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Mathematics

The roots of the 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.

Quadratic Equations

ICSE Sp 2025

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Answer

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 that p + r = 2q.

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