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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 11 Sep 2025, 01:36
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Erm, conversely x can be -10 and then option B will not be valid ?
Bunuel


You are wrong. It's the other way around. Again, we have \(x < -5\) or \(-2 < x < 2\). ANY \(x\) from these possible ranges will for sure be less than 2 (option B). E, x < -5, on the other hand, is NOT always true, because x can be say 0, and in this case x < -5 will not be true.

To understand the underline concept better practice other Trickiest Inequality Questions Type: Confusing Ranges.
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 11 Sep 2025, 01:40
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Erm, conversely x can be -10 and then option B will not be valid ?

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Yes, x can be -10 as it is in the valid range (\(x < -5\) or \(-2 < x < 2\)). But in this case how is B (x < 2) not correct? Isn't -10 less than 2? Please invest more time in the question and the discussion. Follow the links to the similar questions for more.
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 11 Sep 2025, 01:45
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Sorry.

Help me understand, I am new to GMAT.

If x = -2, then it doesn't satisfy the given inequality. As, Mod(-2) - 2 will result in 0. And 0 is not less than 0 ? Hence, B cannot be the answer for sure.

Thoughts ??
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Yes, x can be -10 as it is in the valid range (\(x < -5\) or \(-2 < x < 2\)). But in this case how is B (x < 2) not correct? Isn't -10 less than 2? Please invest more time in the question and the discussion. Follow the links to the similar questions for more.
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 11 Sep 2025, 01:51
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Brian_1
Sorry.

Help me understand, I am new to GMAT.

If x = -2, then it doesn't satisfy the given inequality. As, Mod(-2) - 2 will result in 0. And 0 is not less than 0 ? Hence, B cannot be the answer for sure.

Thoughts ??

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The valid ranges for x are \(x < -5\) or \(-2 < x < 2\). So, x CANNOT be -2! Please, invest time and study the discussion carefully!
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 11 Sep 2025, 01:57
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Nvm, I understood.

X can take a lot of values and it has to less than 2. Gotcha. I got caught up with the option choices. Thanks
Bunuel


The valid ranges for x are \(x < -5\) or \(-2 < x < 2\). So, x CANNOT be -2! Please, invest time and study the discussion carefully!
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 25 Nov 2025, 23:39
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If (|x| - 2)(x + 5) < 0, then which of the following must be true?

Case 1: |x| - 2 < 0 & x + 5>0
|x| < 2; -2 < x < 2
x > -5

-2 < x < 2 (1)

Case 2: |x| - 2 > 0 & x + 5 < 0
x < -5
x < -2 or x > 2

x < -5 (2)

Combining (1) & (2)

-2 < x < 2 or
x < -5

x < 2 must be TRUE

IMO B
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 24 Apr 2026, 22:54
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For the first factor, mod(x) = 2 is the critical point.
-2 to 2 gives negative values for mod(x)-2 expression.

Now for each of this range from -2 to 2 we see that second factor is always positive.
So it remains negative through and satisfies.

Between -2 to 5 we see it is positive.

For making second factor to be less than 0 we need x<-5.
And for each of these values first expression remains positive and satisfies expression.

So basically all values x<2 satisfies except -2 to 5 range and x<2 is a guarantee this way.
If x>2 then both equations become positive, this not valid.

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

A. x > 2
B. x < 2
C. -2 < x < 2
D. -5 < x < 2
E. x < -5


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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 24 Apr 2026, 23:24
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Official question???
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 24 Apr 2026, 23:42
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Official question???
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This is a GMAT Club Tests questions. One of the most trickiest and hardest for the students.
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 25 Apr 2026, 01:58
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Option B also includes values like -3 & -4. At these values of x the inequality becomes positive. Does anyone can explain why B is answer?
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If (|x| - 2)(x + 5) < 0, then which of the following must be true?

A. x > 2
B. x < 2
C. -2 < x < 2
D. -5 < x < 2
E. x < -5

Option B also includes values like -3 & -4. At these values of x the inequality becomes positive. Does anyone can explain why B is answer?
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 25 Apr 2026, 04:37
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Bunuel
Option B also includes values like -3 & -4. At these values of x the inequality becomes positive. Does anyone can explain why B is answer?

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Your doubt is addressed in the thread. Please review.

To understand the underline concept better practice other Trickiest Inequality Questions Type: Confusing Ranges.
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 13 Jul 2026, 11:47
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i got the same answer but why are my ranges different from others? cant spot my mistake?
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 13 Jul 2026, 22:27
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i got the same answer but why are my ranges different from others? cant spot my mistake?
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In the x <= 0 section,
when you get (-x-2)(x+5) < 0, you multiply both sides by -1 to get (x+2)(x+5) > 0
The inequality sign flips so you need to look at the positive regions on the number line, not the negative region.
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 13 Jul 2026, 23:06
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Hi Bunnel ,
what if x=-3 it satisfies x<2 but don't fall in the range.
Bunuel
Official Solution:

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

A. \(x > 2\)
B. \(x < 2\)
C. \(-2 < x < 2\)
D. \(-5 < x < 2\)
E. \(x < -5\)


\((|x| - 2)(x + 5) < 0\) means that \(|x| - 2\) and \(x + 5\) must have the opposite signs.

CASE 1: \(|x| - 2 > 0\) and \(x + 5 < 0\):

\(|x| - 2 > 0\) means that \(x < -2\) or \(x > 2\);

\(x + 5 < 0\) means that \(x < -5\).

Intersection of these ranges is \(x< -5\).

CASE 2: \(|x| - 2 < 0\) and \(x + 5 > 0\):

\(|x| - 2 < 0\) means that \(-2 < x < 2\);

\(x + 5 > 0\) means that \(x > -5\).

Intersection of these ranges is \(-2 < x < 2\).

So, we have that \((|x| - 2)(x + 5) < 0\) means that \(x < -5\) or \(-2 < x < 2\). ANY \(x\) from these possible ranges will for sure be less than 2 (option B).

To explaining other options:

\(x > 2\) (A) is not true because \(x\) could say be 0.

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

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

\(x < -5\) (E) is not true because \(x\) could say be 0.


Answer: B
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 14 Jul 2026, 02:33
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ZetaHope
Hi Bunnel ,
what if x=-3 it satisfies x<2 but don't fall in the range.

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Please review the previous pages.
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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 14 Jul 2026, 10:59
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Hi ZetaHope,

You've spotted the exact spot where this question trips almost everyone, so let me line it up with what chetan2u kept repeating in the thread: "we are not looking at the RANGE of x, but what MUST be true."

The key is the direction of the check. The valid values of x - the ones that actually make the inequality true - are only:

- x < -5, or
- -2 < x < 2

Your x = -3 is not in either of those ranges, so it is not a valid value of x in the first place. It never enters the picture.

The question is asking: for every x that does satisfy the inequality, which choice is always true? So you take the real solutions and check the choice - not the other way around.

- Every x < -5 is less than 2. ✓
- Every x with -2 < x < 2 is less than 2. ✓

Since all valid x are less than 2, choice B must be true. B is not claiming "every number below 2 solves the inequality" - it only has to cover the actual solutions, which it does.

A simpler version to lock it in

Say a problem's real solutions turned out to be just x = 1 and x = 4. Now ask: must "x < 10" be true?

- 1 < 10 ✓ and 4 < 10 ✓ - yes, it must be true.

The fact that x = 7 is also below 10 but is not one of our solutions is irrelevant - we only test the values that actually solve the problem. That's the same move: check the answer choice against the real solution set, never plug a non-solution back in to break it.

Answer: B

ZetaHope
Hi Bunnel ,
what if x=-3 it satisfies x<2 but don't fall in the range.

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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 14 Jul 2026, 22:45
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How would the solution go if i were to open up the mod for two cases x>0 and X<0 and then employ the wavey curve method?
Bunuel
If (|x| - 2)(x + 5) < 0, then which of the following must be true?

A. x > 2
B. x < 2
C. -2 < x < 2
D. -5 < x < 2
E. x < -5


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If (|x| - 2)(x + 5) < 0, then which of the following must be true? 15 Jul 2026, 04:23
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Hi dignissimosminima,

Good instinct. The wavy curve method plugs right in once you've opened the mod, but there's one spot in the negative case where people slip. Let me walk both cases exactly the way KarishmaB set them up.

Case 1: x ≥ 0, so |x| = x

The inequality becomes (x − 2)(x + 5) < 0.

Mark the roots −5 and 2 on a number line. For a product of two linear factors with positive leading coefficient, the wavy curve is + on the far right, then alternates: + above 2, − between −5 and 2, + below −5.

We want < 0, so take the negative stretch: −5 < x < 2. But this case only allows x ≥ 0, so we keep 0 ≤ x < 2.

Case 2: x < 0, so |x| = −x

Now it's (−x − 2)(x + 5) < 0. Factor out the −1 to make the wavy curve usable:

−(x + 2)(x + 5) < 0 → (x + 2)(x + 5) > 0

This is the step to watch: multiplying by −1 flips the sign, so you're now solving > 0, not < 0. (This is exactly the flip KarishmaB flagged for NetOrb.)

Roots are −5 and −2. Wavy curve: + outside the roots, − between them. Since we want > 0, take the outside pieces: x < −5 or x > −2. This case only allows x < 0, so we keep x < −5 or −2 < x < 0.

Combine

Union of both cases: x < −5 or −2 < x < 2. Every value here is less than 2 - answer B.

Quick habit to lock in the flip: solve −(x − 1)(x − 3) < 0. Multiply by −1 first → (x − 1)(x − 3) > 0 → x < 1 or x > 3. If you forget to flip, you'd wrongly grab 1 < x < 3. Always flip the inequality the moment you multiply by a negative.

Answer: B

dignissimosminima
How would the solution go if i were to open up the mod for two cases x>0 and X<0 and then employ the wavey curve method?

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