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

BrushMyQuant could you pls help me to get this question solved with the help of wavy line method? thanks in advance !

General Approach (Wavy Line Method with Cases):
1. Find the critical points where each factor equals zero.
2. Split the problem into cases based on the definition of the absolute value.
3. For each case, solve the resulting inequality using the Wavy Line Method (by identifying critical points for that case and testing intervals).
4. Combine the solutions from all cases.
5. Determine which of the given options must be true based on the combined solution set.

Step 1: Find all critical points.
Set each part of the expression to zero:
- $|x|-2=0 \Longrightarrow|x|=2 \Longrightarrow x=2$ or $x=-2$.
- $x+5=0 \Longrightarrow x=-5$.

The critical points are $\{-5,-2,2\}$. These points divide the number line into intervals.

Step 2: Handle the Absolute Value by Cases.
Case 1: $x \geq 0$
In this case, $|x|=x$. The inequality becomes:

$$
\((x-2)(x+5)<0\)
$$

- Critical points for this case: $x=2$ and $x=-5$.
- Consider only the domain $x \geq 0$ : The relevant critical point in this domain is $x=2$.
- Test intervals in the domain $x \geq 0$ :
- Interval $0 \leq x<2$ (e.g., test $x=1$ ): $(1-2)(1+5)=(-1)(6)=-6$. Since $-6<$ 0 , this interval is part of the solution.
- Interval $x>2$ (e.g., test $x=3$ ): $(3-2)(3+5)=(1)(8)=8$. Since $8 \nless 0$, this interval is NOT part of the solution.
- Solution for Case 1: $0 \leq x<2$.

Case 2: $x<0$
In this case, $|x|=-x$. The inequality becomes:

$$
\((-x-2)(x+5)<0\)
$$


Factor out -1 from the first term:

$$
\(-(x+2)(x+5)<0\)
$$


Multiply both sides by -1 and reverse the inequality sign:

$$
\((x+2)(x+5)>0\)
$$

- Critical points for this case: $x=-2$ and $x=-5$.
- Consider only the domain $x<0$ : Both critical points are in this domain.
- Test intervals in the domain $x<0$ :

- Interval $x<-5$ (e.g., test $x=-6$ ): $(-6+2)(-6+5)=(-4)(-1)=4$. Since $4>0$, this interval is part of the solution.
- Interval $-5<x<-2$ (e.g., test $x=-3$ ): $(-3+2)(-3+5)=(-1)(2)=-2$. Since $\(-2 \ngtr 0\)$, this interval is NOT part of the solution.
- Interval $-2<x<0$ (e.g., test $x=-1$ ): $(-1+2)(-1+5)=(1)(4)=4$. Since $4>0$, this interval is part of the solution.
- Solution for Case 2: $x<-5$ or $-2<x<0$.

Step 3: Combine the solutions from all cases.
The complete solution set for the inequality $(|x|-2)(x+5)<0$ is the union of the solutions from Case 1 and Case 2:

$$
\((0 \leq x<2) \text { OR }(x<-5 \text { OR }-2<x<0)\)
$$


Combining these intervals on a single number line:
Solution Set $\(=(-\infty,-5) \cup(-2,2)\)$

Step 4: Determine which of the given options must be true.
We need to find the option that is a superset of our solution set, meaning if ' $x$ ' is a solution, then the statement in the option must also be true.

Our solution is $x<-5$ OR $(-2<x<2)$.
- A. $x>2$ : False. For example, $x=-6$ is a solution, but $-6 \ngtr 2$.
- B. $x<2$ : True.
- If $x<-5$, then it is certainly true that $x<2$.
- If $-2<x<2$, then it is certainly true that $x<2$.
- Since all solutions are either $x<-5$ or $-2<x<2$, all solutions satisfy $x<2$.

- C. $-2<x<2$ : False. It excludes the solutions where $x<-5$. For example, $x=-6$ is a solution, but it's not in this interval.
- D. $-5<x<2$ : False. It excludes the solutions where $x<-5$. For example, $x=-6$ is a solution, but it's not in this interval.
- E. $x<-5$ : False. It excludes the solutions where $-2<x<2$. For example, $x=1$ is a solution, but $\(1 \nless-5\)$.

Therefore, the only statement that must be true is $x<2$.
The final answer is B