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Asked by manoj255 | 20th Dec, 2009, 04:17: PM

Expert Answer:

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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