If (|x| - 2)(x + 5) < 0, then which of the following must be true?

Given: (|x| - 2)(x + 5) < 0

Case 1: when (|x| - 2) is negative and (x + 5) is positive

i.e. (|x| - 2) < 0 if -2 < x < +2 and x > -5

i.e. -2 < x < +2

Case 2: when (|x| - 2) is Positive and (x + 5) is Negative

i.e. (|x| - 2) > 0 if x < -2 or x > 2 and (x + 5) < 0 i.e. x < -5

x < -5

Looking at case 1 and 2 both we know that x is definitely less than 2

Answer: Option B

If (|x| - 2)(x + 5) < 0, then which of the following must be true?

If x is positive:

(x-2)(x+5)<0
=> x belongs to (-5,2)
but x is positive so x belongs to (0,2)

If x is negative:

(x+2)(x+5)>0
=> x belongs to (-infi,-5) U (-2,infi)
but x is negative, so x belongs to (-infi,-5) U (-2,0)

If x is 0:

(-2)(5)= -10 which is <0. So, Zero is included.

Therefore, final range is X belongs to (-infi,-5) U (-2,2)

So answer options C and E work.

Let me try to address the confusion here. Personally, the question does not seem to be a very good quality question, though I would give merit to what it is testing us on.

It tests us on the basic understanding of what would ALWAYS be true, and NOT the range of X.

X does not satisfy the equation at various points, but it will definitely never be greater than 2. This is the reasoning behind the correct answer.

The question is absolutely correct and the answer is B only.

(|x| - 2)(x + 5) < 0 is true when x < -5 or -2 < x < 2. ANY x from these possible ranges will for sure be less than 2 (option B).

-2 < x < 2 (C) is not true because x could say be -10.
x < -5 (D) is not true because x could say be 0.

To understand the underline concept better practice other Trickiest Inequality Questions Type: Confusing Ranges.

STRATEGY: Upon reading any GMAT Problem Solving question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can easily test x-values that satisfy the given inequality.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also attempt to solve the inequality, but the absolute value part of the inequality looks tricky. So, I'm pretty sure testing values is going to be a lot faster and much easier

Let's find an x-value that satisfies the inequality (|x| - 2)(x + 5) < 0.
x = 0 is an obvious solution, which means x = 0 must also be a solution to the correct answer choice.

Now plug x = 0 into each answer choice to get:
(A) 0 > 2. Not true. Eliminate.
(B) 0 < 2. True. KEEP.
(C) -2 < 0 < 2. True. KEEP.
(D) -5 < 0 < 2. True. KEEP.
(E) 0 < -5. Not true. Eliminate.

Now let's find another x-value that satisfies the inequality (|x| - 2)(x + 5) < 0.
I can see that x = -10 is a solution, since we get (|-10| - 2)((-10) + 5) < 0, which simplifies to be (8)(-5) < 0, which is true.

Now plug x = -10 into the three remaining answer choices:
(B) -10 < 2. True. KEEP.
(C) -2 < -10 < 2. Not true. Eliminate.
(D) -5 < -10 < 2. Not true. Eliminate.

By the process of elimination, the correct answer is B.

Bunuel

Let's take x= -2 (which should satisfy the equation as per the option B)

(|x| - 2)(x + 5)
= (|-2| - 2)(-2 + 5)
= (2 - 2)(3)
= 0*3 = 0

It doesn't satisfy the equation.

You have to work the other way.

Fit a value in the equation and if that value fits in, the option should contain that.
Here, a value is in the option but does not fit in equation. It is ok as we are not looking at RANGE of x but what MUST be true.
So, a x<100 in option will also be an answer.

­It says must be true, based on that I think E should be the answer, because no matter what the value, E will always be true. Even though we are missing out on some values such as 0 and -1. For must be true cases, E is correct I believe.­

If X can be between -2 and 2, how can it always be true that X<-5 ?

­
We have that \(x < -5\) or \(-2 < x < 2\). ANY \(x\) from these possible ranges will for sure be less than 2 (option B).

\(x < -5\), option (E), is not true because \(x\) could say be 0 and if it's 0 then it's not less than -5.

Check other similar questions from Trickiest Inequality Questions Type: Confusing Ranges (part of our Special Questions Directory).

Hope it helps.­

But B says x<2, meaning x could be -2 as well. If x is -2, then |x| gives 2 as the output making (|x| - 2) as zero. Now zero X (x+5) can’t be < 0. Then why is B correct and not C? C definitely rules out any of the brackets to become zero.

Please explain.

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­x cannot be -2, becasue we know that \(x < -5\) or \(-2 < x < 2\). -2 is not in these ranges.

We have that \(x < -5\) or \(-2 < x < 2\). ANY \(x\) from these possible ranges will for sure be less than 2 (option B).

\(-2 < x < 2\), option (C), is not true because \(x\) could say be -10 and if it's -10 then it's not less than -5.

Check other similar questions from Trickiest Inequality Questions Type: Confusing Ranges (part of our Special Questions Directory).

Hope it helps.­

 
Fun Question which has a quick and easy solution!

Note that it is a "must be true" kind of question. It is not "complete range"; it is not "could be true"; it is "must be true." (Link given below)

Given: (|x| - 2)(x + 5) < 0

When we want to remove the absolute value sign, there are two cases: x >= 0 or x < 0 (link given below)

Case 1: x >= 0
(x - 2)(x + 5) < 0
means -5 < x < 2 (Wavy line method link given below)
Since x is positive, 0 <= x < 2 
Hence x is certainly less than 2 in all these cases.

Now, do we need to consider Case 2 at all in which x < 0? No. Whatever solutions we get will in every case be less than 2 since x will be negative. 
Hence x less than 2 will ALWAYS be true for all valid values of x. 

Answer (B)

Post on difference between must be true, complete range and could be true: https://anaprep.com/algebra-must-be-true-could-be-true-complete-range-questions/
Post on why there are two cases with absolute value sign: 
https://anaprep.com/algebra-the-why-beh ... questions/
Video on wavy line method: 
https://youtu.be/PWsUOe77__E

­

B encompasses C, D, E

So the answer is either A or B

We can clearly see that B is the right choice. On the real GMAT I don't think the answer choices will be overlapping in this way, but it makes for a very easy and quick solution here