1I dont think that will be of much help....
Hint: f(x)=a(x-α)(x-β) where α,β ε (0,1)
so, f(0)f(1)=a2αβ(1-α)(1-β)
AM≥GM will be helpful...now proceed...
If ax2 - bx + c = 0 have two distinct roots lying in the interval (0,1), a, b, c ε N, then prove that log5(abc) ≥ 2.
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12 Answers
1sum of the roots = b/a
prod. of roots = c/a
AM>=GM>=HM
SO
4ac>=2c/b
a+c/2>=4ac
this is just the funda i cud reach
help me out!!!!
11then i suppose
f(0)f(1)<0 so that it may hv 2 real roots btwn (0,1)
1totally wrong....read the question carefully...
both the roots here lie between 0 and 1. So, f(0)f(1) > 0 always.
1srry!! i mistook the funda --- the pt. at which the tangent =0 & then looking for sign change..for the orig. function....
1I'm trying hard but not getting it......any small hint ?
I'm getting c > 0
and
a-b+c > 0
1read the question again...it is already given that a,b,c are natural numbers. Try it a little longer , i think u will get it...
0<α<1
so, 0<(1-α)<1
now proceed...use AM≥GM
341See this only if you are desperate: http://goiit.com/posts/list/algebra-problem-no-3-67916.htm#336240
But there is a much easier way i saw later. So prob is still open
66Since the polynomial f(x)=ax^2-bx+c has real and distinct roots,
b2 > 4ac
Also, f(x)=a(x-\alpha)(x-\beta)
Since the roots of f lie strictly between 0 and 1 so
f(0)f(1)>0
But being an integer, f(0)f(1) ≥1
Also
f(0)f(1)=
a^2\alpha(1-\alpha)\beta(1-\beta)<\dfrac{a^2}{16}
In this last step, I used AM-GM and the fact that equality cannot hold.
Combining the inequalities we get
a^2>16\Rightarrow a\geq 5
Thus,
b^2>20c\geq 20
which gives us
b\geq 5
I have taken c≥1.
Hence,
abc\geq 25
With a=5, b=5, c=1, the conditions of the problem are indeed satisfied, so abc has a minimum value 25.