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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:01:32 IST
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if f(x) is a continuous function and f(x) is an element of Q where Q is a rational number then find the roots of the equation: f(2)x^2 + f(6)x + f(9) = 0 given f(10)=5 plz try and solve or at least reply
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:07:23 IST
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r there no brainies in goiit ??
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:14:58 IST
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Well, is the answer -1, 4/3
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:16:30 IST
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how ??
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:19:02 IST
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i don't hav answer with me
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:20:24 IST
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Well, its just a guess !!!! I have used the hit and trial method. Yhe only expression for f(x) through which i was getting real roots was f(x) = x-5 And thats how i got the answer !!!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:22:56 IST
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experts, where are you ??
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:25:16 IST
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put it on expert poll ,its not on it
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:26:39 IST
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how can i put it there ??
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:31:01 IST
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this question is not on expert poll and ask in expert option,this is in community option ask again the same question
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:32:44 IST
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okay
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:35:08 IST
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as f(x) can have only rational values it will be constant function.f(2),f(6),f(9)=5.Eqn. will be reduced to x^2+x+1 whose roots will come out to be imaginary..plz tell if i'm right.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:39:43 IST
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mr popat
f(x) is not constant, it is continuous also tell how u assumed it constant
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Who says nothing is impossible.
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I know KUNG FU KARATE
and 47 other dangerous words.............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Sep 2007 20:47:04 IST
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if f(x) belongs to rational numbers it means that it is a constant function(equal to 5 which is ratinal) because if the function depends on x then it will give you rational as well as irrational values.
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