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

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

(I) Make use of choices.

a) If x=0, (|x| - 2)(x + 5) < 0...... => (0 - 2)(0 + 5) < 0....-10<0...YES
So the range should be containing 0..
Eliminate A and E

a) If x=-10, (|x| - 2)(x + 5) < 0...... => (10 - 2)(-10 + 5) < 0....-40<0...YES
So the range should be containing -10..
Eliminate C and D

B

(II) Algebraic way
\((|x| - 2)(x + 5) < 0\)
If (|x| - 2)>0 or |x|>2, that is x<-2 and x>2, then (x + 5) < 0 or x<-5.....Common range x<-5
If (|x| - 2)<0 or |x|<2, that is -2<x<2, then (x + 5) > 0 or x>-5.....Common range -2<x<2

Combined range...x<-5 and -2<x<2
Look at the choices that contain both these ranges....x<2

B


NOTE : Do not get confused that -5<x<-2 does not satisfy the inequality, we are looking at what must be true and NOT the range of x.
Even if it was x<100, it would be correct as an answer as it would contain both the ranges x<-5 and -2<x<2
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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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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
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Is B really the answer?
If we insert X=-4, the result will be 2*1=2>0

Case 1: (|x|-2)>0 and x+5<0
→x<-5
Case 2: (|x|-2)<0 and x+5>0
→-2<x<2

X can be x<-5 or -2<x<2

So actually both C and E stand...

Posted from my mobile device

-4 and even -3 will not hold. But as I wrote above we are not looking for the EXACT range of x, but what is true of x?
Even if it was given x<10, it would be true as all possible answers to the inequality will fall in this range
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OMG !! That is one wrong answer.
y= ( |x|-2)(x+5)
Let us analyze cases
case 1. ) x=3;y=8 >0
case 2. ) x=2 ;y=0
case 3. ) x= 1 ; y=-6 <0
case 4. ) x=0 ; y= -10 <0
case 5. x=-1; y= -4 <0
case 6.) x=-2 ;y=0
case 7.) x= -3;y=2 >0
Case 8 )y= -4;y= 2 >0
case 9 ) y=-6; y=-4<0

NOW for Option B case 3, case 4, case 5 are right answer, but case 7 & 8 are WRONG.
but for Option C is absolute for the expression "y" to be true i.e less than zero, case 3 , 4 & 5 which is within the range,
-2<x<2; you may also include any fraction or decimal as well to check the correctness of the expression
Interested in any suggestion.

i think its c) because -4 and -3 stands out from the range x<2 and provides a +ve answer.

Please correct this question, if B MUST BE TRUE, then also C, D & E MUST BE TRUE. This question is flawed

The question is perfectly fine. Please go through the discussion above. Your query is already answered.

Take an easier thing.

Say
If x is a positive odd integer, what must be true?
1) x>5...
No, what if x=1, 3 or 5.

2) x is an integer.....
Yes, x has to be an integer for it to be positive odd integer.

3) x>0....
Yes again

4) x is odd....
yes again.

Your options C, D and E are similar to case 1.
And B is similar to other cases, that is 2, 3 and 4.
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for the range x<2, if i take x to be -5 then L.H.S becomes zero, then how is B a correct option

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.
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Now THAT is one great explanation.

Able to understand now.

Posted from my mobile device
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ans is E because if you take x as negative you get two solutions i.e, x>-2 and x<-5 as x>-2 doest not satisfy the equation if x=2.....so if x<-5 all the numbers will satisfy the equation so the ans is E

C is the correct answer.

Please correct me if I am wrong

if x>0
eq=>
(x-2)(x+5)<0
=>
0<x<2

if x<0
eq=>
(x+2)(x+5)>0
=>
0>x>-2 & x<-5

Hence 2>x>-2 & x<-5

Both C and E are correct.

put 1, -1 and -6 as sample values to check. They all satisfy original equation.

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.
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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.