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

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."

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

Solution:

We see that if x = -3 or -2, then the value of the expression (on the left hand side of the inequality) will be 0, which is greater than or EQUAL to 0.

If x is -4 or less, then the numerator of the expression is positive (because the two factors are negative) and the denominator is negative. However, the value of the entire expression will be negative, which does not satisfy the inequality.

If x = -1, 0, or 1, then the numerator of the expression is positive (because the two factors are positive) and the denominator is negative. However, the value of the expression will be negative, which does not satisfy the inequality.

We see that x can’t be 2 since the denominator will then be 0 and we can’t divide by 0.

If x is 3 or greater, then the numerator of the expression is positive (because the two factors are negative) and the denominator is also positive. Therefore, the value of the expression will be positive, which does satisfy the inequality.

However, since we are looking for the values of x that are less than 5, so there are only 4 such values (namely, -3, -2, 3, and 4).

Answer: D

We can see that, when x = -2 and x = -3, the fraction evaluates to be ZERO.
So, we already have two values that satisfy the given inequality: x = -2 and x = -3
Also, when x = 2, the fraction is UNDEFINED

So, we have three critical points: x = -2, x = -3 and x = 2
These 3 points divide the number line into 4 regions:
Region i: x < -3
Region ii: -3 < x < -2
Region iii: -2 < x < 2
Region iv: x > 2

Let's see what happens with the values of x in each region...
Region i: x < -3
If x < -3, then (x+2), (x+3), and (x-2) are all NEGATIVE, which means the fraction evaluates to be a NEGATIVE value
Since we want the fraction to be greater than or equal to 0, none of the values in this region satisfy the inequality.

Region ii: -3 < x < -2
If -3 < x < -2, then (x+2) is NEGATIVE, (x+3) is POSITIVE, and (x-2) is NEGATIVE, which means the fraction evaluates to be a POSITIVE value
However, since there are no integers within this range, there's nothing to count here.

Region iii: -2 < x < 2
If -2 < x < 2, then (x+2) is POSITIVE, (x+3) is POSITIVE, and (x-2) is NEGATIVE, which means the fraction evaluates to be a NEGATIVE value
Since we want the fraction to be greater than or equal to 0, none of the values in this region satisfy the inequality.

Region iv: x > 2
If x > 2, then (x+2) is POSITIVE, (x+3) is POSITIVE, and (x-2) is POSITIVE, which means the fraction evaluates to be a POSITIVE value
In other words, ALL x-values greater than 2 satisfy the given inequality.
However, since we're looking for integer values less than 5, the only x values that satisfy the given inequality are x = 3 and x = 4

In total, there are FOUR solutions

Answer: D

Cheers,
Brent

Solution:

First, let's find the values of x which make the given expression 0. Setting the numerator equal to 0, we see that if x = -2 or x = -3, then the expression becomes 0 and thus, the inequality is satisfied.

If x is less than -3, then all of x + 3, x + 2, and x - 2 are negative; thus, [(x + 2)(x + 3)]/(x - 2) is also negative. There are no values of x less than -3 that satisfy the given inequality.

If x = -1, 0, or 1, then the numerator is positive while the denominator is negative; hence, these values will not satisfy the inequality.

Notice that the expression is undefined for x = 2.

Finally, if x = 3 or x = 4, then all of x + 3, x + 2, and x - 2 are positive; thus, the expression itself is positive as well.

We see that the inequality holds for -3, -2, 3, and 4. Thus, four integers that satisfy the inequality are less than 5.

Answer: D

Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Solution:

The question stem can be rephrased as

Is (x+2)(x+3)(x-2)>=0 ? (Multiply (x-2)^2 on both sides)

The roots of the quadratic are -3,-2 and 2

Using the wavy line concept, with a greater than sign the range would be

-3<=x<=-2

=>Integral solutions would be x = -3, -2, 3, 4 (option d)

Devmitra Sen
GMAT SME

Yes, just test numbers

x < 5
x=4 Yes
x=3 Yes
x=2 (no)(Undefined)
x= 1 (no)
x = 0 (no)
x = -1 (No)
x = -2 (Yes)
x= -3 (Yes)
x = -4 (No)
x = -5 (No)
x = -6 (No)
Pattern shows us that we're done.

Hi TooLong150,

Yes - that's perfect! Once you get down to X = -4, you might recognize that a Number Property pattern emerges: the numerator will be the product of two negative numbers, so that product will STAY POSITIVE as X gets smaller AND the denominator will STAY NEGATIVE as X gets smaller. Dividing a positive number by a negative number ALWAYS yields a negative result, meaning that there's no reason to keep looking at integer values for X that are smaller than -4 (because they will ALL lead to total numbers that are negative numbers - which is NOT what the question is asking us for).

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

Bunuel Can you please help me understand as to how we can directly equate this to zero initially to find roots? This is an inequality right? Either 2 of them have to be negative or all of them have to be positive for the numbers >0?

Also after finding roots, can we solve this type by the wavy line method? Or is it limited to only basic quadratic equations/inequations?

Inequality with a very different technique used by mbahouse

(x+2)(x+3)>=0 gives -2 and -3 as the roots. Therefore, between -2 and -3 the value is negative and for x<-3 and x>-2, the values are positive.
Now, (x-2) is positive for x>2 and negative for x<2.
We know, for the whole equation to be positive both the numerator and denominator has to be both positive or negative.
The above condition is only possible for x>2 and x lies between -2 and -3.

The question says integers less than 5, so that definitely loops 3 and 4 in since both are greater than 2 but less than 5.
Now there's no integer between -2 and -3. Therefore 2 is supposed to be the answer.

The trick we often miss is that is says : equation>=0. So we have to loop in the integers which yields 0 too. So x=-2,-3 are those 2 integers which makes the equation 0.

Therefore the total no. of integers is 4 : -2,-3,3,4

Want more help? Here you go: http://linktr.ee/thegmatstrategy

Hi,
Solved using the shortcut method-
link- https://gmatclub.com/forum/inequalities ... 91482.html
1. Identify the number of roots- 3 roots in this qs: 2,-2,-3. Arrange in ascending order: -3,-2,2
2. Plot the roots in the diagram at the points the curve cuts the line
The rightmost will be positive infinity, then plot the roots in a descending way.
(See image)
Take any value>2 to test the sign for the rightmost area(over the line). Check what will be the test value. Here it is positive, thus the region will be positive. The points in this region will be positive. At the roots it will be 0. The signs in the regions to the left will be alternate.
(the logic is explained in the link above)
Alternatively negative values will be observed in the regions marked '-'
3. Suitable values under 5: 3,4,-2,-3
Since at x=2, denominator will become 0 and thus not possible

Voila!

Where did you find the -3 and -2 inequality?