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macjas
Joined: 09 May 2012
Last visit: 30 Jul 2015
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Difficulty:
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Question Stats:
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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
OG 2019 PS14203
A. 1
B. 2
C. 3
D. 4
E. 5
OG 2019 PS14203
Solving this question helps.
Taking a timed set of similar questions in
GMAT Club Forum Quiz →
is even better.
macjas
Given: \(\frac{(x+2)(x+3)}{x-2}\geq{0}\).
The roots of the expression \(\frac{(x+2)(x+3)}{x-2}\) are -3, -2, and 2 (equate the expressions to zero to find the roots and list them in ascending order). This gives us four ranges: \(x<-3\), \(-3\leq{x}\leq{-2}\), \(-2<x<2\), and \(x>2\). Note that since we have the \(\geq\) sign, we should include -3 and -2 in the ranges but not 2, as \(x=2\) would make the denominator zero, and we cannot divide by zero.
Now, test an extreme value: for example, if \(x\) is a very large number, then all three terms will be positive, which gives a positive result for the whole expression. Therefore, when \(x>2\), the expression is positive. Here’s the trick: if in the 4th range the expression is positive, then in the 3rd range it will be negative, in the 2nd range positive again, and finally negative in the 1st range: - + - +. Thus, 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 \(x>2\) mean that the only four integers less than 5 that satisfy the given inequality are: -3, -2, 3, and 4
Answer: D.
Solving inequalities:
https://gmatclub.com/forum/x2-4x-94661.html#p731476
https://gmatclub.com/forum/inequalities ... 91482.html
https://gmatclub.com/forum/everything-i ... me#p868863
https://gmatclub.com/forum/xy-plane-714 ... ic#p841486
Hope it helps.
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cyberjadugar
Joined: 28 Mar 2012
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Posts: 264
Updated on: 19 Jun 2012, 04:56
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.
Last edited by cyberjadugar on 19 Jun 2012, 04:56, edited 1 time in total.
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
Answer is (D)
Regards,
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
Answer is (D)
Regards,
Attachments
nline.jpg [ 4.21 KiB | Viewed 336657 times ]
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.
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.
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.
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.
General Discussion
macjas
Joined: 09 May 2012
Last visit: 30 Jul 2015
Posts: 306
Affiliations: UWC
Location: Canada
Schools: Rotman '15 (A) Schulich '15 (M)
GMAT 1: 620 Q42 V33
GMAT 2: 680 Q44 V38
GPA: 3.43
WE:Engineering (Media/Entertainment)
Products:
Bunuel
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!
Bunuel
Bunuel,
Could you explain this graphical method you use or direct me to a post which does the same.
Your help is much appreciated.
eaakbari
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
xy-plane-71492.html?hilit=solving%20quadratic#p841486
Hope it helps.
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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
\((x+2)(x+3)=0\)
\(x+2=0, x=-2\)
\(x+3=0, x=-3\)
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?
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?
Bunuel
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.
mp2469
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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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>
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>
nave81
\(\frac{(x+2)(x+3)}{x-2}\geq{0}\) 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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I don't see why 2 isn't a solution? Maybe someone can clarify!
sunchild
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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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)
1.png [ 2.87 KiB | Viewed 163030 times ]
Representing (x+3) on number line
x+3 is +ve (when x>-3)
x+3 is -ve (when x<-3)
2.png [ 2.92 KiB | Viewed 162931 times ]
Representing (x-2) on number line
x-2 is +ve (when x>2)
x-2 is -ve (when x<2)
3.png [ 2.79 KiB | Viewed 162819 times ]
Combining all these equations.
The representation of these equations on the number line is shown in the figure attached.
4.png [ 1.8 KiB | Viewed 162684 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.
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 163030 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 162931 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 162819 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 162684 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.
EMPOWERgmatRichC
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Expert reply
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,
Rich
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,
Rich
macjas
MATH REVOLUTION VIDEO SOLUTION:
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ScottTargetTestPrep
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Expert reply
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macjas
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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if I take x=2 then >=0 !
why not its satisfied ?
why not its satisfied ?
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