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 (SQP)2025

Ans:

Step 1: Given the polynomial: (q – r) x ² + (r – p) x + (p – q) = 0

When we compare it standard quadratic equation, a x ² + b x + c = 0, we get:

a = (q – r), b = ( r – p) and c = (p – q)

Step 2: We know that when roots of a polynomial are equal, then its discriminant is zero.

∴ D = 0

∴ b ² – 4 a c = 0

∴ (r – p) ² – 4 (q – r) (p – q) = 0

∴ r ² + p ²  – 2 r p – 4 (q p – q ² – r p +  r q) = 0

∴ r ² + p ²  – 2 r p – 4 q p + 4 q ² + 4 r p –  4 r q = 0

∴ (r ² + p ²  + 2 r p) + 4 q ² – 4 q p –  4 r q = 0

∴ (p + r) ²  + (2 q) ² – 4 q (p + r) = 0

∴ (p + r) ²  + (2 q) ² – 2 (2 q) (p + r) = 0

∴ (p + r – 2 q) ² = 0

∴ p + r – 2 q =0

∴ 2 q = p + r

Therefore, p, q, r are in AP.   

(∵ common difference d =  r – q = q – p)

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