If zeroes of the polynomials f(x)=x^3-3px^2+qx-r are in AP then what is the relation between p, q and r?

1 Answer
Jan 13, 2018

2p^3+pq+r = 0

Explanation:

Denoting the three zeros by alpha, beta, gamma, we have:

f(x) = x^3-3px^2+qx-r

color(white)(f(x)) = (x-alpha)(x-beta)(x-gamma)

color(white)(f(x)) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma

In particular, equating the coefficient of x^2 we have:

alpha+beta+gamma = 3p

If alpha, beta and gamma are in arithmetic progression with common difference delta, then:

alpha = beta-delta

gamma = beta+delta

So:

3 beta = (beta-delta)+beta+(beta+delta) = alpha+beta+gamma = 3p

So:

beta = p

That is: p is one of the zeros of f(x)

So:

0 = f(p) = color(blue)(p^3)-3pcolor(blue)(p)^2-qcolor(blue)(p)-r = -2p^3-pq-r

Inverting the signs, that is:

2p^3+pq+r = 0