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How many of the integers that satisfy the inequality (x+2)(x [#permalink]
 09 Jun 2012, 00:00
Question Stats:
29% (01:55) correct
70% (01:15) wrong based on 11 sessions
How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5? A. 1 B. 2 C. 3 D. 4 E. 5
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Re: How many of the integers that satisfy the inequality (x+2) ( [#permalink]
09 Jun 2012, 00:35
How many of the integers that satisfy the inequality ((x+2)(x+3)) / (x-2) >= 0 are less than 5? Just start testing numbers: 4,3,2,1,0,-1,-2,-3,-4 etc 4 - yep 3 - yep 2 - no 1 - no 0 - no -1 - no -2 - yes -3 - yes -4 and below - no 4,3,-2,-3, so D.
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Re: How many of the integers that satisfy the inequality (x+2) ( [#permalink]
09 Jun 2012, 00:38
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Yeah, you could test numbers.
Alternatively, just find the solutions to the inequality:
This solves to x>2 and -3<=x<=-2
So X can be 3, 4, ... or -2 or -3.
So 4 integers.
Answer = D.
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Re: How many of the integers that satisfy the inequality (x+2) ( [#permalink]
09 Jun 2012, 01:03
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Hi, General method: {(x+2)(x+3)}/(x-2) \geq 0if we plot it on number line, we have, -3 \leq x \leq -2& x > 2, since x-2 \neq 0 (no equality). Also, it is given x < 5Thus integral solutions would be x = -3, -2, 3, 4 Answer is (D) Regards,
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
09 Jun 2012, 02:51
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macjas wrote: How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5 \frac{(x+2)(x+3)}{x-2}\geq{0} --> the roots are -3, -2, and 2 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-3, -3\leq{x}\leq{-2}, -2<x<2 and x>2 (notice that we have \geq sign, so, we should include -3 and -2 in the ranges but not 2, since if x=2 then the denominator becomes zero and we cannot divide by zero). Now, test some extreme value: for example if x is very large number then all three terms will be positive which gives the positive result for the whole expression, so when x>2 the expression is positive. Now the trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative: - + - +. So, the ranges when the expression is positive are: -3\leq{x}\leq{-2}, (2nd range) and x>2 (4th range). -3\leq{x}\leq{-2} and x>2 means that only 4 integers that are less than 5 satisfy given inequality: -3, -2, 3, and 4. Answer: D. Solving inequalities: x2-4x-94661.html#p731476inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486Hope it helps.
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
09 Jun 2012, 03:20
Bunuel wrote: macjas wrote: How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5 \frac{(x+2)(x+3)}{x-2}\geq{0} --> the roots are -3, -2, and 2 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-3, -3\leq{x}\leq{-2}, -2<x<2 and x>2 (notice that we have \geq sign, so, we should include -3 and -2 in the ranges but not 2, since if x=2 then the denominator becomes zero and we cannot divide by zero). Now, test some extreme value: for example if x is very large number then all three terms will be positive which gives the positive result for the whole expression, so when x>2 the expression is positive. Now the trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative: - + - +. So, the ranges when the expression is positive are: -3\leq{x}\leq{-2}, (2nd range) and x>2 (4th range). -3\leq{x}\leq{-2} and x>2 means that only 4 integers that are less than 5 satisfy given inequality: -3, -2, 3, and 4. Answer: D. Solving inequalities: x2-4x-94661.html#p731476inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486Hope it helps. Thanks Bunuel, while I could easily solve this one using numbers, I couldn't get the algebraic approach. You explanation with the graphical approach is bang on. Thanks!
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
02 Dec 2012, 06:13
Bunuel wrote: macjas wrote: How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5 \frac{(x+2)(x+3)}{x-2}\geq{0} --> the roots are -3, -2, and 2 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-3, -3\leq{x}\leq{-2}, -2<x<2 and x>2 (notice that we have \geq sign, so, we should include -3 and -2 in the ranges but not 2, since if x=2 then the denominator becomes zero and we cannot divide by zero). Now, test some extreme value: for example if x is very large number then all three terms will be positive which gives the positive result for the whole expression, so when x>2 the expression is positive. Now the trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative: - + - +. So, the ranges when the expression is positive are: -3\leq{x}\leq{-2}, (2nd range) and x>2 (4th range). -3\leq{x}\leq{-2} and x>2 means that only 4 integers that are less than 5 satisfy given inequality: -3, -2, 3, and 4. Answer: D. Solving inequalities: x2-4x-94661.html#p731476inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486Hope it helps. Bunuel, Could you explain this graphical method you use or direct me to a post which does the same. Your help is much appreciated.
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
02 Dec 2012, 06:15
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eaakbari wrote: Bunuel wrote: macjas wrote: How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5 \frac{(x+2)(x+3)}{x-2}\geq{0} --> the roots are -3, -2, and 2 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-3, -3\leq{x}\leq{-2}, -2<x<2 and x>2 (notice that we have \geq sign, so, we should include -3 and -2 in the ranges but not 2, since if x=2 then the denominator becomes zero and we cannot divide by zero). Now, test some extreme value: for example if x is very large number then all three terms will be positive which gives the positive result for the whole expression, so when x>2 the expression is positive. Now the trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative: - + - +. So, the ranges when the expression is positive are: -3\leq{x}\leq{-2}, (2nd range) and x>2 (4th range). -3\leq{x}\leq{-2} and x>2 means that only 4 integers that are less than 5 satisfy given inequality: -3, -2, 3, and 4. Answer: D. Solving inequalities: x2-4x-94661.html#p731476inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486Hope it helps. Bunuel, Could you explain this graphical method you use or direct me to a post which does the same. Your help is much appreciated. Solving inequalities: x2-4x-94661.html#p731476 (check this one first) inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486Hope it helps.
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Re: How many of the integers that satisfy the inequality [#permalink]
16 Jan 2013, 05:44
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
19 Jan 2013, 03:19
Bunuel wrote: macjas wrote: How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5 \frac{(x+2)(x+3)}{x-2}\geq{0} --> the roots are -3, -2, and 2 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-3, -3\leq{x}\leq{-2}, -2<x<2 and x>2 (notice that we have \geq sign, so, we should include -3 and -2 in the ranges but not 2, since if x=2 then the denominator becomes zero and we cannot divide by zero). Now, test some extreme value: for example if x is very large number then all three terms will be positive which gives the positive result for the whole expression, so when x>2 the expression is positive. Now the trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative: - + - +. So, the ranges when the expression is positive are: -3\leq{x}\leq{-2}, (2nd range) and x>2 (4th range). -3\leq{x}\leq{-2} and x>2 means that only 4 integers that are less than 5 satisfy given inequality: -3, -2, 3, and 4. Answer: D. Solving inequalities: x2-4x-94661.html#p731476inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486Hope it helps. Bunuel, Why do we take only +ve values of the inequlaities? Is that because in the question >= 0 is given??? What if in the question <= 0 was given???? Do we take -ve values of the inequaliteis from number line...?? Help appriciated...
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
30 Jan 2013, 07:06
@bhavinshah5685
The question would have had an option as infinte if the inequality had a less than equal to sign!
It is so because then the two ranges would have been [-2,2) and [-infinity,-3].
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
23 Feb 2013, 21:08
Can someone please explain why we take the denominator (x-2) as one of the roots of this inequality? I thought when you set the equation to = 0 and bring the denominator to the right side it becomes 0.
For example (x^2+ 5x-6)/(x^2- 4x+3)=0 we would only consider the solutions of the numerator NOT the denominator.
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Re: 700 Level PS OG13 229 [#permalink]
30 Mar 2013, 06:36
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How many of the integers that satisfy the inequality (x+2)(x+3)/(x-2)>=0 are less than 5? A. 1 B. 2 C. 3 D. 4 E. 5 We can analize the numerator >=0 (x+2)(x+3)=0x+2=0, x=-2x+3=0, x=-3Since we have a ">=" we take the external values x>=-2 and x<=-3Then we analyze the denominator >0 (it can't be =0) x-2>0, x>2~~~~~~~(-3)~~~~~(-2)~~~~~~~(+2) negative, negative,negative| positive For the Dpositive | negative| positive , positive For the NYou sum up the sign of the values and obtain: negative | positive | negative | positive We are looking for >=0 value, so we keep the positive intervals and discard the negative ones. -3>=x>=-2 ( in this we have also the =) and x>2 ( no = here) The values less than 5 are : -3,-2,3,4 Is it clear?
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Re: 700 Level PS OG13 229 [#permalink]
01 Apr 2013, 16:06
Bunuel wrote: rakeshd347 wrote: How many of the integers that satisfy the inequality (x+2)(x+3)/(x-2)>=0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5
I am not really good with inequalities to be honest. I have solved this question and found the answer but It took me 4minutes. Is there any short approach please. Merging similar topics. Please refer to the solutions above. I still don't understand how -2 and -3 are solutions. Don't they make the numerator = to 0? I kind of understand the theory, but i'm having trouble reconciling the number picking strategy with the theory.
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Re: 700 Level PS OG13 229 [#permalink]
01 Apr 2013, 21:33
mp2469 wrote: Bunuel wrote: rakeshd347 wrote: How many of the integers that satisfy the inequality (x+2)(x+3)/(x-2)>=0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5
I am not really good with inequalities to be honest. I have solved this question and found the answer but It took me 4minutes. Is there any short approach please. Merging similar topics. Please refer to the solutions above. I still don't understand how -2 and -3 are solutions. Don't they make the numerator = to 0? I kind of understand the theory, but i'm having trouble reconciling the number picking strategy with the theory. We are given (x+2)(x+3)/(x-2)>=0 Now we can not cross multiply (x-2) as we don't about its sign. All we know from the problem is that x can not be equal to 2 as because that will make the expression undefined. Now, as know that (x-2)^2 is a positive quantity. Safely multiply it on both sides, thus we get, (x-2)(x+2)(x+3)>=0. AS because there is an equality sign in the given inequality, we can say that x=-2 and x=-3 are two valid solutions, for which the expression assumes the value of zero. X can't be equal to 2, as stated before.
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Re: 700 Level PS OG13 229 [#permalink]
09 Apr 2013, 07:32
Zarrolou wrote: How many of the integers that satisfy the inequality (x+2)(x+3)/(x-2)>=0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5
We can analize the numerator >=0
(x+2)(x+3)=0
x+2=0, x=-2
x+3=0, x=-3
Since we have a ">=" we take the external values x>=-2 and x<=-3
Sorry to bump this old thread, but I have a question. How is the solution for (x+2)(x+3) >= 0 x>=-2 and x<=-3 and not x>=-2 and x>=-3Like: (x+2) >= 0 => x>= -2and (x+3) >=0 => x>= -3I guess inputting numbers [-4, -5 etc] will make the inequality true but when solving practice questions, instinctively, I am missing this range. Is this something I can get good at only by practice?  any tips?
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Re: 700 Level PS OG13 229 [#permalink]
09 Apr 2013, 07:46
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bcrawl wrote: Sorry to bump this old thread, but I have a question. How is the solution for (x+2)(x+3) >= 0 x>=-2 and x<=-3 and not x>=-2 and x>=-3Like: (x+2) >= 0 => x>= -2and (x+3) >=0 => x>= -3I guess inputting numbers [-4, -5 etc] will make the inequality true but when solving practice questions, instinctively, I am missing this range. Is this something I can get good at only by practice? any tips? To solve this : (x+2)(x+3) \geq{0}, we can use an old method. Think it this way (x+2)(x+3) = 0 the solutions are x=-2 and x=-3; now I use an old trick: if the sign of x^2 and the operator are " the same" ie (+,>) or (-,<) we take the external values : x\leq{-3} and x\geq{-2}. In the other two cases (+,<) (-,>) we take the internal values. If the sign was < ( (x+2)(x+3) \leq{0}) the solution would be -3\leq{x}\leq{-2}. Let me know if it's clear now
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Re: 700 Level PS OG13 229
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