Question

If zeros of the polynomial f (x) = x³ – 3px² + qx – r are in A.P., then
A. 2p³ = pq – r
B. 2p³ = pq + r
C. p³ = pq – r
D. None of these

Answer 1

If the zeros of the polynomial are in A.P.

Let the roots be a,b,c

We know,

1.a+b+c = 3p

2.a.b.c= r

3. ab+bc+ca = q

•As the roots are in A.P we can assume roots as a-k , a , a+k (where k is the common difference)

i.e a-k +a + a+k = 3p

3a= 3p which implies a=p

•And (a-k)(a)(a+k) = r

•We get a^2 - k^2 =r/3p

•And a-k(a) + a(a+k) + (a-k)(a+k) =q

2a^2 + (a^2- k^2) = q

•Substituting values of a and a^2- k^2,

•We get -2p^3 + pq + r =0

i.e. 2p^3 = pq+r.

•Hence , option B is the right answer.

Answer 2

Step-by-step explanation:

option b is the right answer bro