Correct Answer - D
Since `alpha, beta` are the roots of `x^(2) - px + r = 0`
`therefore" "(alpha)/(2)+2beta = q and (alpha)/(2) xx 2 beta r`
Solving `alpha + beta = p and (alpha)/(2)+2 beta = q`, we get
`alpha =(2)/(3)(2p-q) and beta = (1)/(3)(2q-p)`
`therefore" "alpha beta = r rArr r = (2)/(9) (2p-q) (2q-p)`