x=-3 doesn't work for B, right?

I think that the answer should be C, not B. Can you please correct me if I am wrong?
I did it in the following way:
I multiplied (|x| - 2)(x + 5)
x|x| + 5|x| -2x - 10 < 0

x|x| + 5|x| < 2x + 10

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

I see that |x| should be smaller than 2.
|x| < 2
-2 < x < 2

One thing that I am not very confident about, can I multiply by |x|?
For example |x|(a + b), can I write it as (a|x| + b|x|)?

I think the anwer shd be 'e'

cause taking +x and -x conditions to break the modulus,

you would get equation: -5 < x < 2 (+x)
and

equation: x>-2 & x<-5 (-x)

since the question mentions "must" x shd be x<-5 cause all values in this range satify the condition

example, x=-6
(|-6|-2)(-6+5)=(4)(-1)<0 whereas in -2<x<2 we can take x=0 which does not satisfy the equation.

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

how can you take 0 when x<-5 shouldn't the possible range be [-inf, -5)

Can you highlight specifically the part you are referring to sir ?

if x>2 how can it be x can take the value of 0

similarly, if x<-5 how can it take value of 0

we are dealing with RANGE of x

Moreover, we must assess if the option MUST be true

We also tested various scenarios in which , in the end, the only reasonable value is that x <2

So for example A is NOT true because it says that x >2 NO

It is x < 2 and can be also 0, of course

I was stating option E not A as a potential answer since values where x<-5 would make the function negative thereby <0

We are asked to determine which condition must be true. Another way to interpret this is to imagine which condition is satisfied for every x that is a solution to the equation. Since x = 0 is a solution, if E was the answer, 0 < -5 should be true (which it isn't). Thus, the answer is B because all the possible solutions satisfy it.

Is 0 < -5. I think 0 > -5

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