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Asked by manoj255 | 20th Dec, 2009, 04:17: PM
Let p, q and r be the zeroes of the polynomial ax3+3bx2+3cx -d
Since the zeroes are in AP, 2q = p + r ..........(1)
Sum of the zeroes of a cubic polynomial , p + q + r = -3b/a ................(2)
Sumof zeroes taking two at a time, pq + qr + rp = 3c/a .................(3)
Product of zeroes, pqr = d/a .....................(4)
From (1) and (2), we get
3q = -3b/ a
⇒ q = -b/a .....................(5)
Now from equation (4)
pr = d/ aq = -d/b ....................(6)
Now, from equation (3)
pq + qr + rp = 3c/a
⇒ q (p+r) + rp = 3c/a
⇒ 2(-b/a)2 - (d/b) = 3c/a
Solving the above you will get the answer as 2b3-3abc - a2d = 0
Answered by | 5th Jan, 2010, 01:55: PM
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