The polynomial formed is x²-(p+q)x +pq
The polynomial formed with zeroes being the reciprocal of the zeroes of the above polynomials.
x²-(1/p+1/q)x+1/pq
x² - (p+q)x/pq +1/pq
x² -(-b/a)x/(-c/a)+1/(-c/a)
x²-bx/c - a/c
= cx²-bx-a

Answer 2

hey mate here is ur ans