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#