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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 19 Apr 2018, 02:46
My simple way to solve: Experts can suggest if its flawed.

(x+2)(x+3)/(x-2) ≥ 0

Try make a square in the denominator by multiplying both numerator and denominator by (x-2).

Then equation would like like this: (x+2)(x-2)(x+3)/(x-2)^2 ≥ 0

The denominator shall be +ive for all values of x, other than 2, where the expression will become undefined. Remember to exclude x=2 in all solutions.

Coming to numerator, (x+2)(x-2) makes (x^2-2^2) => (x^2-4). This will be = 0 or +ive for for x>= 2 and x<= -2. So list of possible solutions of x = 2,3,4..... and -2, -3, -4.... Remember to exclude x=2 here.

Coming to the other remaining term in the numerator (x+3) => for this to be 0 or +ive, x can be any value of x >= -3. i.e. -3, -2, -1.....

Now observe what values are common across all solution = -3, -2 and 3, 4, for the question only wants values less than 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 Aug 2018, 21:04
cyberjadugar wrote:
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)


Hi CyberJadugar,

Could you please explain how you got 3 & 4 as the integral solutions.

I understand how you get -3 & -2 as the answers but 3 & 4 does not make sense to me. As the question is asking for the final value not to be greater than 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 26 Sep 2018, 00:39
First lets eliminate the obvious \(x\neq{2}\) as that would result in a division by 0.

We'll get our possible solutions when denominator and numerator are of the same sign.

(1) Lets say both the values (x+2) & (x+3) in the numerator are positive, which means denominator has to be positive to satisfy the equation.
That implies x > 2 (remember \(x\neq{2}\)), the limit as per the question x < 5 => x = 3 or 4

(2) As the equation can be 0, the numerator can be 0 => x = -2 or -3

(3) \(x\neq{0}\) as that would result in a \(\frac{+ve}{-ve}\), resulting in a negative value

(4) From the equation its clear that choosing x < -3 will result in a \(\frac{-ve*-ve}{-ve}\) and thats a negative.

Hence all possible values of x < 5 are -3, -2, 3 & 4 (= 4 possible values)

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

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New post 17 Nov 2018, 06:43
pike wrote:
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
4,3,-2,-3, so D.



Hello
Please help
Why when I test 4,3, -3,-4 I receive wrong answer? I mean

(4+2) (4+3) / (4-2) = 21
(3+2) (3+3) / (3-2) = 30

21 and 30 are bigger than 5


Where is my mistake???
Thank you
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Re: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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New post 17 Nov 2018, 15:31
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Hi Ross7it,

With certain questions, you have to be careful about confusing the information that you've been given with the question that you are asked to answer. Your solution is correct - but I think you lost track of the question that you attempting to answer.

Your work shows that there are 4 VALUES FOR X that fit the given inequality. When you plug in each of those values for X, the resulting calculation IS greater than 0. The question is NOT asking for the end calculation though; it's asking about the possible values of X. Each of those 4 values for X ARE less than 5, so the answer to the question is 4 values.

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

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New post 17 Nov 2018, 16:24
EMPOWERgmatRichC wrote:
Hi Ross7it,

With certain questions, you have to be careful about confusing the information that you've been given with the question that you are asked to answer. Your solution is correct - but I think you lost track of the question that you attempting to answer.

Your work shows that there are 4 VALUES FOR X that fit the given inequality. When you plug in each of those values for X, the resulting calculation IS greater than 0. The question is NOT asking for the end calculation though; it's asking about the possible values of X. Each of those 4 values for X ARE less than 5, so the answer to the question is 4 values.

GMAT assassins aren't born, they're made,
Rich



Dear Rich,
Thanks for answering, but it still not clear.

So, how the testing method works here? Could explain me better, please? Some guys on the first page of the topic tested numbers from -4 to 4, but it was not the same to substitute those numbers for X. Am I correct?

Many thanks
Ross

P.S. I am refering to this one:

"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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New post 18 Nov 2018, 12:18
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Hi Ross7it,

The inequality (x+2)(x+3)/(x-2) >= 0 actually has an unlimited number of integer solutions for X.

The question is NOT asking us for ALL of the solutions though; it's only asking for the ones that are LESS than 5. Based on the five answer choices, there will be at least 1 value, but no more than 5 values, that fit what we're looking for. Thus, it should not be too difficult to find those 1-5 values; we can just 'plug in' until we find them all.

When you TEST X=4, you will end up with (6)(7)/(2). This equals 42/2 = 21, which is >= 0. Thus, X=4 IS one of the examples of what we are looking for (an integer value of X that 'fits' the given inequality AND is LESS than 5). That math is relatively easy to do, so you should be able to find the other values without too much trouble (by 'plugging in' 3, 2, 1, 0, -1, etc.) and paying attention to the results.

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

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New post 18 Nov 2018, 13:19
EMPOWERgmatRichC wrote:
Hi Ross7it,

The inequality (x+2)(x+3)/(x-2) >= 0 actually has an unlimited number of integer solutions for X.

The question is NOT asking us for ALL of the solutions though; it's only asking for the ones that are LESS than 5. Based on the five answer choices, there will be at least 1 value, but no more than 5 values, that fit what we're looking for. Thus, it should not be too difficult to find those 1-5 values; we can just 'plug in' until we find them all.

When you TEST X=4, you will end up with (6)(7)/(2). This equals 42/2 = 21, which is >= 0. Thus, X=4 IS one of the examples of what we are looking for (an integer value of X that 'fits' the given inequality AND is LESS than 5). That math is relatively easy to do, so you should be able to find the other values without too much trouble (by 'plugging in' 3, 2, 1, 0, -1, etc.) and paying attention to the results.

GMAT assassins aren't born, they're made,
Rich


Dear Rich,

Thank you very much, finally I got it, you have provided a perfect and clear explanation !!! Thank you for your time!

Cheers,
Ross
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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 Feb 2019, 23:08
Why x=2 is not valid?

The equation still holds true if x=2.
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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 Feb 2019, 18:03
landle2693 wrote:
Why x=2 is not valid?

The equation still holds true if x=2.


The equation does not hold when x = 2. When x = 2, the denominator becomes zero and thus, the expression is undefined.
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Re: How many of the integers that satisfy the inequality (x+2)(x   [#permalink] 25 Feb 2019, 18:03

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