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प्रश्न
If the roots of the equation (q – r)x2 + (r – p)x + p – q = 0 are equal, then show that p, q and r are in AP
उत्तर
The roots are equal ⇒ ∆ = 0
(i.e.) b2 – 4ac = 0
Hence, a = q – r, b = r – p, c = p – q
b2 – 4ac = 0
⇒ (r – p)2 – 4(q – r)(p – q) = 0
r2 + p2 – 2pr – 4[qr – q2 – pr + pq] = 0
r2 + p2 – 2pr – 4qr + 4q2 + 4pr – 4pq = 0
(i.e.) p2 + 4q2 + r2 – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)2 = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.
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