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

ICSE Specimen Question Paper – 2026

Step 1: Let’s start comparing the given quadratic equation with standard equation ax² + bx + c = 0
We get, a = (q – r), b = (r – p), c = (p – q)

Step 2: Now, since the roots are equal, hence discriminant D = 0
b² = 4 a c
(r – p)² = 4(q – r) (p – q)
r² + p² – 2pr = 4 (pq -pr – q² + qr)
r² + p² – 2pr = 4 pq – 4pr – 4q² + 4qr
r² + p² + 2pr = 4pq – 4q² + 4qr
(p + r)² = 4q (p+r) – 4q²
(p + r)² – 4q (p+r) + 4q² = 0
(p + r)² – 2 (2q) (p+r) + (2q)² = 0
(p + r – 2q)² = 0
(p + r – 2q) = 0
p + r = 2q
r – q = q – p
Therefore, p, q and r are in AP.

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