Question

The roots of equation (q – r)x² + (r – p)x + (p – q) = 0 are equal.

Prove that 2q = p + r; i.e., p, q, and r are in A.P.

Answer 1

Given the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² – 4ac = 0

⇒ (r – p)² – 4 × (q – r) × (p – q) = 0

⇒ r² + p² – 2pr – 4[qp – q² – rp + qr] = 0

⇒ r² + p² – 2pr – 4qp + 4q² + 4rp – 4qr = 0

⇒ r² + p² + 2pr – 4qp – 4qr + 4q² = 0

⇒ (p + r)² – 4q(p + r) + 4q² = 0

Let (p + r) = y

⇒ y² – 4qy + 4q² = 0

⇒ (y – 2q)² = 0

⇒ y – 2q = 0

⇒ y = 2q

⇒ p + r = 2q

Hence proved.