Question

If the roots of the equation (q – r)x² + (r – p)x + p – q = 0 are equal, then show that p, q and r are in AP

Answer 1

The roots are equal ⇒ ∆ = 0

(i.e.) b² – 4ac = 0

Hence, a = q – r, b = r – p, c = p – q

b² – 4ac = 0

⇒ (r – p)² – 4(q – r)(p – q) = 0

r² + p² – 2pr – 4[qr – q² – pr + pq] = 0

r² + p² – 2pr – 4qr + 4q² + 4pr – 4pq = 0

(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0

(i.e.) (p – 2q + r)² = 0

⇒ p – 2q + r = 0

⇒ p + r = 2q

⇒ p, q, r are in A.P.