Question

if p(q-r)x 2 + q(r-p)x + r(p-q) = 0 has equal roots prove that 2/q = 1/p + 1/r (cyclic order) i.e. p,q,r are in h.p

Answer 1

p (q-r) x²  + q (r-p) x + r (p-q) = 0

Roots equal => discriminant = 0.

   q²(r-p)² = 4 (pq - pr)(pr - qr)
   q² r² + q²p² - 2 q² rp = 4 p²qr - 4 pq²r - 4 p²r² + 4 pq r²

Divide the whole by p²q²r²
    1/p² + 1/r² + 2/pr = 4/qr - 4/q² + 4/pq
    (1/p + 1/r)² - (1/r +1/p) *4/q + 4/q² = 0
    (1/p + 1/r -2/q)² = 0
     1/p + 1/r = 2/q