Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

Show Tags

09 Jun 2012, 02:20

1

This post received KUDOS

1

This post was BOOKMARKED

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.

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!

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

Show Tags

02 Dec 2012, 05:13

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.

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.

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

Show Tags

19 Jan 2013, 02:19

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.

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

Show Tags

23 Feb 2013, 20:08

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.

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?
_________________

It is beyond a doubt that all our knowledge that begins with experience.

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.

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.
_________________

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 \(x>=-3\)

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?

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 \(x>=-3\)

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
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: How many of the integers that satisfy the inequality [#permalink]

Show Tags

09 Jun 2013, 06:16

samara15000 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

The answer is [D] indeed. There are only 4 values for which the equation will hold considering x < 5

Consider \((x+2)(x+3)/(x-2) >=0\) for values like {0, 5}, we can clearly see that x =2 is not a acceptable value. Further x = 1,0 will also not hold. Only 3,4 are acceptable values. Now considering the negative range, {-infinity, 0}, we can see that -3 and -2 are the values for which the inequality is equal to 0. For all other values under the range, the values are -ve. Hence 4 values with the solution set as {-3,-2,3,4}.

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

Show Tags

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.

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.
_________________

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...