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

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How many of the integers that satisfy the inequality (x+2)(x [#permalink] New post 08 Jun 2012, 23:00
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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
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Re: How many of the integers that satisfy the inequality (x+2) ( [#permalink] New post 08 Jun 2012, 23:35
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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] New post 08 Jun 2012, 23:38
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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] New post 09 Jun 2012, 00:03
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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 givenx < 5
Thus integral solutions would be x = -3, -2, 3, 4

Answer is (D)

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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink] New post 09 Jun 2012, 01:51
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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#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.
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink] New post 09 Jun 2012, 02:20
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#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.


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] New post 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.

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,

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] New post 02 Dec 2012, 05:15
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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#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,

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.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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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,
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] New post 30 Jan 2013, 06:06
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@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] New post 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.
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Re: 700 Level PS OG13 229 [#permalink] New post 30 Mar 2013, 05:36
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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)=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: 700 Level PS OG13 229 [#permalink] New post 01 Apr 2013, 15:06
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] New post 01 Apr 2013, 20:33
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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] New post 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: 700 Level PS OG13 229 [#permalink] New post 09 Apr 2013, 06:46
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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>=-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 [#permalink] New post 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}.

Hope it helps!

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Re: How many of the integers that satisfy the inequality [#permalink] New post 09 Jun 2013, 06:42
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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


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

:arrow: <apologies for the pedantry>
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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink] New post 25 Aug 2013, 10:08
Expert's post
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.
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Re: How many of the integers that satisfy the inequality (x+2)(x   [#permalink] 25 Aug 2013, 10:08
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