Question
Sat July 14, 2012 By:
 

if one of the zeroes of ax2+bx+c is triple the other,show that 3b2=16ac.

Expert Reply
Sat July 14, 2012
Let the zeroes be p and 3p.
Sum of zeroes = 4p = -b/a
Product of zeroes = 3p2 = c/a
This gives,
3(-b/4a)2 = c/a
3b2/16a2 = c/a
3b2/16a = c
3b2 = 16ac
Hence proved.
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