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Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
26 Aug 2013, 06:48

Bunuel wrote:

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.

Indeed, but I was not talking about the official answer and hence my disclaimer and the apology

Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
03 Dec 2013, 22:51

It may be infinite but not sure (undefined) if it is +infinity or negative infinity. if it is 1/(+0.0000000001) it is +big number (i.e. tends to +infinity) but if it is 1/ (-0.0000000001) it is -big number (i.e. tends to -infinity) so at the junction of minus tends to 0 and plus tends to zero (i.e. Zero itself), it is undefined whether it is positive or negative but sure it is infinity. In this question we need this inequality to be greater than zero (so we need a positive number) so x=2 is ruled out Hope this helps.

Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
16 Aug 2014, 08:14

Hi Bunuel,

my Solution: -

Step 1 : - Given that (X+2)(X+3)/(x-2)>=0 therefore x=2 is not possible and that X has to be less than 5 Step 2:- x+2 >0 , X+3> 0 and X+2= 0 and X+3 = 0 Step 3 : - The above equations tell me that x=-2, x=-3 and x>-2 and "X>-3", so i have two values with me , x=-2 and -3 which satisfy x<5. Can you please tell me where am i wrong in creating my inequality equations?? i can see in some posts that there isa range X<=-3 but i am unable to get that . Can you please explain me using basic principles of Inequality.

Step 4 :- Now i know that i have following values to work with , X= -1,0,1,3,4, they are all less than 5. Step 5 : - On manual calculation i get that only 3 and 4 satisfy the equation as (X+2)(X+3)/(X-2) >= 0, ( -1 and 0 result in negative values).

therefore possible values are : - 3, 4 -3 and -4 so answer is D

Please please let me know where am i wrong in step 3 , i am struggling at other Complex forms of Inequality

Re: How many of the integers that satisfy the inequality (x+2)(x [#permalink]
22 Aug 2014, 05:29

bunuel can you please tell me what is wrong with Step 3 in my solution.

thanks.

Jerry1982 wrote:

Hi Bunuel,

my Solution: -

Step 1 : - Given that (X+2)(X+3)/(x-2)>=0 therefore x=2 is not possible and that X has to be less than 5 Step 2:- x+2 >0 , X+3> 0 and X+2= 0 and X+3 = 0 Step 3 : - The above equations tell me that x=-2, x=-3 and x>-2 and "X>-3", so i have two values with me , x=-2 and -3 which satisfy x<5. Can you please tell me where am i wrong in creating my inequality equations?? i can see in some posts that there isa range X<=-3 but i am unable to get that . Can you please explain me using basic principles of Inequality.

Step 4 :- Now i know that i have following values to work with , X= -1,0,1,3,4, they are all less than 5. Step 5 : - On manual calculation i get that only 3 and 4 satisfy the equation as (X+2)(X+3)/(X-2) >= 0, ( -1 and 0 result in negative values).

therefore possible values are : - 3, 4 -3 and -4 so answer is D

Please please let me know where am i wrong in step 3 , i am struggling at other Complex forms of Inequality

How many of the integers that satisfy the inequality (x+2)(x [#permalink]
23 Jun 2015, 08:59

Bunuel wrote:

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.

Algebraic approach is too long here. I've just picked numbers <5 and stopped by -3 (considering some restrictions as X>2 and X<5 and also testing +/- cases. But I'm nevertheless interested in the algebraic approach - Bunuel can you check it please.

Expression = 0 Solutions x= -2; -3 and x=2 can not be a solution here - we can not divide by 0

Expression > 0 Case 1 Denominator/Numerator are both positive +/+ x>-2; -3; 2 ---> x>2 which gives us two solutions 3 and 4 (because x must be < 5)

Case 2 Denominator/Numerator are both negativ -/- Case A (x+2)>0; (x+3)<0; (x-2)<0 x>-2, x<-3, x<2

Case b (x+2)<0; (x+3)>0; (x-2)<0 x<-2, x>-3, x<2

There is no solution for Case A & B - no intersections, so we can't make this expression negativ _________________

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