Question

The roots of equation (q – r)x² + (r – p)x + (p – q) = 0 are equal. Prove that: 2q = p + r, that is, p, q and r are in A.P.

Answer 1

a = q – r, b = r – p and c = p – q

For equal roots,

b² = 4ac

`\implies` (r – p)² = 4(q – r)(p – q)

r² + p² – 2pr = 4[pq – q² – pr + qr]

r² + p² – 2pr + 4pr = 4[pq – q² + qr]

(p + r)² = 4[q(p + r) – q²]

(p + r)² – 4q(p + r) + 4q² = 0

Let (p + r) = y

y² – 4qy + 4q² = 0

(y – 2q)² = 0

y – 2q = 0

or

p + r = 2q

Hence proved