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705-805 (Hard)|   Inequalities|                              
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financebro28
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How many of the integers that satisfy the inequality (x+2)(x 09 Jan 2025, 07:13
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How does one solve this using Wavy Line method egmat ?
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Navyashrivastava
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How many of the integers that satisfy the inequality (x+2)(x 11 Jan 2025, 23:20
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How are you getting 3& 4? isn't answer greater than 5 in that case?
pike
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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How many of the integers that satisfy the inequality (x+2)(x 12 Jan 2025, 01:08
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We need integers that are less than 5, when plugged into ((x+2)(x+3)) / (x-2) are equal to or greater than 0.

3 and 4, which are less than 5, when plugged into the above equation give 30 and 21 which is greater than 0. Hope it helps.

Navyashrivastava
How are you getting 3& 4? isn't answer greater than 5 in that case?
pike
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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How many of the integers that satisfy the inequality (x+2)(x 12 Jan 2025, 01:52
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Mark 2, -2, and -3 on number line, every number right to 2 will give the equation positive, and satisfy the equation - we need to find integers less than 5 - so we get 3 and 4 from there.

Numbers between -2 and 2 will be negative, and we cannot take 2 as it will make denominator 0.

Numbers from -3 and -2 will be positive or 0 which satisfies the equation, we get -3 and -2.

Anything below -3 will be negative.

So we got -3, -2, 3, and 4; 4 integers.


financebro28
How does one solve this using Wavy Line method egmat ?
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How many of the integers that satisfy the inequality (x+2)(x 16 May 2025, 23:49
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Denominator can't be 0
mkumar26
if I take x=2 then >=0 !
why not its satisfied ?
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How many of the integers that satisfy the inequality (x+2)(x 16 May 2025, 23:52
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I solved it like this :

critical points where denominator is not 0,

x can be = -2,-3

now we if we keep x as > -3 say -4
Numerator = -ve * -ve = +ve
Denominator = -ve + (- ve) = -ve
thereby not > 0

so now we take values of x > 2 ( so that denominator is not 0) and < 5 (given)
hence possible are 3 & 4

thereby D, -3,-2,3,4
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How many of the integers that satisfy the inequality (x+2)(x 09 Jun 2025, 11:45
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in gmat anything divide by zero is taken undefined quantity ?

since here when x=2 the value will be infinity/ undefined

how to reject x=2 as possible solution
Bunuel

macjas
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
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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How many of the integers that satisfy the inequality (x+2)(x 09 Jun 2025, 11:52
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MacT750
in gmat anything divide by zero is taken undefined quantity ?

since here when x=2 the value will be infinity/ undefined

how to reject x=2 as possible solution
Bunuel

macjas
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
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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If you check carefully the solution you quote, you'll notice that x = 2 is excluded from the ranges.
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How many of the integers that satisfy the inequality (x+2)(x 02 Jul 2025, 04:58
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If the 4th range was negative, then the third is positive, the second is negative and the fist positive? Does this rule apply when you have 3 ranges also?
Bunuel

macjas
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
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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How many of the integers that satisfy the inequality (x+2)(x Updated on: 20 Jul 2026, 11:16
Originally posted by Bunuel on 02 Jul 2025, 07:54.
Last edited by Bunuel on 20 Jul 2026, 11:16, edited 1 time in total.
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jerothomas
If the 4th range was negative, then the third is positive, the second is negative and the fist positive? Does this rule apply when you have 3 ranges also?

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Yes, the rule applies for 3 ranges too, signs alternate across intervals. Also, check the links shared for more clarity.
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How many of the integers that satisfy the inequality (x+2)(x 20 Jul 2026, 10:41
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But anything divided by 0 is infinity, which is >0, so why can't we consider 2 also in the solution?
Bunuel


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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How many of the integers that satisfy the inequality (x+2)(x 20 Jul 2026, 11:15
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MistyNova
But anything divided by 0 is infinity, which is >0, so why can't we consider 2 also in the solution?

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Anything divided by 0 is not infinity, it is undefined. The expression has no value at x = 2, so x = 2 cannot be included in the solution.
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