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प्रश्न
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.
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उत्तर
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.
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