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How many of the integers that satisfy the inequality (x+2)(x

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 26 Aug 2013, 07:48
Bunuel wrote:
nave81 wrote:
Disclaimer: The below is not going to be helpful for your GMAT

Strictly speaking when x takes the value of 2, the value of the expression leads to infinity. I know GMAT is way too scared of infinity but the question asks if the expression leads to equal to or greater than zero, and since numerator is positive, the infinity is in the positive direction which indeed meets the inequality criteria.

:arrow: <apologies for the pedantry>


\(\frac{(x+2)(x+3)}{x-2}\geq{0}\) holds true for \(-3\leq{x}\leq{-2}\) and \(x>2\). Thus four integers that are less than 5 satisfy given inequality: -3, -2, 3, and 4. Notice that 2 is not among these four integers.



Indeed, but I was not talking about the official answer and hence my disclaimer and the apology :wink:

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 03 Dec 2013, 23:51
It may be infinite but not sure (undefined) if it is +infinity or negative infinity.
if it is 1/(+0.0000000001) it is +big number (i.e. tends to +infinity) but if it is 1/ (-0.0000000001) it is -big number (i.e. tends to -infinity) so at the junction of minus tends to 0 and plus tends to zero (i.e. Zero itself), it is undefined whether it is positive or negative but sure it is infinity.
In this question we need this inequality to be greater than zero (so we need a positive number) so x=2 is ruled out
Hope this helps.

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 24 Apr 2014, 10:43
I don't see why 2 isn't a solution? Maybe someone can clarify!

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 25 Apr 2014, 01:55
sunchild wrote:
I don't see why 2 isn't a solution? Maybe someone can clarify!


If x=2, then \(\frac{(x+2)(x+3)}{x-2}\) would be undefined and not \(\geq{0}\) because the denominator would become zero. We cannot divide by zero.

Does this make sense?
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 16 Aug 2014, 09:14
Hi Bunuel,

my Solution: -

Step 1 : - Given that (X+2)(X+3)/(x-2)>=0 therefore x=2 is not possible and that X has to be less than 5
Step 2:- x+2 >0 , X+3> 0 and X+2= 0 and X+3 = 0
Step 3 : - The above equations tell me that x=-2, x=-3 and x>-2 and "X>-3", so i have two values with me , x=-2 and -3 which satisfy x<5.
Can you please tell me where am i wrong in creating my inequality equations?? i can see in some posts that there isa range X<=-3 but i am unable to get that . Can you please explain me using basic principles of Inequality.

Step 4 :- Now i know that i have following values to work with , X= -1,0,1,3,4, they are all less than 5.
Step 5 : - On manual calculation i get that only 3 and 4 satisfy the equation as (X+2)(X+3)/(X-2) >= 0, ( -1 and 0 result in negative values).

therefore possible values are : - 3, 4 -3 and -4 so answer is D


Please please let me know where am i wrong in step 3 , i am struggling at other Complex forms of Inequality

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 23 Jun 2015, 09:59
Bunuel wrote:
nave81 wrote:
Disclaimer: The below is not going to be helpful for your GMAT

Strictly speaking when x takes the value of 2, the value of the expression leads to infinity. I know GMAT is way too scared of infinity but the question asks if the expression leads to equal to or greater than zero, and since numerator is positive, the infinity is in the positive direction which indeed meets the inequality criteria.

:arrow: <apologies for the pedantry>


\(\frac{(x+2)(x+3)}{x-2}\geq{0}\) holds true for \(-3\leq{x}\leq{-2}\) and \(x>2\). Thus four integers that are less than 5 satisfy given inequality: -3, -2, 3, and 4. Notice that 2 is not among these four integers.



Algebraic approach is too long here. I've just picked numbers <5 and stopped by -3 (considering some restrictions as X>2 and X<5 and also testing +/- cases.
But I'm nevertheless interested in the algebraic approach - Bunuel can you check it please.

Expression = 0
Solutions x= -2; -3 and x=2 can not be a solution here - we can not divide by 0

Expression > 0
Case 1 Denominator/Numerator are both positive +/+
x>-2; -3; 2 ---> x>2 which gives us two solutions 3 and 4 (because x must be < 5)

Case 2 Denominator/Numerator are both negativ -/-
Case A (x+2)>0; (x+3)<0; (x-2)<0
x>-2, x<-3, x<2

Case b (x+2)<0; (x+3)>0; (x-2)<0
x<-2, x>-3, x<2

There is no solution for Case A & B - no intersections, so we can't make this expression negativ
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 30 Aug 2015, 14:02
Hi can anybody explain how did we conclude 1<x<3 ?
I am still not clear with this part! :-(

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 29 Nov 2015, 04:55
For these inequality questions, the best way is to represent numbers on a number line. By representing on a number line, we will be easily able to compare the sign of the equations and will result in faster solving of inequalities.

Representing (x+2) on number line

x+2 is +ve (when x>-2)
x+2 is -ve (when x<-2)

Attachment:
1.png
1.png [ 2.87 KiB | Viewed 1945 times ]


Representing (x+3) on number line

x+3 is +ve (when x>-3)
x+3 is -ve (when x<-3)

Attachment:
2.png
2.png [ 2.92 KiB | Viewed 1943 times ]


Representing (x-2) on number line

x-2 is +ve (when x>2)
x-2 is -ve (when x<2)

Attachment:
3.png
3.png [ 2.79 KiB | Viewed 1942 times ]


Combining all these equations.
The representation of these equations on the number line is shown in the figure attached.

Attachment:
4.png
4.png [ 1.8 KiB | Viewed 1937 times ]


From the figure, we can easily deduce that the given equation \((x+2)(x+3)/(x-2) >= 0\) will be satisfied when
1) All three equations are positive
2) Two equations are negative and one equation is positive.
3) Equations give the result as zero.

Such a condition is satisfied for the following integers.

At integers 3 and 4 all the equations are positive. Hence satisfy the inequality
At integers -2 and -3 the value of the equation is zero, hence satisfies the inequality.

So overall 4 integers satisfy the inequality. So option (D) is correct.

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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Hi All,

While this question looks a bit complex, it's not as difficult as it might appear. This question emphasizes a particular part of the process that is so important for GMAT questions of all types: you have to take notes and do work in an organized way.

In this prompt, we're asked to focus on integer solutions that are LESS than 5. From the answers, we know that there is at least one solution, but no more than five solutions. This means that there aren't that many options and they shouldn't be too hard to find.

If you were "stuck" on this question, then here's how you can go about solving it quickly - Just start plugging in integers until you've "found" all of the ones that "fit." Start with the number 4, then 3, then 2, etc. You'd be amazed how often you can use what's called "brute force" against a Quant question; plug in numbers and pound on the question until you've found the solution.

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 10 Mar 2016, 10:48
The key to solving these kind of questions to realize the critical points here being => -2,-3,2
and we need to abide the inequality => x<5
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 21 Jul 2016, 10:34
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


Before diving into this problem we want to first analyze the given inequality:

(x+2)(x+3)/(x-2) ≥ 0

This means that (x+2)(x+3) divided by x-2 is equal to or greater than zero. This provides us with 3 options for the signs of the numerator and denominator to yield a final answer that is either positive or 0:

1) (+)/(+) = positive

2) (-)/(-) = positive

3) 0/any nonzero number = 0

The above options will allow the inequality to hold true. We must be strategic in the numbers that we test. The easiest course of action is to start with option 3. We know that when n = -2 or when n = -3, the numerator of our inequality, (x+2)(x+3), will be zero. Thus, we have found two integer values for x less than 5 that fulfill the inequality. Next let’s focus our attention on option 1.

Option 1 tells us that both the numerator and denominator of (x+2)(x+3)/(x-2), must be positive. We see that when x = 4 or x = 3, we have a positive numerator and a positive denominator. At this point we should notice that we have already found 4 values of x that fulfill the inequality and, at most (according to the answer choices), there could be 5. So let’s consider option 2 to see whether we can find any values for x that make both the numerator and denominator negative. To do this we concentrate on the denominator of the fraction, x-2. We can see right away that the only integers that will make “x – 2” negative are 1, 0, -1, -2, -3, -4, and so on. That is, if x is an integer less than 2, then the denominator will be negative. However, when plugging 1, 0, or -1, into the numerator, we see that the numerator will remain positive and thus the entire fraction will be a negative value. When we plug in -2 or -3 for x, the numerator is 0 and entire fraction is 0. (This is actually option 3 above and we have already included -2 and -3 as part of the solutions.) Lastly, when we plug in -4 or integers less than -4, the numerator will be again positive and thus the entire fraction will result in a negative value. So with this information we can sufficiently determine that there are only 4 integer values less than 5 that fulfill the inequality.

The answer is D.
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 15 Dec 2016, 22:59
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#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.


Bunuel, important question here - why is the first range not x<=3 but x<3? Please help.

Thanks,
Oreo

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 31 Mar 2017, 09:35
if I take x=2 then >=0 !
why not its satisfied ?

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 01 Apr 2017, 05:53
mkumar26 wrote:
if I take x=2 then >=0 !
why not its satisfied ?


x = 2 does not satisfy (x+2)(x+3)/(x-2) >= 0. In this case the denominator (x-2) becomes 0 and the whole expression (x+2)(x+3)/(x-2) becomes undefined - recall that we cannot divide by 0.
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 14 Apr 2017, 17:55
Hello,

Here is a short video to explain a smart way of solving this question. If you get the concept explained in the video, the question can be solved in 60 seconds or less.

https://youtu.be/ZIgKyR7oN88

I have covered this concept in detail on my conceptual post today-
https://gmatclub.com/forum/gmat-quant-o ... l#p1836874

All the best!
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 27 Apr 2017, 10:03
Hi,
Sorry to bump this old thread, but I have a question. How is the solution for
(x+2)(x+3)>=0
(x+2)(x+3)>=0 is
x>=−2 and x<=−3

and not x>=−2 and x>=−3

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]

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New post 20 May 2017, 10:16
Here I find it easier to use a method of picking numbers. We have the following range
-4, -3, -2, -1, 0, 1, 2, 3, 4.

1. x= - 4, => (x+2)*(x+3)/(x-2) => (-4+2)(-4+3)/-6 = - 1/3 < 0, doesn’t work

2. x=- 3, => x= - 4, => (x+2)*(x+3)/(x-2) => 0/… = 0, works.

3. x= -2 => 0/-4, => works

4. x = -1 => - 2/3 < 0, doesn’t work

5. x = 0, = > -3, < 0, doesn’t work

6. x = 1, => (-1 +2)*(-1 + 3)/-3 = - 2/3 < 0, doesn’t work

7. x = 2, => (4)*(5)/0 => doesn’t work

8. x = 3, => 5*6/1 = 30 > 0, works,

9. x = 4, => 6*7/2 = 21 > 0, works

So here we have 4 integers (-3, -2, 3 and 4) that are less than 5 and satisfying the inequality.

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Re: How many of the integers that satisfy the inequality (x+2)(x   [#permalink] 20 May 2017, 10:16

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