Let a,b,c be natural numbers and the quadratic equation ax2-bx+c=0 have two distinct roots lying in the interval (0,1)...
Now..
a) the least value of a
b) the least value of b
c) the least value of c
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4 Answers
62also, b2>4ac because of distinct roots
also, f(0) is +ve so c>0
also f(1) is +ve so a+c>b
f'(0).f'(1)<0
(2a.0-b)(2a1-b)<0
so b(2a-b)>0. b>0 so 2a>b
so in all we have
a+c>b
2a>b
b2>4ac
33ye sab to maine bhi laya... par what are there min values... [12]
341I have done this earlier in goiit. So i'll just post the link.
See the last post in the thread as the first post does not have a rigorous solution.
http://www.goiit.com/posts/list/algebra-problem-no-3-67916.htm#447898
So a=b=5 and c=1 are the respective minima