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

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

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08 Jun 2012, 23:00
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How many of the integers that satisfy the inequality $$\frac{(x+2)(x+3)}{(x-2)} \geq 0$$ 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)(x  [#permalink]

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09 Jun 2012, 01:51
40
2
95
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.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

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

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Updated on: 19 Jun 2012, 04:56
40
1
28
Hi,

General method:

$${(x+2)(x+3)}/(x-2) \geq 0$$

if 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$$x < 5$$
Thus integral solutions would be x = -3, -2, 3, 4

Regards,
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nline.jpg [ 4.21 KiB | Viewed 179561 times ]

Originally posted by cyberjadugar on 09 Jun 2012, 00:03.
Last edited by cyberjadugar on 19 Jun 2012, 04:56, edited 1 time in total.
##### General Discussion
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Re: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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08 Jun 2012, 23:35
18
8
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)(x  [#permalink]

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08 Jun 2012, 23:38
9
3
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.

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

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09 Jun 2012, 02:20
3
2
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.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope 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]

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02 Dec 2012, 05:13
1
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.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Bunuel,

Could you explain this graphical method you use or direct me to a post which does the same.

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

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02 Dec 2012, 05:15
2
3
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.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Bunuel,

Could you explain this graphical method you use or direct me to a post which does the same.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

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

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30 Mar 2013, 05:36
7
1
3
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 $$x<=-3$$
Then we analyze the denominator >0 (it can't be =0)
$$x-2>0, x>2$$

~~~~~~~(-3)~~~~~(-2)~~~~~~~(+2)
negative, negative,negative|positive For the D
positive | negative| positive , positive For the N
You 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: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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01 Apr 2013, 15:06
1
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: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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Updated on: 09 Jun 2013, 07:00
6
1
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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Originally posted by mau5 on 01 Apr 2013, 20:33.
Last edited by mau5 on 09 Jun 2013, 07:00, edited 1 time in total.
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Re: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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09 Apr 2013, 06:32
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 $$x>=-3$$

Like: $$(x+2) >= 0$$ => $$x>= -2$$
and $$(x+3) >=0$$ => $$x>= -3$$

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

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09 Apr 2013, 06:46
2
4
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 $$x>=-3$$

Like: $$(x+2) >= 0$$ => $$x>= -2$$
and $$(x+3) >=0$$ => $$x>= -3$$

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

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25 Aug 2013, 10:05
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.

<apologies for the pedantry>
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Re: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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25 Aug 2013, 10:08
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.

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

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24 Apr 2014, 09:43
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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25 Apr 2014, 00:55
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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29 Nov 2015, 03:55
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 [ 2.87 KiB | Viewed 25310 times ]

Representing (x+3) on number line

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

Attachment:

2.png [ 2.92 KiB | Viewed 25314 times ]

Representing (x-2) on number line

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

Attachment:

3.png [ 2.79 KiB | Viewed 25291 times ]

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

Attachment:

4.png [ 1.8 KiB | Viewed 25287 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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29 Nov 2015, 10:09
1
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.

GMAT assassins aren't born, they're made,
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Re: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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22 Jan 2016, 09:01
5
1
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

MATH REVOLUTION VIDEO SOLUTION:

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Re: How many of the integers that satisfy the inequality (x+2)(x   [#permalink] 22 Jan 2016, 09:01

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