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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jul 2007 13:46:29 IST
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20 + 10 +1 > 0 (x-3)(x-4) (x-4)
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29 Jul 2007 13:48:34 IST
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20+10x-30+x^2-7x+12/(x-3)(x-4)>0
therefore
(x+1)(x+2)/(x-3)(x-4)>0....
therefore
(x+1)(x+2)>0.....for x not equal to 3 and not equal to 4....
x (- ,-2) U (-1,3) U (3,4) U (4,)
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31 Jul 2007 13:34:10 IST
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20/(x-3)(x-4) + 10/(x-4) + 1 > 0
{20+10(x-3)+(x-3)(x-4)} / (x-3)(x-4) > 0
(x2 + 3x + 2) / (x-3)(x-4) > 0
(x+1)(x+2) / (x-3)(x-4) > 0
Now check in intervals (-,-2) , (-2,-1) , (-1,3) , (3,4) , (4,) where the above expression is positive.
You will get x  (-,-2)  (-1,3) (4,)
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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31 Jul 2007 14:27:06 IST
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i agree
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
   
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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2 Aug 2007 00:45:14 IST
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we get {20 + 10(x-3) + (x-4)(x-3) }/ (x-4)(x-3) > 0
cancel x - 4 and x - 3 applying cond that x not equal to 4, 3
we get x^2 + 3x + 2>0
solving we get
x E (-inf, -1) U ( -2, +inf) ~ {3,4}
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