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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jul 2007 19:16:23 IST
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20 + 10 + 1 > 0 (x-3)(x-4) (x-4)
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29 Jul 2007 19:45:53 IST
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the answer is x > - 2
Soln.
20 + 10 + 1 > 0 (x-3)(x-4) (x-4)
taking lcm
20 + 10(x-3) + 1(x-3)(x-4)> 0
20 + 10x - 30 + x^2- 7x + 12 > 0
x^2 + 3x + 2 > 0
(x+1)(x+2) > 0
i.e. x > - 2
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29 Jul 2007 20:43:44 IST
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by wavy curve method we get
x<-2  -1<x<3 x>4 .
rate me if the answer is correct
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santom |
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29 Jul 2007 20:51:01 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,)
there is no closed interval.....................i am sure....for example substitute -2 in the answer.........................the result will be less than 0.........
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29 Jul 2007 20:55:06 IST
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nivedh_89 is correct even i got the same ans.
@ dhwanitmunshi we cant cross muliplty the denominator wid 0 as we do not know if it is +ve or not
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29 Jul 2007 22:14:32 IST
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ans given here
http://www.goiit.com/posts/list/algebra-linear-inequalities-22317.htm
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30 Jul 2007 22:10:10 IST
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Here is the answer : .'.[20/(x-3)(x-4)] + [10/(x-4)] + [1] > 0 .'.[20 + 10(x-3) + (x-3)(x-4)]/(x-3)(x-4) > 0 On Solving this we will get : (x2+3x+2)/(x-3)(x-4) > 0 (x+1)(x+2)/(x-3)(x-4) > 0 From above the critical points will be [-1,-2,3,4] Now draw the wavy curve as I have given below, So the answer will be: x  (- ,-2) U (-1,3) U (3,4) U (4,) (3,4) for only (x+1)(x+3) > 0 Hope you find it useful. plZ rate me for my efforts. Cheers!!!!!!!!!!!!!
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3 Aug 2007 14:45:09 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....
(x+1)(x+2)>0.....for x not equal to 3 and 4....
apply wave curve and u get
x E (-inf,-2) U (-1,+inf) ~ {3,4}
closed brackets r not there as equation has no equal to sign.
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* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
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3 Aug 2007 14:47:29 IST
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dhwanitmunshi oh my god.. how can u make such a big mistake in the last step... its (x-1)(x-2) > 0 just HOW did u get x >-2 ?
apply way curve
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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3 Aug 2007 14:48:47 IST
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@shohita we can cross multiply the denominator with 0 after we take the condition x not equal to 3, 4 ... thts y these points r not included in interval of x
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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3 Aug 2007 14:49:28 IST
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waterdemons graph will help u to see clearly :)
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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7 Aug 2007 21:01:56 IST
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wavy curve method most easy
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7 Aug 2007 21:38:16 IST
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First of all, u cannot simply take (x-3)(x-4) as LCM. U cannot guess the sign... Remember this is an inequality. So, if you are multiplying by a negative sign, the inequality gets reversed... Secondly, i dont understand how is everyone making the same mistake... We'll have to do the question by taking cases... Case 1 x<3 The equation becomes 20/(x-3)(x-4) - 10/(x-4) + 1 > 0 Now wen LCM is taken, we actually multiply each term of the expression with the LCM. The LCM is (x-3)(x-4) is +ve for the first case, hence the inequality sign is preserved. Now the solution for this case would be the intersection of the solution and the condition taken... Next Case2 3<x<4 Case 3 x>4
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HOPE U GOT IT... |
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7 Aug 2007 22:43:36 IST
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Moderation Team
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